Stetige Funktionen sind riemannintegrierbar
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Anschaulich ist das Integral einer Funktion
f
:
[
a
,
b
]
→
R
{\displaystyle f:[a,b]\to \mathbb {R} }
x
{\displaystyle x}
∫
a
b
f
(
x
)
d
x
{\displaystyle \int _{a}^{b}f(x)\mathrm {d} x}
Beweis (Stetige Funktionen sind riemannintegrierbar)
f
{\displaystyle f}
[
a
,
b
]
{\displaystyle [a,b]}
f
{\displaystyle f}
ϵ
>
0
{\displaystyle \epsilon >0}
δ
>
0
{\displaystyle \delta >0}
x
,
y
∈
[
a
,
b
]
{\displaystyle x,y\in [a,b]}
|
x
−
y
|
<
δ
{\displaystyle |x-y|<\delta }
|
f
(
x
)
−
f
(
y
)
|
<
ϵ
b
−
a
{\displaystyle |f(x)-f(y)|<{\tfrac {\epsilon }{b-a}}}
Δ
=
(
x
0
,
…
,
x
n
)
{\displaystyle \Delta =(x_{0},\ldots ,x_{n})}
[
a
,
b
]
{\displaystyle [a,b]}
|
Δ
|
<
δ
{\displaystyle |\Delta |<\delta }
k
∈
{
0
,
…
,
n
−
1
}
{\displaystyle k\in \{0,\ldots ,n-1\}}
sup
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
−
inf
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
↓
f
ist stetig
=
max
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
−
min
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
↓
Für alle
x
,
y
∈
[
x
k
,
x
k
+
1
]
gilt
|
x
−
y
|
<
δ
<
ϵ
b
−
a
{\displaystyle {\begin{aligned}&\sup _{x\in [x_{k},x_{k+1}]}f(x)-\inf _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ f{\text{ ist stetig}}\right.}\\[0.3em]=\ &\max _{x\in [x_{k},x_{k+1}]}f(x)-\min _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{Für alle }}x,y\in [x_{k},x_{k+1}]{\text{ gilt }}|x-y|<\delta \right.}\\[0.3em]<\ &{\frac {\epsilon }{b-a}}\end{aligned}}}
Folglich gilt
O
(
Δ
,
f
)
−
U
(
Δ
,
f
)
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
sup
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
−
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
inf
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
(
sup
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
−
inf
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
)
<
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
ϵ
b
−
a
=
(
b
−
a
)
ϵ
b
−
a
=
ϵ
{\displaystyle {\begin{aligned}&O(\Delta ,f)-U(\Delta ,f)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}f(x)-\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\left(\sup _{x\in [x_{k},x_{k+1}]}f(x)-\inf _{x\in [x_{k},x_{k+1}]}f(x)\right)\\[0.3em]<\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k}){\frac {\epsilon }{b-a}}\\[0.3em]=\ &(b-a){\frac {\epsilon }{b-a}}\\[0.3em]=\ &\epsilon \end{aligned}}}
Damit ist
f
{\displaystyle f}
Monotonie des Riemannintegrals
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Nun betrachten wir zwei riemannintegrierbare Funktionen
f
:
[
a
,
b
]
→
R
{\displaystyle f:[a,b]\to \mathbb {R} }
g
:
[
a
,
b
]
→
R
{\displaystyle g:[a,b]\to \mathbb {R} }
f
≤
g
{\displaystyle f\leq g}
f
(
x
)
≤
g
(
x
)
{\displaystyle f(x)\leq g(x)}
x
∈
[
a
,
b
]
{\displaystyle x\in [a,b]}
a
,
b
∈
R
{\displaystyle a,b\in \mathbb {R} }
a
≤
b
{\displaystyle a\leq b}
To-Do:
Bild von
f
{\displaystyle f}
g
{\displaystyle g}
Anschaulich macht es Sinn, dass
∫
a
b
f
(
x
)
d
x
≤
∫
a
b
g
(
x
)
d
x
{\displaystyle \int _{a}^{b}f(x)\mathrm {d} x\leq \int _{a}^{b}g(x)\mathrm {d} x}
f
{\displaystyle f}
g
{\displaystyle g}
Dass
∫
a
b
f
(
x
)
d
x
≤
∫
a
b
g
(
x
)
d
x
{\displaystyle \int _{a}^{b}f(x)\mathrm {d} x\leq \int _{a}^{b}g(x)\mathrm {d} x}
Wir betrachten die Ober- und die Untersumme für eine beliebige Zerlegung
Δ
~
{\displaystyle {\tilde {\Delta }}}
[
a
,
b
]
{\displaystyle [a,b]}
To-Do:
Bild von davor mit Ober- und Untersummen
Wir sehen, dass
O
(
Δ
~
,
f
)
≤
O
(
Δ
~
,
g
)
U
(
Δ
~
,
f
)
≤
U
(
Δ
~
,
g
)
{\displaystyle {\begin{aligned}&O({\tilde {\Delta }},f)\leq O({\tilde {\Delta }},g)\\[0.3em]&U({\tilde {\Delta }},f)\leq U({\tilde {\Delta }},g)\\[0.3em]\end{aligned}}}
Da dies für alle Zerlegungen
Δ
~
{\displaystyle {\tilde {\Delta }}}
I
+
(
f
,
[
a
,
b
]
)
≤
I
+
(
g
,
[
a
,
b
]
)
{\displaystyle I_{+}(f,[a,b])\leq I_{+}(g,[a,b])}
I
−
(
f
,
[
a
,
b
]
)
≤
I
−
(
g
,
[
a
,
b
]
)
{\displaystyle I_{-}(f,[a,b])\leq I_{-}(g,[a,b])}
∫
a
b
f
(
x
)
d
x
≤
∫
a
b
g
(
x
)
d
x
{\displaystyle \int _{a}^{b}f(x)\mathrm {d} x\leq \int _{a}^{b}g(x)\mathrm {d} x}
Wir haben uns gerade anschaulich überlegt, warum der folgende Satz gilt. Nun werden wir diesen auch beweisen.
Beweis (Monotonie des Riemannintegrals)
Es sei
Δ
~
=
(
x
0
,
…
,
x
n
)
{\displaystyle {\tilde {\Delta }}=(x_{0},\ldots ,x_{n})}
[
a
,
b
]
{\displaystyle [a,b]}
O
(
Δ
~
,
f
)
{\displaystyle O({\tilde {\Delta }},f)}
O
(
Δ
~
,
g
)
{\displaystyle O({\tilde {\Delta }},g)}
O
(
Δ
~
,
f
)
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
sup
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
↓
f
(
x
)
≤
g
(
x
)
für alle
x
∈
[
a
,
b
]
≤
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
sup
x
∈
[
x
k
,
x
k
+
1
]
g
(
x
)
=
O
(
Δ
~
,
g
)
{\displaystyle {\begin{aligned}&O({\tilde {\Delta }},f)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ f(x)\leq g(x){\text{ für alle }}x\in [a,b]\right.}\\[0.3em]\leq \ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}g(x)\\[0.3em]=\ &O({\tilde {\Delta }},g)\\[0.3em]\end{aligned}}}
Folglich gilt
I
+
(
f
,
[
a
,
b
]
)
=
inf
Δ
Zerlegung
O
(
Δ
,
f
)
↓
O
(
Δ
,
f
)
≤
O
(
Δ
,
g
)
für alle Zerlegungen
Δ
≤
inf
Δ
Zerlegung
O
(
Δ
,
g
)
=
I
+
(
g
,
[
a
,
b
]
)
{\displaystyle {\begin{aligned}&I_{+}(f,[a,b])\\[0.3em]=\ &\inf _{\Delta {\text{ Zerlegung}}}O(\Delta ,f)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ O(\Delta ,f)\leq O(\Delta ,g){\text{ für alle Zerlegungen }}\Delta \right.}\\[0.3em]\leq \ &\inf _{\Delta {\text{ Zerlegung}}}O(\Delta ,g)\\[0.3em]=\ &I_{+}(g,[a,b])\end{aligned}}}
Somit ist
∫
a
b
f
(
x
)
d
x
↓
f
riemannintegrierbar
=
I
+
(
f
,
[
a
,
b
]
)
≤
I
+
(
g
,
[
a
,
b
]
)
↓
g
riemannintegrierbar
=
∫
a
b
g
(
x
)
d
x
{\displaystyle {\begin{aligned}&\int _{a}^{b}f(x)\mathrm {d} x\\[0.3em]&{\color {OliveGreen}\left\downarrow \ f{\text{ riemannintegrierbar}}\right.}\\[0.3em]=\ &I_{+}(f,[a,b])\\[0.3em]\leq \ &I_{+}(g,[a,b])\\[0.3em]&{\color {OliveGreen}\left\downarrow \ g{\text{ riemannintegrierbar}}\right.}\\[0.3em]=\ &\int _{a}^{b}g(x)\mathrm {d} x\\[0.3em]\end{aligned}}}
Die Summe zweier riemannintegrierbarer Funktionen ist riemannintegrierbar
Bearbeiten
Beweis
Seien
Δ
1
{\displaystyle \Delta _{1}}
Δ
2
{\displaystyle \Delta _{2}}
[
a
,
b
]
{\displaystyle [a,b]}
Δ
~
=
(
x
0
,
x
1
,
…
,
x
n
)
{\displaystyle {\tilde {\Delta }}=(x_{0},x_{1},\ldots ,x_{n})}
Δ
1
{\displaystyle \Delta _{1}}
Δ
2
{\displaystyle \Delta _{2}}
O
(
Δ
1
,
f
)
+
O
(
Δ
2
,
g
)
≥
O
(
Δ
~
,
f
)
+
O
(
Δ
~
,
g
)
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
sup
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
+
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
sup
x
∈
[
x
k
,
x
k
+
1
]
g
(
x
)
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
(
sup
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
+
sup
x
∈
[
x
k
,
x
k
+
1
]
g
(
x
)
)
≥
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
sup
x
∈
[
x
k
,
x
k
+
1
]
(
f
(
x
)
+
g
(
x
)
)
=
O
(
Δ
~
,
f
+
g
)
≥
I
+
(
f
+
g
,
[
a
,
b
]
)
{\displaystyle {\begin{aligned}&O(\Delta _{1},f)+O(\Delta _{2},g)\\[0.3em]\geq \ &O({\tilde {\Delta }},f)+O({\tilde {\Delta }},g)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}f(x)+\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}g(x)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\left(\sup _{x\in [x_{k},x_{k+1}]}f(x)+\sup _{x\in [x_{k},x_{k+1}]}g(x)\right)\\[0.3em]\geq \ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}(f(x)+g(x))\\[0.3em]=\ &O({\tilde {\Delta }},f+g)\\[0.3em]\geq \ &I_{+}(f+g,[a,b])\end{aligned}}}
sowie
U
(
Δ
1
,
f
)
+
U
(
Δ
2
,
g
)
≤
U
(
Δ
~
,
f
)
+
U
(
Δ
~
,
g
)
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
inf
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
+
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
inf
x
∈
[
x
k
,
x
k
+
1
]
g
(
x
)
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
(
inf
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
+
inf
x
∈
[
x
k
,
x
k
+
1
]
g
(
x
)
)
≤
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
inf
x
∈
[
x
k
,
x
k
+
1
]
(
f
(
x
)
+
g
(
x
)
)
=
U
(
Δ
~
,
f
+
g
)
≤
I
−
(
f
+
g
,
[
a
,
b
]
)
{\displaystyle {\begin{aligned}&U(\Delta _{1},f)+U(\Delta _{2},g)\\[0.3em]\leq \ &U({\tilde {\Delta }},f)+U({\tilde {\Delta }},g)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}f(x)+\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}g(x)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\left(\inf _{x\in [x_{k},x_{k+1}]}f(x)+\inf _{x\in [x_{k},x_{k+1}]}g(x)\right)\\[0.3em]\leq \ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}(f(x)+g(x))\\[0.3em]=\ &U({\tilde {\Delta }},f+g)\\[0.3em]\leq \ &I_{-}(f+g,[a,b])\end{aligned}}}
Folglich gilt
I
+
(
f
,
[
a
,
b
]
)
+
I
+
(
g
,
[
a
,
b
]
)
=
inf
Δ
1
Zerlegung
O
(
Δ
1
,
f
)
+
inf
Δ
2
Zerlegung
O
(
Δ
2
,
g
)
=
inf
Δ
1
,
Δ
2
Zerlegungen
(
O
(
Δ
1
,
f
)
+
O
(
Δ
2
,
g
)
)
↓
O
(
Δ
1
,
f
)
+
O
(
Δ
2
,
g
)
≥
I
+
(
f
+
g
,
[
a
,
b
]
)
für alle Zerlegungen
Δ
1
,
Δ
2
≥
I
+
(
f
+
g
,
[
a
,
b
]
)
{\displaystyle {\begin{aligned}&I_{+}(f,[a,b])+I_{+}(g,[a,b])\\[0.3em]=\ &\inf _{\Delta _{1}{\text{ Zerlegung}}}O(\Delta _{1},f)+\inf _{\Delta _{2}{\text{ Zerlegung}}}O(\Delta _{2},g)\\[0.3em]=\ &\inf _{\Delta _{1},\Delta _{2}{\text{ Zerlegungen}}}\left(O(\Delta _{1},f)+O(\Delta _{2},g)\right)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ O(\Delta _{1},f)+O(\Delta _{2},g)\geq I_{+}(f+g,[a,b]){\text{ für alle Zerlegungen }}\Delta _{1},\Delta _{2}\right.}\\[0.3em]\geq \ &I_{+}(f+g,[a,b])\end{aligned}}}
Genauso zeigen wir
I
−
(
f
+
g
,
[
a
,
b
]
)
≥
I
−
(
f
,
[
a
,
b
]
)
+
I
−
(
g
,
[
a
,
b
]
)
{\displaystyle I_{-}(f+g,[a,b])\geq I_{-}(f,[a,b])+I_{-}(g,[a,b])}
I
−
(
f
,
[
a
,
b
]
)
+
I
−
(
g
,
[
a
,
b
]
)
=
sup
Δ
1
Zerlegung
U
(
Δ
1
,
f
)
+
sup
Δ
2
Zerlegung
U
(
Δ
2
,
g
)
=
sup
Δ
1
,
Δ
2
Zerlegungen
(
U
(
Δ
1
,
f
)
+
U
(
Δ
2
,
g
)
)
↓
U
(
Δ
1
,
f
)
+
U
(
Δ
2
,
g
)
≤
I
−
(
f
+
g
,
[
a
,
b
]
)
für alle Zerlegungen
Δ
1
,
Δ
2
≤
I
−
(
f
+
g
,
[
a
,
b
]
)
{\displaystyle {\begin{aligned}&I_{-}(f,[a,b])+I_{-}(g,[a,b])\\[0.3em]=\ &\sup _{\Delta _{1}{\text{ Zerlegung}}}U(\Delta _{1},f)+\sup _{\Delta _{2}{\text{ Zerlegung}}}U(\Delta _{2},g)\\[0.3em]=\ &\sup _{\Delta _{1},\Delta _{2}{\text{ Zerlegungen}}}\left(U(\Delta _{1},f)+U(\Delta _{2},g)\right)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ U(\Delta _{1},f)+U(\Delta _{2},g)\leq I_{-}(f+g,[a,b]){\text{ für alle Zerlegungen }}\Delta _{1},\Delta _{2}\right.}\\[0.3em]\leq \ &I_{-}(f+g,[a,b])\end{aligned}}}
Bisher haben wir damit folgendes bewiesen:
I
+
(
f
+
g
,
[
a
,
b
]
)
≤
I
+
(
f
,
[
a
,
b
]
)
+
I
+
(
g
,
[
a
,
b
]
)
I
−
(
f
+
g
,
[
a
,
b
]
)
≥
I
−
(
f
,
[
a
,
b
]
)
+
I
−
(
g
,
[
a
,
b
]
)
{\displaystyle {\begin{aligned}I_{+}(f+g,[a,b])&\leq I_{+}(f,[a,b])+I_{+}(g,[a,b])\\[0.3em]I_{-}(f+g,[a,b])&\geq I_{-}(f,[a,b])+I_{-}(g,[a,b])\\[0.3em]\end{aligned}}}
Nach Voraussetzung sind die Funktionen
f
{\displaystyle f}
g
{\displaystyle g}
I
+
(
f
+
g
,
[
a
,
b
]
)
≤
I
+
(
f
,
[
a
,
b
]
)
+
I
+
(
g
,
[
a
,
b
]
)
↓
I
+
(
f
,
[
a
,
b
]
)
=
I
−
(
f
,
[
a
,
b
]
)
und
I
+
(
g
,
[
a
,
b
]
)
=
I
−
(
g
,
[
a
,
b
]
)
=
I
−
(
f
,
[
a
,
b
]
)
+
I
−
(
g
,
[
a
,
b
]
)
≤
I
−
(
f
+
g
,
[
a
,
b
]
)
{\displaystyle {\begin{aligned}&I_{+}(f+g,[a,b])\\[0.3em]\leq \ &I_{+}(f,[a,b])+I_{+}(g,[a,b])\\[0.3em]&{\color {OliveGreen}\left\downarrow \ I_{+}(f,[a,b])=I_{-}(f,[a,b]){\text{ und }}I_{+}(g,[a,b])=I_{-}(g,[a,b])\right.}\\[0.3em]=\ &I_{-}(f,[a,b])+I_{-}(g,[a,b])\\[0.3em]\leq \ &I_{-}(f+g,[a,b])\\[0.3em]\end{aligned}}}
Außerdem wissen wir, dass
I
+
(
f
+
g
,
[
a
,
b
]
)
≥
I
−
(
f
+
g
,
[
a
,
b
]
)
{\displaystyle I_{+}(f+g,[a,b])\geq I_{-}(f+g,[a,b])}
I
+
(
f
+
g
,
[
a
,
b
]
)
=
I
−
(
f
+
g
,
[
a
,
b
]
)
{\displaystyle I_{+}(f+g,[a,b])=I_{-}(f+g,[a,b])}
f
+
g
{\displaystyle f+g}
∫
a
b
(
f
(
x
)
+
g
(
x
)
)
d
x
=
I
+
(
f
+
g
,
[
a
,
b
]
)
=
I
+
(
f
,
[
a
,
b
]
)
+
I
+
(
g
,
[
a
,
b
]
)
=
∫
a
b
f
(
x
)
d
x
+
∫
a
b
g
(
x
)
d
x
{\displaystyle \int _{a}^{b}\left(f(x)+g(x)\right)\mathrm {d} x=I_{+}(f+g,[a,b])=I_{+}(f,[a,b])+I_{+}(g,[a,b])=\int _{a}^{b}f(x)\mathrm {d} x+\int _{a}^{b}g(x)\mathrm {d} x}
Beweis (Faktorregel)
Sei
Δ
~
=
(
x
0
,
…
,
x
n
)
{\displaystyle {\tilde {\Delta }}=(x_{0},\ldots ,x_{n})}
[
a
,
b
]
{\displaystyle [a,b]}
λ
≥
0
{\displaystyle \lambda \geq 0}
λ
<
0
{\displaystyle \lambda <0}
Fall 1:
λ
≥
0
{\displaystyle \lambda \geq 0}
Es gilt:
O
(
Δ
~
,
λ
f
)
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
sup
x
∈
[
x
k
,
x
k
+
1
]
λ
⋅
f
(
x
)
↓
λ
≥
0
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
⋅
λ
⋅
sup
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
=
λ
⋅
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
sup
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
=
λ
⋅
O
(
Δ
~
,
f
)
{\displaystyle {\begin{aligned}&O({\tilde {\Delta }},\lambda f)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}\lambda \cdot f(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ \lambda \geq 0\right.}\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\cdot \lambda \cdot \sup _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]=\ &\lambda \cdot \sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]=\ &\lambda \cdot O({\tilde {\Delta }},f)\\[0.3em]\end{aligned}}}
und
U
(
Δ
~
,
λ
f
)
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
inf
x
∈
[
x
k
,
x
k
+
1
]
λ
⋅
f
(
x
)
↓
λ
≥
0
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
⋅
λ
⋅
inf
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
=
λ
⋅
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
inf
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
=
λ
⋅
U
(
Δ
~
,
f
)
{\displaystyle {\begin{aligned}&U({\tilde {\Delta }},\lambda f)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}\lambda \cdot f(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ \lambda \geq 0\right.}\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\cdot \lambda \cdot \inf _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]=\ &\lambda \cdot \sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]=\ &\lambda \cdot U({\tilde {\Delta }},f)\\[0.3em]\end{aligned}}}
Somit gilt
I
+
(
λ
f
,
[
a
,
b
]
)
=
inf
Δ
Zerlegung
O
(
Δ
,
λ
f
)
↓
O
(
Δ
~
,
λ
f
)
=
λ
O
(
Δ
~
,
f
)
für alle Zerlegungen
Δ
~
=
inf
Δ
Zerlegung
λ
O
(
Δ
,
f
)
=
λ
inf
Δ
Zerlegung
O
(
Δ
,
f
)
=
λ
⋅
I
+
(
f
,
[
a
,
b
]
)
{\displaystyle {\begin{aligned}&I_{+}(\lambda f,[a,b])\\[0.3em]=\ &\inf _{\Delta {\text{ Zerlegung}}}O(\Delta ,\lambda f)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ O({\tilde {\Delta }},\lambda f)=\lambda O({\tilde {\Delta }},f){\text{ für alle Zerlegungen }}{\tilde {\Delta }}\right.}\\[0.3em]=\ &\inf _{\Delta {\text{ Zerlegung}}}\lambda O(\Delta ,f)\\[0.3em]=\ &\lambda \inf _{\Delta {\text{ Zerlegung}}}O(\Delta ,f)\\[0.3em]=\ &\lambda \cdot I_{+}(f,[a,b])\end{aligned}}}
und
I
−
(
λ
f
,
[
a
,
b
]
)
=
sup
Δ
Zerlegung
U
(
Δ
,
λ
f
)
↓
U
(
Δ
~
,
λ
f
)
=
λ
U
(
Δ
~
,
f
)
für alle Zerlegungen
Δ
~
=
sup
Δ
Zerlegung
λ
U
(
Δ
,
f
)
=
λ
sup
Δ
Zerlegung
U
(
Δ
,
f
)
=
λ
⋅
I
−
(
f
,
[
a
,
b
]
)
{\displaystyle {\begin{aligned}&I_{-}(\lambda f,[a,b])\\[0.3em]=\ &\sup _{\Delta {\text{ Zerlegung}}}U(\Delta ,\lambda f)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ U({\tilde {\Delta }},\lambda f)=\lambda U({\tilde {\Delta }},f){\text{ für alle Zerlegungen }}{\tilde {\Delta }}\right.}\\[0.3em]=\ &\sup _{\Delta {\text{ Zerlegung}}}\lambda U(\Delta ,f)\\[0.3em]=\ &\lambda \sup _{\Delta {\text{ Zerlegung}}}U(\Delta ,f)\\[0.3em]=\ &\lambda \cdot I_{-}(f,[a,b])\end{aligned}}}
Also:
I
+
(
λ
f
,
[
a
,
b
]
)
=
λ
⋅
I
+
(
f
,
[
a
,
b
]
)
↓
f
ist riemannintegrierbar, d.h.
I
+
(
f
,
[
a
,
b
]
)
=
I
−
(
f
,
[
a
,
b
]
)
=
λ
⋅
I
−
(
f
,
[
a
,
b
]
)
=
I
−
(
λ
f
,
[
a
,
b
]
)
{\displaystyle {\begin{aligned}&I_{+}(\lambda f,[a,b])\\[0.3em]=\ &\lambda \cdot I_{+}(f,[a,b])\\[0.3em]&{\color {OliveGreen}\left\downarrow \ f{\text{ ist riemannintegrierbar, d.h. }}I_{+}(f,[a,b])=I_{-}(f,[a,b])\right.}\\[0.3em]=\ &\lambda \cdot I_{-}(f,[a,b])\\[0.3em]=\ &I_{-}(\lambda f,[a,b])\\[0.3em]\end{aligned}}}
Damit ist die Funktion
λ
f
{\displaystyle \lambda f}
∫
a
b
λ
f
(
x
)
d
x
=
I
+
(
λ
f
,
[
a
,
b
]
)
=
λ
⋅
I
+
(
f
,
[
a
,
b
]
)
=
λ
∫
a
b
f
(
x
)
d
x
{\displaystyle {\begin{aligned}&\int _{a}^{b}\lambda f(x)\mathrm {d} x\\[0.3em]=\ &I_{+}(\lambda f,[a,b])\\[0.3em]=\ &\lambda \cdot I_{+}(f,[a,b])\\[0.3em]=\ &\lambda \int _{a}^{b}f(x)\mathrm {d} x\end{aligned}}}
Fall 2:
λ
<
0
{\displaystyle \lambda <0}
Es gilt:
O
(
Δ
~
,
λ
f
)
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
sup
x
∈
[
x
k
,
x
k
+
1
]
λ
⋅
f
(
x
)
↓
λ
<
0
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
⋅
λ
⋅
inf
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
=
λ
⋅
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
inf
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
=
λ
⋅
U
(
Δ
~
,
f
)
{\displaystyle {\begin{aligned}&O({\tilde {\Delta }},\lambda f)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}\lambda \cdot f(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ \lambda <0\right.}\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\cdot \lambda \cdot \inf _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]=\ &\lambda \cdot \sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]=\ &\lambda \cdot U({\tilde {\Delta }},f)\\[0.3em]\end{aligned}}}
und
U
(
Δ
~
,
λ
f
)
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
inf
x
∈
[
x
k
,
x
k
+
1
]
λ
⋅
f
(
x
)
↓
λ
<
0
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
⋅
λ
⋅
sup
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
=
λ
⋅
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
sup
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
=
λ
⋅
O
(
Δ
~
,
f
)
{\displaystyle {\begin{aligned}&U({\tilde {\Delta }},\lambda f)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}\lambda \cdot f(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ \lambda <0\right.}\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\cdot \lambda \cdot \sup _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]=\ &\lambda \cdot \sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]=\ &\lambda \cdot O({\tilde {\Delta }},f)\\[0.3em]\end{aligned}}}
Somit gilt
I
+
(
λ
f
,
[
a
,
b
]
)
=
inf
Δ
Zerlegung
O
(
Δ
,
λ
f
)
↓
O
(
Δ
~
,
λ
f
)
=
λ
U
(
Δ
~
,
f
)
für alle Zerlegungen
Δ
~
=
inf
Δ
Zerlegung
λ
U
(
Δ
,
f
)
↓
λ
<
0
=
λ
sup
Δ
Zerlegung
U
(
Δ
,
f
)
=
λ
⋅
I
−
(
f
,
[
a
,
b
]
)
{\displaystyle {\begin{aligned}&I_{+}(\lambda f,[a,b])\\[0.3em]=\ &\inf _{\Delta {\text{ Zerlegung}}}O(\Delta ,\lambda f)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ O({\tilde {\Delta }},\lambda f)=\lambda U({\tilde {\Delta }},f){\text{ für alle Zerlegungen }}{\tilde {\Delta }}\right.}\\[0.3em]=\ &\inf _{\Delta {\text{ Zerlegung}}}\lambda U(\Delta ,f)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ \lambda <0\right.}\\[0.3em]=\ &\lambda \sup _{\Delta {\text{ Zerlegung}}}U(\Delta ,f)\\[0.3em]=\ &\lambda \cdot I_{-}(f,[a,b])\end{aligned}}}
und
I
−
(
λ
f
,
[
a
,
b
]
)
=
sup
Δ
Zerlegung
U
(
Δ
,
λ
f
)
↓
U
(
Δ
~
,
λ
f
)
=
λ
O
(
Δ
~
,
f
)
für alle Zerlegungen
Δ
~
=
sup
Δ
Zerlegung
λ
O
(
Δ
,
f
)
↓
λ
<
0
=
λ
inf
Δ
Zerlegung
O
(
Δ
,
f
)
=
λ
⋅
I
+
(
f
,
[
a
,
b
]
)
{\displaystyle {\begin{aligned}&I_{-}(\lambda f,[a,b])\\[0.3em]=\ &\sup _{\Delta {\text{ Zerlegung}}}U(\Delta ,\lambda f)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ U({\tilde {\Delta }},\lambda f)=\lambda O({\tilde {\Delta }},f){\text{ für alle Zerlegungen }}{\tilde {\Delta }}\right.}\\[0.3em]=\ &\sup _{\Delta {\text{ Zerlegung}}}\lambda O(\Delta ,f)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ \lambda <0\right.}\\[0.3em]=\ &\lambda \inf _{\Delta {\text{ Zerlegung}}}O(\Delta ,f)\\[0.3em]=\ &\lambda \cdot I_{+}(f,[a,b])\end{aligned}}}
Also:
I
+
(
λ
f
,
[
a
,
b
]
)
=
λ
⋅
I
−
(
f
,
[
a
,
b
]
)
↓
f
ist riemannintegrierbar, d.h.
I
+
(
f
,
[
a
,
b
]
)
=
I
−
(
f
,
[
a
,
b
]
)
=
λ
⋅
I
+
(
f
,
[
a
,
b
]
)
=
I
−
(
λ
f
,
[
a
,
b
]
)
{\displaystyle {\begin{aligned}&I_{+}(\lambda f,[a,b])\\[0.3em]=\ &\lambda \cdot I_{-}(f,[a,b])\\[0.3em]&{\color {OliveGreen}\left\downarrow \ f{\text{ ist riemannintegrierbar, d.h. }}I_{+}(f,[a,b])=I_{-}(f,[a,b])\right.}\\[0.3em]=\ &\lambda \cdot I_{+}(f,[a,b])\\[0.3em]=\ &I_{-}(\lambda f,[a,b])\\[0.3em]\end{aligned}}}
Damit ist die Funktion
λ
f
{\displaystyle \lambda f}
∫
a
b
λ
f
(
x
)
d
x
=
I
+
(
λ
f
,
[
a
,
b
]
)
=
λ
⋅
I
−
(
f
,
[
a
,
b
]
)
=
λ
∫
a
b
f
(
x
)
d
x
{\displaystyle {\begin{aligned}&\int _{a}^{b}\lambda f(x)\mathrm {d} x\\[0.3em]=\ &I_{+}(\lambda f,[a,b])\\[0.3em]=\ &\lambda \cdot I_{-}(f,[a,b])\\[0.3em]=\ &\lambda \int _{a}^{b}f(x)\mathrm {d} x\end{aligned}}}
Additivität der Grenzen beim Riemannintegral
Bearbeiten
Satz
Seien
a
,
b
,
c
∈
R
{\displaystyle a,b,c\in \mathbb {R} }
a
≤
c
≤
b
{\displaystyle a\leq c\leq b}
f
:
[
a
,
b
]
→
R
{\displaystyle f:[a,b]\to \mathbb {R} }
f
{\displaystyle f}
[
a
,
b
]
{\displaystyle [a,b]}
f
{\displaystyle f}
[
a
,
c
]
{\displaystyle [a,c]}
[
c
,
b
]
{\displaystyle [c,b]}
∫
a
b
f
(
x
)
d
x
=
∫
a
c
f
(
x
)
d
x
+
∫
c
b
f
(
x
)
d
x
{\displaystyle \int _{a}^{b}f(x)\mathrm {d} x=\int _{a}^{c}f(x)\mathrm {d} x+\int _{c}^{b}f(x)\mathrm {d} x}
Beweis
Wir beweisen zunächst
I
+
(
f
,
[
a
,
c
]
)
+
I
+
(
f
,
[
c
,
b
]
)
≥
I
+
(
f
,
[
a
,
b
]
)
{\displaystyle I_{+}(f,[a,c])+I_{+}(f,[c,b])\geq I_{+}(f,[a,b])}
I
−
(
f
,
[
a
,
c
]
)
+
I
−
(
f
,
[
c
,
b
]
)
≤
I
−
(
f
,
[
a
,
b
]
)
{\displaystyle I_{-}(f,[a,c])+I_{-}(f,[c,b])\leq I_{-}(f,[a,b])}
Δ
1
=
(
x
0
,
x
1
,
…
,
x
n
)
{\displaystyle \Delta _{1}=(x_{0},x_{1},\ldots ,x_{n})}
[
a
,
c
]
{\displaystyle [a,c]}
Δ
2
=
(
w
0
,
…
,
w
l
)
{\displaystyle \Delta _{2}=(w_{0},\ldots ,w_{l})}
[
c
,
b
]
{\displaystyle [c,b]}
a
=
x
0
{\displaystyle a=x_{0}}
x
n
=
c
=
w
0
{\displaystyle x_{n}=c=w_{0}}
w
l
=
b
{\displaystyle w_{l}=b}
Δ
:=
(
x
0
,
x
1
,
…
,
x
n
,
w
1
,
…
,
w
l
)
{\displaystyle \Delta :=(x_{0},x_{1},\ldots ,x_{n},w_{1},\ldots ,w_{l})}
[
a
,
b
]
{\displaystyle [a,b]}
O
(
Δ
1
,
f
)
+
O
(
Δ
2
,
f
)
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
sup
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
+
∑
k
=
0
l
−
1
(
w
k
+
1
−
w
k
)
sup
x
∈
[
w
k
,
w
k
+
1
]
f
(
x
)
=
O
(
Δ
,
f
)
≥
I
+
(
f
,
[
a
,
b
]
)
{\displaystyle {\begin{aligned}&O(\Delta _{1},f)+O(\Delta _{2},f)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}f(x)+\sum _{k=0}^{l-1}(w_{k+1}-w_{k})\sup _{x\in [w_{k},w_{k+1}]}f(x)\\[0.3em]=\ &O(\Delta ,f)\\[0.3em]\geq \ &I_{+}(f,[a,b])\end{aligned}}}
sowie
U
(
Δ
1
,
f
)
+
U
(
Δ
2
,
f
)
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
inf
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
+
∑
k
=
0
l
−
1
(
w
k
+
1
−
w
k
)
inf
x
∈
[
w
k
,
w
k
+
1
]
f
(
x
)
=
U
(
Δ
,
f
)
≤
I
−
(
f
,
[
a
,
b
]
)
{\displaystyle {\begin{aligned}&U(\Delta _{1},f)+U(\Delta _{2},f)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}f(x)+\sum _{k=0}^{l-1}(w_{k+1}-w_{k})\inf _{x\in [w_{k},w_{k+1}]}f(x)\\[0.3em]=\ &U(\Delta ,f)\\[0.3em]\leq \ &I_{-}(f,[a,b])\end{aligned}}}
Folglich gilt
I
+
(
f
,
[
a
,
c
]
)
+
I
+
(
f
,
[
c
,
b
]
)
=
(
inf
Δ
1
Zerlegung
O
(
Δ
1
,
f
)
)
+
(
inf
Δ
2
Zerlegung
O
(
Δ
2
,
f
)
)
=
inf
Δ
1
,
Δ
2
Zerlegungen
(
O
(
Δ
1
,
f
)
+
O
(
Δ
2
,
f
)
)
↓
O
(
Δ
1
,
f
)
+
O
(
Δ
2
,
f
)
≥
I
+
(
f
,
[
a
,
b
]
)
für alle Zerlegungen
Δ
1
,
Δ
2
von
[
a
,
c
]
bzw.
[
c
,
b
]
≥
I
+
(
f
,
[
a
,
b
]
)
{\displaystyle {\begin{aligned}&I_{+}(f,[a,c])+I_{+}(f,[c,b])\\[0.3em]=\ &\left(\inf _{\Delta _{1}{\text{ Zerlegung}}}O(\Delta _{1},f)\right)+\left(\inf _{\Delta _{2}{\text{ Zerlegung}}}O(\Delta _{2},f)\right)\\[0.3em]=\ &\inf _{\Delta _{1},\Delta _{2}{\text{ Zerlegungen}}}\left(O(\Delta _{1},f)+O(\Delta _{2},f)\right)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ O(\Delta _{1},f)+O(\Delta _{2},f)\geq I_{+}(f,[a,b]){\text{ für alle Zerlegungen }}\Delta _{1},\Delta _{2}{\text{ von }}[a,c]{\text{ bzw. }}[c,b]\right.}\\[0.3em]\geq \ &I_{+}(f,[a,b])\end{aligned}}}
und
I
−
(
f
,
[
a
,
c
]
)
+
I
−
(
f
,
[
c
,
b
]
)
=
(
sup
Δ
1
Zerlegung
U
(
Δ
1
,
f
)
)
+
(
sup
Δ
2
Zerlegung
U
(
Δ
2
,
f
)
)
=
sup
Δ
1
,
Δ
2
Zerlegungen
(
U
(
Δ
1
,
f
)
+
U
(
Δ
2
,
f
)
)
↓
U
(
Δ
1
,
f
)
+
U
(
Δ
2
,
f
)
≤
I
−
(
f
,
[
a
,
b
]
)
für alle Zerlegungen
Δ
1
,
Δ
2
von
[
a
,
c
]
bzw.
[
c
,
b
]
≤
I
−
(
f
,
[
a
,
b
]
)
{\displaystyle {\begin{aligned}&I_{-}(f,[a,c])+I_{-}(f,[c,b])\\[0.3em]=\ &\left(\sup _{\Delta _{1}{\text{ Zerlegung}}}U(\Delta _{1},f)\right)+\left(\sup _{\Delta _{2}{\text{ Zerlegung}}}U(\Delta _{2},f)\right)\\[0.3em]=\ &\sup _{\Delta _{1},\Delta _{2}{\text{ Zerlegungen}}}\left(U(\Delta _{1},f)+U(\Delta _{2},f)\right)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ U(\Delta _{1},f)+U(\Delta _{2},f)\leq I_{-}(f,[a,b]){\text{ für alle Zerlegungen }}\Delta _{1},\Delta _{2}{\text{ von }}[a,c]{\text{ bzw. }}[c,b]\right.}\\[0.3em]\leq \ &I_{-}(f,[a,b])\end{aligned}}}
Als Nächstes zeigen wir
I
+
(
f
,
[
a
,
b
]
)
≥
I
+
(
f
,
[
a
,
c
]
)
+
I
+
(
f
,
[
c
,
b
]
)
{\displaystyle I_{+}(f,[a,b])\geq I_{+}(f,[a,c])+I_{+}(f,[c,b])}
I
−
(
f
,
[
a
,
b
]
)
≤
I
−
(
f
,
[
a
,
c
]
)
+
I
−
(
f
,
[
c
,
b
]
)
{\displaystyle I_{-}(f,[a,b])\leq I_{-}(f,[a,c])+I_{-}(f,[c,b])}
Δ
=
(
x
0
,
x
1
,
…
,
x
n
)
{\displaystyle \Delta =(x_{0},x_{1},\ldots ,x_{n})}
[
a
,
b
]
{\displaystyle [a,b]}
Δ
~
=
(
y
0
,
y
1
,
…
,
y
m
)
{\displaystyle {\tilde {\Delta }}=(y_{0},y_{1},\ldots ,y_{m})}
Δ
{\displaystyle \Delta }
c
{\displaystyle c}
c
=
y
i
{\displaystyle c=y_{i}}
0
≤
i
≤
m
{\displaystyle 0\leq i\leq m}
c
{\displaystyle c}
Δ
{\displaystyle \Delta }
Δ
~
=
Δ
{\displaystyle {\tilde {\Delta }}=\Delta }
0
≤
i
<
n
{\displaystyle 0\leq i<n}
x
i
<
c
<
x
i
+
1
{\displaystyle x_{i}<c<x_{i+1}}
Δ
~
=
(
x
0
,
x
1
,
…
,
x
i
,
c
,
x
i
+
1
,
…
,
x
n
)
{\displaystyle {\tilde {\Delta }}=(x_{0},x_{1},\ldots ,x_{i},c,x_{i+1},\ldots ,x_{n})}
m
=
n
+
1
{\displaystyle m=n+1}
Δ
~
=
(
y
0
,
y
1
,
…
,
y
m
)
{\displaystyle {\tilde {\Delta }}=(y_{0},y_{1},\ldots ,y_{m})}
Δ
{\displaystyle \Delta }
c
=
y
i
{\displaystyle c=y_{i}}
Δ
1
:=
(
y
0
,
y
1
,
…
,
y
i
)
{\displaystyle \Delta _{1}:=(y_{0},y_{1},\ldots ,y_{i})}
Δ
2
:=
(
y
i
,
y
i
+
1
,
…
,
y
m
)
{\displaystyle \Delta _{2}:=(y_{i},y_{i+1},\ldots ,y_{m})}
[
a
,
c
]
{\displaystyle [a,c]}
[
c
,
b
]
{\displaystyle [c,b]}
Δ
~
{\displaystyle {\tilde {\Delta }}}
Δ
{\displaystyle \Delta }
O
(
Δ
,
f
)
≥
O
(
Δ
~
,
f
)
=
∑
k
=
0
m
−
1
(
y
k
+
1
−
y
k
)
sup
x
∈
[
y
k
,
y
k
+
1
]
f
(
x
)
=
∑
k
=
0
i
−
1
(
y
k
+
1
−
y
k
)
sup
x
∈
[
y
k
,
y
k
+
1
]
f
(
x
)
+
∑
k
=
i
m
−
1
(
y
k
+
1
−
y
k
)
sup
x
∈
[
y
k
,
y
k
+
1
]
f
(
x
)
=
O
(
Δ
1
,
f
)
+
O
(
Δ
2
,
f
)
≥
I
+
(
f
,
[
a
,
c
]
)
+
I
+
(
f
,
[
c
,
b
]
)
{\displaystyle {\begin{aligned}&O(\Delta ,f)\\[0.3em]\geq \ &O({\tilde {\Delta }},f)\\[0.3em]=\ &\sum _{k=0}^{m-1}(y_{k+1}-y_{k})\sup _{x\in [y_{k},y_{k+1}]}f(x)\\[0.3em]=\ &\sum _{k=0}^{i-1}(y_{k+1}-y_{k})\sup _{x\in [y_{k},y_{k+1}]}f(x)+\sum _{k=i}^{m-1}(y_{k+1}-y_{k})\sup _{x\in [y_{k},y_{k+1}]}f(x)\\[0.3em]=\ &O(\Delta _{1},f)+O(\Delta _{2},f)\\[0.3em]\geq \ &I_{+}(f,[a,c])+I_{+}(f,[c,b])\end{aligned}}}
Genauso ist
U
(
Δ
,
f
)
≤
U
(
Δ
~
,
f
)
=
∑
k
=
0
m
−
1
(
y
k
+
1
−
y
k
)
inf
x
∈
[
y
k
,
y
k
+
1
]
f
(
x
)
=
∑
k
=
0
i
−
1
(
y
k
+
1
−
y
k
)
inf
x
∈
[
y
k
,
y
k
+
1
]
f
(
x
)
+
∑
k
=
i
m
−
1
(
y
k
+
1
−
y
k
)
inf
x
∈
[
y
k
,
y
k
+
1
]
f
(
x
)
=
U
(
Δ
1
,
f
)
+
U
(
Δ
2
,
f
)
≤
I
−
(
f
,
[
a
,
c
]
)
+
I
−
(
f
,
[
c
,
b
]
)
{\displaystyle {\begin{aligned}&U(\Delta ,f)\\[0.3em]\leq \ &U({\tilde {\Delta }},f)\\[0.3em]=\ &\sum _{k=0}^{m-1}(y_{k+1}-y_{k})\inf _{x\in [y_{k},y_{k+1}]}f(x)\\[0.3em]=\ &\sum _{k=0}^{i-1}(y_{k+1}-y_{k})\inf _{x\in [y_{k},y_{k+1}]}f(x)+\sum _{k=i}^{m-1}(y_{k+1}-y_{k})\inf _{x\in [y_{k},y_{k+1}]}f(x)\\[0.3em]=\ &U(\Delta _{1},f)+U(\Delta _{2},f)\\[0.3em]\leq \ &I_{-}(f,[a,c])+I_{-}(f,[c,b])\end{aligned}}}
Folglich gilt
I
+
(
f
,
[
a
,
b
]
)
=
inf
Δ
Zerlegung
O
(
Δ
,
f
)
↓
O
(
Δ
,
f
)
≥
I
+
(
f
,
[
a
,
c
]
)
+
I
+
(
f
,
[
c
,
b
]
)
für alle Zerlegungen
Δ
von
[
a
,
b
]
≥
I
+
(
f
,
[
a
,
c
]
)
+
I
+
(
f
,
[
c
,
b
]
)
{\displaystyle {\begin{aligned}&I_{+}(f,[a,b])\\[0.3em]=\ &\inf _{\Delta {\text{ Zerlegung}}}O(\Delta ,f)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ O(\Delta ,f)\geq I_{+}(f,[a,c])+I_{+}(f,[c,b]){\text{ für alle Zerlegungen }}\Delta {\text{ von }}[a,b]\right.}\\[0.3em]\geq \ &I_{+}(f,[a,c])+I_{+}(f,[c,b])\end{aligned}}}
und
I
−
(
f
,
[
a
,
b
]
)
=
sup
Δ
Zerlegung
U
(
Δ
,
f
)
↓
U
(
Δ
,
f
)
≤
I
−
(
f
,
[
a
,
c
]
)
+
I
−
(
f
,
[
c
,
b
]
)
für alle Zerlegungen
Δ
von
[
a
,
b
]
≤
I
−
(
f
,
[
a
,
c
]
)
+
I
−
(
f
,
[
c
,
b
]
)
{\displaystyle {\begin{aligned}&I_{-}(f,[a,b])\\[0.3em]=\ &\sup _{\Delta {\text{ Zerlegung}}}U(\Delta ,f)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ U(\Delta ,f)\leq I_{-}(f,[a,c])+I_{-}(f,[c,b]){\text{ für alle Zerlegungen }}\Delta {\text{ von }}[a,b]\right.}\\[0.3em]\leq \ &I_{-}(f,[a,c])+I_{-}(f,[c,b])\end{aligned}}}
Insgesamt haben wir
I
+
(
f
,
[
a
,
b
]
)
=
I
+
(
f
,
[
a
,
c
]
)
+
I
+
(
f
,
[
c
,
b
]
)
{\displaystyle I_{+}(f,[a,b])=I_{+}(f,[a,c])+I_{+}(f,[c,b])}
I
−
(
f
,
[
a
,
b
]
)
=
I
−
(
f
,
[
a
,
c
]
)
+
I
−
(
f
,
[
c
,
b
]
)
{\displaystyle I_{-}(f,[a,b])=I_{-}(f,[a,c])+I_{-}(f,[c,b])}
Ist nun
f
{\displaystyle f}
[
a
,
c
]
{\displaystyle [a,c]}
[
c
,
b
]
{\displaystyle [c,b]}
I
+
(
f
,
[
a
,
c
]
)
=
I
−
(
f
,
[
a
,
c
]
)
{\displaystyle I_{+}(f,[a,c])=I_{-}(f,[a,c])}
I
+
(
f
,
[
c
,
b
]
)
=
I
−
(
f
,
[
c
,
b
]
)
{\displaystyle I_{+}(f,[c,b])=I_{-}(f,[c,b])}
I
+
(
f
,
[
a
,
b
]
)
=
I
−
(
f
,
[
a
,
b
]
)
{\displaystyle I_{+}(f,[a,b])=I_{-}(f,[a,b])}
f
{\displaystyle f}
[
a
,
b
]
{\displaystyle [a,b]}
f
{\displaystyle f}
[
a
,
c
]
{\displaystyle [a,c]}
[
c
,
b
]
{\displaystyle [c,b]}
I
+
(
f
,
[
a
,
c
]
)
>
I
−
(
f
,
[
a
,
c
]
)
{\displaystyle I_{+}(f,[a,c])>I_{-}(f,[a,c])}
I
+
(
f
,
[
c
,
b
]
)
>
I
−
(
f
,
[
c
,
b
]
)
{\displaystyle I_{+}(f,[c,b])>I_{-}(f,[c,b])}
I
+
(
f
,
[
a
,
c
]
)
≥
I
−
(
f
,
[
a
,
c
]
)
{\displaystyle I_{+}(f,[a,c])\geq I_{-}(f,[a,c])}
I
+
(
f
,
[
c
,
b
]
)
≥
I
−
(
f
,
[
c
,
b
]
)
{\displaystyle I_{+}(f,[c,b])\geq I_{-}(f,[c,b])}
I
+
(
f
,
[
a
,
b
]
)
>
I
−
(
f
,
[
a
,
b
]
)
{\displaystyle I_{+}(f,[a,b])>I_{-}(f,[a,b])}
f
{\displaystyle f}
[
a
,
b
]
{\displaystyle [a,b]}
Im Falle der Integrierbarkeit gilt zudem
∫
a
b
f
(
x
)
d
x
=
I
+
(
f
,
[
a
,
b
]
)
=
I
+
(
f
,
[
a
,
c
]
)
+
I
+
(
f
,
[
c
,
b
]
)
=
∫
a
c
f
(
x
)
d
x
+
∫
c
b
f
(
x
)
d
x
{\displaystyle \int _{a}^{b}f(x)\mathrm {d} x=I_{+}(f,[a,b])=I_{+}(f,[a,c])+I_{+}(f,[c,b])=\int _{a}^{c}f(x)\mathrm {d} x+\int _{c}^{b}f(x)\mathrm {d} x}
Dreiecksungleichung für das Riemannintegral
Bearbeiten
Beweis
Um zu zeigen, dass
|
f
|
{\displaystyle |f|}
ϵ
{\displaystyle \epsilon }
ϵ
>
0
{\displaystyle \epsilon >0}
Δ
~
{\displaystyle {\tilde {\Delta }}}
O
(
Δ
~
,
|
f
|
)
−
U
(
Δ
~
,
|
f
|
)
<
ϵ
{\displaystyle O({\tilde {\Delta }},|f|)-U({\tilde {\Delta }},|f|)<\epsilon }
f
{\displaystyle f}
Δ
~
=
(
x
0
,
x
1
,
…
,
x
n
)
{\displaystyle {\tilde {\Delta }}=(x_{0},x_{1},\ldots ,x_{n})}
O
(
Δ
~
,
f
)
−
U
(
Δ
~
,
f
)
<
ϵ
{\displaystyle O({\tilde {\Delta }},f)-U({\tilde {\Delta }},f)<\epsilon }
Δ
~
{\displaystyle {\tilde {\Delta }}}
O
(
Δ
~
,
|
f
|
)
−
U
(
Δ
~
,
|
f
|
)
<
ϵ
{\displaystyle O({\tilde {\Delta }},|f|)-U({\tilde {\Delta }},|f|)<\epsilon }
sup
x
∈
[
c
,
d
]
|
f
(
x
)
|
−
inf
x
∈
[
c
,
d
]
|
f
(
x
)
|
≤
sup
x
∈
[
c
,
d
]
f
(
x
)
−
inf
x
∈
[
c
,
d
]
f
(
x
)
{\displaystyle \sup _{x\in [c,d]}|f(x)|-\inf _{x\in [c,d]}|f(x)|\leq \sup _{x\in [c,d]}f(x)-\inf _{x\in [c,d]}f(x)}
[
c
,
d
]
⊆
[
a
,
b
]
{\displaystyle [c,d]\subseteq [a,b]}
c
≤
d
{\displaystyle c\leq d}
A
:=
inf
x
∈
[
c
,
d
]
f
(
x
)
{\displaystyle A:=\inf _{x\in [c,d]}f(x)}
B
:=
sup
x
∈
[
c
,
d
]
f
(
x
)
{\displaystyle B:=\sup _{x\in [c,d]}f(x)}
C
:=
inf
x
∈
[
c
,
d
]
|
f
(
x
)
|
{\displaystyle C:=\inf _{x\in [c,d]}|f(x)|}
D
:=
sup
x
∈
[
c
,
d
]
|
f
(
x
)
|
{\displaystyle D:=\sup _{x\in [c,d]}|f(x)|}
0
≤
A
≤
B
{\displaystyle 0\leq A\leq B}
A
<
0
≤
B
{\displaystyle A<0\leq B}
A
≤
B
<
0
{\displaystyle A\leq B<0}
Zu zeigen ist jeweils
D
−
C
≤
B
−
A
{\displaystyle D-C\leq B-A}
Fall 1:
0
≤
A
≤
B
{\displaystyle 0\leq A\leq B}
Fall 2:
A
<
0
≤
B
{\displaystyle A<0\leq B}
Für alle
x
∈
[
c
,
d
]
{\displaystyle x\in [c,d]}
0
≤
|
f
(
x
)
|
=
max
{
f
(
x
)
,
−
f
(
x
)
}
≤
max
{
B
,
−
A
}
{\displaystyle 0\leq |f(x)|=\max\{f(x),-f(x)\}\leq \max\{B,-A\}}
C
≥
0
{\displaystyle C\geq 0}
D
≤
max
{
B
,
−
A
}
{\displaystyle D\leq \max\{B,-A\}}
B
,
−
A
≥
0
{\displaystyle B,-A\geq 0}
max
{
B
,
−
A
}
≤
B
−
A
{\displaystyle \max\{B,-A\}\leq B-A}
D
−
C
≤
max
{
B
,
−
A
}
−
0
≤
B
−
A
{\displaystyle D-C\leq \max\{B,-A\}-0\leq B-A}
Fall 3:
A
≤
B
<
0
{\displaystyle A\leq B<0}
Damit ist
sup
x
∈
[
c
,
d
]
|
f
(
x
)
|
−
inf
x
∈
[
c
,
d
]
|
f
(
x
)
|
≤
sup
x
∈
[
c
,
d
]
f
(
x
)
−
inf
x
∈
[
c
,
d
]
f
(
x
)
{\displaystyle \sup _{x\in [c,d]}|f(x)|-\inf _{x\in [c,d]}|f(x)|\leq \sup _{x\in [c,d]}f(x)-\inf _{x\in [c,d]}f(x)}
[
c
,
d
]
⊆
[
a
,
b
]
{\displaystyle [c,d]\subseteq [a,b]}
c
≤
d
{\displaystyle c\leq d}
[
x
k
,
x
k
+
1
]
{\displaystyle [x_{k},x_{k+1}]}
Δ
~
{\displaystyle {\tilde {\Delta }}}
O
(
Δ
~
,
|
f
|
)
−
U
(
Δ
~
,
|
f
|
)
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
sup
x
∈
[
x
k
,
x
k
+
1
]
|
f
(
x
)
|
−
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
inf
x
∈
[
x
k
,
x
k
+
1
]
|
f
(
x
)
|
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
(
sup
x
∈
[
x
k
,
x
k
+
1
]
|
f
(
x
)
|
−
inf
x
∈
[
x
k
,
x
k
+
1
]
|
f
(
x
)
|
)
≤
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
(
sup
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
−
inf
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
)
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
sup
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
−
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
inf
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
=
O
(
Δ
~
,
f
)
−
U
(
Δ
~
,
f
)
<
ϵ
{\displaystyle {\begin{aligned}&O({\tilde {\Delta }},|f|)-U({\tilde {\Delta }},|f|)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}|f(x)|-\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}|f(x)|\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\left(\sup _{x\in [x_{k},x_{k+1}]}|f(x)|-\inf _{x\in [x_{k},x_{k+1}]}|f(x)|\right)\\[0.3em]\leq \ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\left(\sup _{x\in [x_{k},x_{k+1}]}f(x)-\inf _{x\in [x_{k},x_{k+1}]}f(x)\right)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}f(x)-\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]=\ &O({\tilde {\Delta }},f)-U({\tilde {\Delta }},f)\\[0.3em]<\ &\epsilon \end{aligned}}}
Nun haben wir gezeigt, dass die Funktion
|
f
|
{\displaystyle |f|}
f
(
x
)
≤
|
f
(
x
)
|
{\displaystyle f(x)\leq |f(x)|}
x
∈
[
a
,
b
]
{\displaystyle x\in [a,b]}
f
{\displaystyle f}
|
f
|
{\displaystyle |f|}
∫
a
b
f
(
x
)
d
x
≤
∫
a
b
|
f
(
x
)
|
d
x
{\displaystyle \int _{a}^{b}f(x)\mathrm {d} x\leq \int _{a}^{b}|f(x)|\mathrm {d} x}
Gemäß der Faktorregel ist die Funktion
−
f
:
[
a
,
b
]
→
R
,
x
↦
−
f
(
x
)
{\displaystyle -f:[a,b]\to \mathbb {R} ,x\mapsto -f(x)}
∫
a
b
−
f
(
x
)
d
x
=
−
∫
a
b
f
(
x
)
d
x
{\displaystyle \int _{a}^{b}-f(x)\mathrm {d} x=-\int _{a}^{b}f(x)\mathrm {d} x}
Für alle
x
∈
[
a
,
b
]
{\displaystyle x\in [a,b]}
−
f
(
x
)
≤
|
f
(
x
)
|
{\displaystyle -f(x)\leq |f(x)|}
−
f
{\displaystyle -f}
|
f
|
{\displaystyle |f|}
∫
a
b
−
f
(
x
)
d
x
≤
∫
a
b
|
f
(
x
)
|
d
x
{\displaystyle \int _{a}^{b}-f(x)\mathrm {d} x\leq \int _{a}^{b}|f(x)|\mathrm {d} x}
Schließlich ist
|
∫
a
b
f
(
x
)
d
x
|
=
max
{
∫
a
b
f
(
x
)
d
x
,
−
∫
a
b
f
(
x
)
d
x
}
=
max
{
∫
a
b
f
(
x
)
d
x
,
∫
a
b
−
f
(
x
)
d
x
}
≤
∫
a
b
|
f
(
x
)
|
d
x
{\displaystyle \left|\int _{a}^{b}f(x)\mathrm {d} x\right|=\max \left\{\int _{a}^{b}f(x)\mathrm {d} x,-\int _{a}^{b}f(x)\mathrm {d} x\right\}=\max \left\{\int _{a}^{b}f(x)\mathrm {d} x,\int _{a}^{b}-f(x)\mathrm {d} x\right\}\leq \int _{a}^{b}|f(x)|\mathrm {d} x}
Hinweis
Aus der Riemannintegrierbarkeit von
|
f
|
{\displaystyle |f|}
f
{\displaystyle f}
Beispiel
Sei
f
:
[
0
,
1
]
→
R
{\displaystyle f:[0,1]\to \mathbb {R} }
f
(
x
)
=
{
+
1
x
∈
Q
−
1
x
∉
Q
{\displaystyle f(x)={\begin{cases}+1&x\in \mathbb {Q} \\-1&x\notin \mathbb {Q} \end{cases}}}
Dann gilt
|
f
(
x
)
|
=
1
{\displaystyle |f(x)|=1}
x
∈
[
0
,
1
]
{\displaystyle x\in [0,1]}
|
f
|
{\displaystyle |f|}
f
{\displaystyle f}
[
c
,
d
]
⊆
[
0
,
1
]
{\displaystyle [c,d]\subseteq [0,1]}
c
<
d
{\displaystyle c<d}
O
(
Δ
,
f
)
=
1
{\displaystyle O(\Delta ,f)=1}
U
(
Δ
,
f
)
=
−
1
{\displaystyle U(\Delta ,f)=-1}
Δ
{\displaystyle \Delta }
[
0
,
1
]
{\displaystyle [0,1]}
Das Produkt zweier riemannintegrierbarer Funktionen ist riemannintegrierbar
Bearbeiten
Beweis
Um zu zeigen, dass
f
g
{\displaystyle fg}
ϵ
{\displaystyle \epsilon }
ϵ
>
0
{\displaystyle \epsilon >0}
Δ
~
{\displaystyle {\tilde {\Delta }}}
O
(
Δ
~
,
f
g
)
−
U
(
Δ
~
,
f
g
)
<
ϵ
{\displaystyle O({\tilde {\Delta }},fg)-U({\tilde {\Delta }},fg)<\epsilon }
f
{\displaystyle f}
g
{\displaystyle g}
S
,
T
>
0
{\displaystyle S,T>0}
|
f
(
x
)
|
≤
S
{\displaystyle |f(x)|\leq S}
|
g
(
x
)
|
≤
T
{\displaystyle |g(x)|\leq T}
x
∈
[
a
,
b
]
{\displaystyle x\in [a,b]}
f
{\displaystyle f}
g
{\displaystyle g}
Δ
1
{\displaystyle \Delta _{1}}
Δ
2
{\displaystyle \Delta _{2}}
O
(
Δ
1
,
f
)
−
U
(
Δ
1
,
f
)
<
ϵ
2
T
{\displaystyle O(\Delta _{1},f)-U(\Delta _{1},f)<{\tfrac {\epsilon }{2T}}}
O
(
Δ
2
,
g
)
−
U
(
Δ
2
,
g
)
<
ϵ
2
S
{\displaystyle O(\Delta _{2},g)-U(\Delta _{2},g)<{\tfrac {\epsilon }{2S}}}
Δ
~
=
(
x
0
,
x
1
,
…
,
x
n
)
{\displaystyle {\tilde {\Delta }}=(x_{0},x_{1},\ldots ,x_{n})}
Δ
1
{\displaystyle \Delta _{1}}
Δ
2
{\displaystyle \Delta _{2}}
O
(
Δ
~
,
f
g
)
−
U
(
Δ
~
,
f
g
)
<
ϵ
{\displaystyle O({\tilde {\Delta }},fg)-U({\tilde {\Delta }},fg)<\epsilon }
sup
x
∈
[
c
,
d
]
f
(
x
)
g
(
x
)
−
inf
x
∈
[
c
,
d
]
f
(
x
)
g
(
x
)
≤
S
(
sup
x
∈
[
c
,
d
]
g
(
x
)
−
inf
x
∈
[
c
,
d
]
g
(
x
)
)
+
T
(
sup
x
∈
[
c
,
d
]
f
(
x
)
−
inf
x
∈
[
c
,
d
]
f
(
x
)
)
{\displaystyle \sup _{x\in [c,d]}f(x)g(x)-\inf _{x\in [c,d]}f(x)g(x)\leq S\left(\sup _{x\in [c,d]}g(x)-\inf _{x\in [c,d]}g(x)\right)+T\left(\sup _{x\in [c,d]}f(x)-\inf _{x\in [c,d]}f(x)\right)}
[
c
,
d
]
⊆
[
a
,
b
]
{\displaystyle [c,d]\subseteq [a,b]}
c
≤
d
{\displaystyle c\leq d}
sup
x
∈
[
c
,
d
]
f
(
x
)
g
(
x
)
−
inf
x
∈
[
c
,
d
]
f
(
x
)
g
(
x
)
=
sup
x
,
y
∈
[
c
,
d
]
(
f
(
x
)
g
(
x
)
−
f
(
y
)
g
(
y
)
)
=
sup
x
,
y
∈
[
c
,
d
]
(
f
(
x
)
(
g
(
x
)
−
g
(
y
)
)
+
g
(
y
)
(
f
(
x
)
−
f
(
y
)
)
)
≤
sup
x
,
y
∈
[
c
,
d
]
(
f
(
x
)
(
g
(
x
)
−
g
(
y
)
)
)
+
sup
x
,
y
∈
[
c
,
d
]
(
g
(
y
)
(
f
(
x
)
−
f
(
y
)
)
)
≤
S
sup
x
,
y
∈
[
c
,
d
]
(
g
(
x
)
−
g
(
y
)
)
+
T
sup
x
,
y
∈
[
c
,
d
]
(
f
(
x
)
−
f
(
y
)
)
=
S
(
sup
x
∈
[
c
,
d
]
g
(
x
)
−
inf
x
∈
[
c
,
d
]
g
(
x
)
)
+
T
(
sup
x
∈
[
c
,
d
]
f
(
x
)
−
inf
x
∈
[
c
,
d
]
f
(
x
)
)
{\displaystyle {\begin{aligned}&\sup _{x\in [c,d]}f(x)g(x)-\inf _{x\in [c,d]}f(x)g(x)\\[0.3em]=\ &\sup _{x,y\in [c,d]}\left(f(x)g(x)-f(y)g(y)\right)\\[0.3em]=\ &\sup _{x,y\in [c,d]}\left(f(x)(g(x)-g(y))+g(y)(f(x)-f(y))\right)\\[0.3em]\leq \ &\sup _{x,y\in [c,d]}\left(f(x)(g(x)-g(y))\right)+\sup _{x,y\in [c,d]}\left(g(y)(f(x)-f(y))\right)\\[0.3em]\leq \ &S\sup _{x,y\in [c,d]}\left(g(x)-g(y)\right)+T\sup _{x,y\in [c,d]}\left(f(x)-f(y)\right)\\[0.3em]=\ &S\left(\sup _{x\in [c,d]}g(x)-\inf _{x\in [c,d]}g(x)\right)+T\left(\sup _{x\in [c,d]}f(x)-\inf _{x\in [c,d]}f(x)\right)\end{aligned}}}
Dies wenden wir jetzt auf die Intervalle
[
x
k
,
x
k
+
1
]
{\displaystyle [x_{k},x_{k+1}]}
Δ
~
{\displaystyle {\tilde {\Delta }}}
O
(
Δ
~
,
f
g
)
−
U
(
Δ
~
,
f
g
)
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
sup
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
g
(
x
)
−
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
inf
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
g
(
x
)
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
(
sup
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
g
(
x
)
−
inf
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
g
(
x
)
)
≤
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
(
S
(
sup
x
∈
[
c
,
d
]
g
(
x
)
−
inf
x
∈
[
c
,
d
]
g
(
x
)
)
+
T
(
sup
x
∈
[
c
,
d
]
f
(
x
)
−
inf
x
∈
[
c
,
d
]
f
(
x
)
)
)
=
S
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
(
sup
x
∈
[
x
k
,
x
k
+
1
]
g
(
x
)
−
inf
x
∈
[
x
k
,
x
k
+
1
]
g
(
x
)
)
+
T
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
(
sup
x
∈
[
c
,
d
]
f
(
x
)
−
inf
x
∈
[
c
,
d
]
f
(
x
)
)
=
S
(
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
sup
x
∈
[
x
k
,
x
k
+
1
]
g
(
x
)
−
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
inf
x
∈
[
x
k
,
x
k
+
1
]
g
(
x
)
)
+
T
(
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
sup
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
−
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
inf
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
)
=
S
(
O
(
Δ
~
,
g
)
−
U
(
Δ
~
,
g
)
)
+
T
(
O
(
Δ
~
,
f
)
−
U
(
Δ
~
,
f
)
)
↓
Δ
~
ist Verfeinerung von
Δ
1
und
Δ
2
≤
S
(
O
(
Δ
2
,
g
)
−
U
(
Δ
2
,
g
)
)
+
T
(
O
(
Δ
1
,
f
)
−
U
(
Δ
1
,
f
)
)
<
S
⋅
ϵ
2
S
+
T
⋅
ϵ
2
T
=
ϵ
{\displaystyle {\begin{aligned}&O({\tilde {\Delta }},fg)-U({\tilde {\Delta }},fg)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}f(x)g(x)-\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}f(x)g(x)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\left(\sup _{x\in [x_{k},x_{k+1}]}f(x)g(x)-\inf _{x\in [x_{k},x_{k+1}]}f(x)g(x)\right)\\[0.3em]\leq \ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\left(S\left(\sup _{x\in [c,d]}g(x)-\inf _{x\in [c,d]}g(x)\right)+T\left(\sup _{x\in [c,d]}f(x)-\inf _{x\in [c,d]}f(x)\right)\right)\\[0.3em]=\ &S\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\left(\sup _{x\in [x_{k},x_{k+1}]}g(x)-\inf _{x\in [x_{k},x_{k+1}]}g(x)\right)+T\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\left(\sup _{x\in [c,d]}f(x)-\inf _{x\in [c,d]}f(x)\right)\\[0.3em]=\ &S\left(\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}g(x)-\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}g(x)\right)\\&\qquad {}+T\left(\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}f(x)-\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}f(x)\right)\\[0.3em]=\ &S\left(O({\tilde {\Delta }},g)-U({\tilde {\Delta }},g)\right)+T\left(O({\tilde {\Delta }},f)-U({\tilde {\Delta }},f)\right)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\tilde {\Delta }}{\text{ ist Verfeinerung von }}\Delta _{1}{\text{ und }}\Delta _{2}\right.}\\[0.3em]\leq \ &S\left(O(\Delta _{2},g)-U(\Delta _{2},g)\right)+T\left(O(\Delta _{1},f)-U(\Delta _{1},f)\right)\\[0.3em]<\ &S\cdot {\tfrac {\epsilon }{2S}}+T\cdot {\tfrac {\epsilon }{2T}}\\[0.3em]=\ &\epsilon \end{aligned}}}
Somit ist
f
g
{\displaystyle fg}
Monotone Funktionen sind riemannintegrierbar
Bearbeiten
Ober- und Untersumme einer monoton steigenden Funktion bei Aufteilung in gleich große Teilintervalle Ist unsere Funktion
f
{\displaystyle f}
h
{\displaystyle h}
h
{\displaystyle h}
f
(
b
)
−
f
(
a
)
{\displaystyle f(b)-f(a)}
h
{\displaystyle h}
f
{\displaystyle f}
Beweis
Wir nehmen an, dass
f
{\displaystyle f}
f
{\displaystyle f}
−
f
{\displaystyle -f}
−
1
{\displaystyle -1}
f
{\displaystyle f}
inf
x
∈
[
c
,
d
]
f
(
x
)
=
f
(
c
)
{\displaystyle \inf _{x\in [c,d]}f(x)=f(c)}
sup
x
∈
[
c
,
d
]
f
(
x
)
=
f
(
d
)
{\displaystyle \sup _{x\in [c,d]}f(x)=f(d)}
[
c
,
d
]
⊆
[
a
,
b
]
{\displaystyle [c,d]\subseteq [a,b]}
c
≤
d
{\displaystyle c\leq d}
a
=
b
{\displaystyle a=b}
f
{\displaystyle f}
Δ
n
:=
(
x
0
,
x
1
,
…
,
x
n
)
{\displaystyle \Delta _{n}:=(x_{0},x_{1},\ldots ,x_{n})}
x
k
:=
a
+
k
b
−
a
n
{\displaystyle x_{k}:=a+k{\tfrac {b-a}{n}}}
n
∈
N
{\displaystyle n\in \mathbb {N} }
[
a
,
b
]
{\displaystyle [a,b]}
[
a
,
b
]
{\displaystyle [a,b]}
n
{\displaystyle n}
O
(
Δ
n
,
f
)
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
sup
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
=
∑
k
=
0
n
−
1
b
−
a
n
f
(
x
k
+
1
)
=
b
−
a
n
(
f
(
x
1
)
+
…
+
f
(
x
n
−
1
)
+
f
(
b
)
)
U
(
Δ
n
,
f
)
=
∑
k
=
0
n
−
1
(
x
k
+
1
−
x
k
)
inf
x
∈
[
x
k
,
x
k
+
1
]
f
(
x
)
=
∑
k
=
0
n
−
1
b
−
a
n
f
(
x
k
)
=
b
−
a
n
(
f
(
a
)
+
f
(
x
1
)
+
…
+
f
(
x
n
−
1
)
)
{\displaystyle {\begin{aligned}O(\Delta _{n},f)&=\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]&=\sum _{k=0}^{n-1}{\tfrac {b-a}{n}}f(x_{k+1})\\[0.3em]&={\tfrac {b-a}{n}}\left(f(x_{1})+\ldots +f(x_{n-1})+f(b)\right)\\[0.3em]U(\Delta _{n},f)&=\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]&=\sum _{k=0}^{n-1}{\tfrac {b-a}{n}}f(x_{k})\\[0.3em]&={\tfrac {b-a}{n}}\left(f(a)+f(x_{1})+\ldots +f(x_{n-1})\right)\end{aligned}}}
Wir stellen fest:
O
(
Δ
n
,
f
)
−
U
(
Δ
n
,
f
)
=
b
−
a
n
(
f
(
b
)
−
f
(
a
)
)
{\displaystyle O(\Delta _{n},f)-U(\Delta _{n},f)={\tfrac {b-a}{n}}\left(f(b)-f(a)\right)}
Daraus folgt die Riemannintegrierbarkeit von
f
{\displaystyle f}
ϵ
{\displaystyle \epsilon }
ϵ
>
0
{\displaystyle \epsilon >0}
n
∈
N
{\displaystyle n\in \mathbb {N} }
b
−
a
n
(
f
(
b
)
−
f
(
a
)
)
<
ϵ
{\displaystyle {\tfrac {b-a}{n}}\left(f(b)-f(a)\right)<\epsilon }
Δ
n
{\displaystyle \Delta _{n}}
O
(
Δ
n
,
f
)
−
U
(
Δ
n
,
f
)
<
ϵ
{\displaystyle O(\Delta _{n},f)-U(\Delta _{n},f)<\epsilon }
Fast überall gleiche Funktionen haben das gleiche Riemannintegral
Bearbeiten
Beweis
Wir dürfen annehmen, dass
f
≡
0
{\displaystyle f\equiv 0}
f
{\displaystyle f}
g
{\displaystyle g}
g
−
f
{\displaystyle g-f}
g
−
f
{\displaystyle g-f}
∫
a
b
(
g
(
x
)
−
f
(
x
)
)
d
x
=
0
{\displaystyle \int _{a}^{b}(g(x)-f(x))\mathrm {d} x=0}
g
=
(
g
−
f
)
+
f
{\displaystyle g=(g-f)+f}
∫
a
b
g
(
x
)
d
x
=
∫
a
b
(
g
(
x
)
−
f
(
x
)
)
d
x
+
∫
a
b
f
(
x
)
d
x
=
∫
a
b
f
(
x
)
d
x
{\displaystyle \int _{a}^{b}g(x)\mathrm {d} x=\int _{a}^{b}(g(x)-f(x))\mathrm {d} x+\int _{a}^{b}f(x)\mathrm {d} x=\int _{a}^{b}f(x)\mathrm {d} x}
Sei nun also
f
≡
0
{\displaystyle f\equiv 0}
g
{\displaystyle g}
f
{\displaystyle f}
x
1
,
x
2
,
…
,
x
n
{\displaystyle x_{1},x_{2},\ldots ,x_{n}}
g
(
x
)
=
{
0
x
≠
x
k
für alle
k
∈
{
1
,
2
,
…
,
n
}
y
k
x
=
x
k
für ein
k
∈
{
1
,
2
,
…
,
n
}
{\displaystyle g(x)={\begin{cases}0&x\neq x_{k}{\text{ für alle }}k\in \{1,2,\ldots ,n\}\\y_{k}&x=x_{k}{\text{ für ein }}k\in \{1,2,\ldots ,n\}\end{cases}}}
wobei
y
k
:=
g
(
x
k
)
≠
0
{\displaystyle y_{k}:=g(x_{k})\neq 0}
g
{\displaystyle g}
n
{\displaystyle n}
g
1
,
…
,
g
n
{\displaystyle g_{1},\ldots ,g_{n}}
g
k
(
x
)
=
{
0
x
≠
x
k
y
k
x
=
x
k
{\displaystyle g_{k}(x)={\begin{cases}0&x\neq x_{k}\\y_{k}&x=x_{k}\end{cases}}}
definiert sind. Indem wir erneut auf die Summenregel zurückgreifen, können wir uns also auf den Fall
k
=
1
{\displaystyle k=1}
g
k
{\displaystyle g_{k}}
x
k
{\displaystyle x_{k}}
0
{\displaystyle 0}
g
{\displaystyle g}
g
=
g
1
{\displaystyle g=g_{1}}
x
1
∈
{
a
,
b
}
{\displaystyle x_{1}\in \{a,b\}}
a
<
x
1
<
b
{\displaystyle a<x_{1}<b}
[
a
,
x
1
]
{\displaystyle [a,x_{1}]}
[
x
1
,
b
]
{\displaystyle [x_{1},b]}
x
1
{\displaystyle x_{1}}
x
1
∈
{
a
,
b
}
{\displaystyle x_{1}\in \{a,b\}}
y
1
>
0
{\displaystyle y_{1}>0}
−
g
{\displaystyle -g}
−
1
{\displaystyle -1}
y
1
>
0
{\displaystyle y_{1}>0}
a
<
b
{\displaystyle a<b}
a
=
b
{\displaystyle a=b}
Δ
=
(
a
)
{\displaystyle \Delta =(a)}
0
{\displaystyle 0}
Fall 1:
x
1
=
a
{\displaystyle x_{1}=a}
Wir betrachten die Zerlegungen
Δ
n
=
(
a
,
a
+
b
−
a
2
n
,
b
)
{\displaystyle \Delta _{n}=(a,a+{\tfrac {b-a}{2^{n}}},b)}
![x\in [a,b]](https://wikimedia.org/api/rest_v1/media/math/render/svg/026357b404ee584c475579fb2302a4e9881b8cce)
![{\displaystyle f:[a,b]\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/b592d102ccd1ba134d401c5b3ea177baaba3ffac)
![[a,b]](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935)
![{\displaystyle [a,c]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/70b45aa26fb924cdd37786af1adcf5d58cfe15af)
![{\displaystyle [c,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/07f3f774df12e18b9f2bb94333ff320bbd2c308d)
![{\displaystyle |f|:[a,b]\to \mathbb {R} ,x\mapsto |f(x)|}](https://wikimedia.org/api/rest_v1/media/math/render/svg/74b287d73debb5305f6a389e84391156221bada6)
![{\displaystyle g:[a,b]\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/8b1d5103d0e36767a27715bcfc57137119294aad)
![{\displaystyle (f\cdot g):[a,b]\to \mathbb {R} ,x\mapsto f(x)\cdot g(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/138a201de67c099c84036ded6d80403912b52168)
![{\displaystyle x,y\in [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e14f38807bc231b9e43d45c8e909d463afb17b5)
![{\displaystyle {\begin{aligned}&\sup _{x\in [x_{k},x_{k+1}]}f(x)-\inf _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ f{\text{ ist stetig}}\right.}\\[0.3em]=\ &\max _{x\in [x_{k},x_{k+1}]}f(x)-\min _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\text{Für alle }}x,y\in [x_{k},x_{k+1}]{\text{ gilt }}|x-y|<\delta \right.}\\[0.3em]<\ &{\frac {\epsilon }{b-a}}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cc6fc56df557bf2ec6f7d808251c0c755fee05b3)
![{\displaystyle {\begin{aligned}&O(\Delta ,f)-U(\Delta ,f)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}f(x)-\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\left(\sup _{x\in [x_{k},x_{k+1}]}f(x)-\inf _{x\in [x_{k},x_{k+1}]}f(x)\right)\\[0.3em]<\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k}){\frac {\epsilon }{b-a}}\\[0.3em]=\ &(b-a){\frac {\epsilon }{b-a}}\\[0.3em]=\ &\epsilon \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4406c4ab41bbdbb263e2049c3c29631342b3a01)
![{\displaystyle {\begin{aligned}&O({\tilde {\Delta }},f)\leq O({\tilde {\Delta }},g)\\[0.3em]&U({\tilde {\Delta }},f)\leq U({\tilde {\Delta }},g)\\[0.3em]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9290f0a4aa4950891a7221eb3c373dec82952a18)
![{\displaystyle I_{+}(f,[a,b])\leq I_{+}(g,[a,b])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee93bcea1c447879c143d70297b7fb364db7ace1)
![{\displaystyle I_{-}(f,[a,b])\leq I_{-}(g,[a,b])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/df316ff8808ff9dfc111605a2d1c94a6a77a38a8)
![{\displaystyle f,g:[a,b]\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/e569de0fc730b7a75b46b6fd57598e0c94712f91)
![{\displaystyle {\begin{aligned}&O({\tilde {\Delta }},f)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ f(x)\leq g(x){\text{ für alle }}x\in [a,b]\right.}\\[0.3em]\leq \ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}g(x)\\[0.3em]=\ &O({\tilde {\Delta }},g)\\[0.3em]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22c8841c4f76d99029d6c15417cd63abb063f0cf)
![{\displaystyle {\begin{aligned}&I_{+}(f,[a,b])\\[0.3em]=\ &\inf _{\Delta {\text{ Zerlegung}}}O(\Delta ,f)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ O(\Delta ,f)\leq O(\Delta ,g){\text{ für alle Zerlegungen }}\Delta \right.}\\[0.3em]\leq \ &\inf _{\Delta {\text{ Zerlegung}}}O(\Delta ,g)\\[0.3em]=\ &I_{+}(g,[a,b])\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b93580e2953e20553961325d123a4b529817dc09)
![{\displaystyle {\begin{aligned}&\int _{a}^{b}f(x)\mathrm {d} x\\[0.3em]&{\color {OliveGreen}\left\downarrow \ f{\text{ riemannintegrierbar}}\right.}\\[0.3em]=\ &I_{+}(f,[a,b])\\[0.3em]\leq \ &I_{+}(g,[a,b])\\[0.3em]&{\color {OliveGreen}\left\downarrow \ g{\text{ riemannintegrierbar}}\right.}\\[0.3em]=\ &\int _{a}^{b}g(x)\mathrm {d} x\\[0.3em]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1cb243650991d3126d04324adfebb54b6a08b6a5)
![{\displaystyle (f+g):[a,b]\to \mathbb {R} ,x\mapsto f(x)+g(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f23c2186122529084996adbff3fe1fafa9312818)
![{\displaystyle {\begin{aligned}&O(\Delta _{1},f)+O(\Delta _{2},g)\\[0.3em]\geq \ &O({\tilde {\Delta }},f)+O({\tilde {\Delta }},g)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}f(x)+\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}g(x)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\left(\sup _{x\in [x_{k},x_{k+1}]}f(x)+\sup _{x\in [x_{k},x_{k+1}]}g(x)\right)\\[0.3em]\geq \ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}(f(x)+g(x))\\[0.3em]=\ &O({\tilde {\Delta }},f+g)\\[0.3em]\geq \ &I_{+}(f+g,[a,b])\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3a9f2c297f24e2da64021c3de6b6f3aee5e91d6f)
![{\displaystyle {\begin{aligned}&U(\Delta _{1},f)+U(\Delta _{2},g)\\[0.3em]\leq \ &U({\tilde {\Delta }},f)+U({\tilde {\Delta }},g)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}f(x)+\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}g(x)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\left(\inf _{x\in [x_{k},x_{k+1}]}f(x)+\inf _{x\in [x_{k},x_{k+1}]}g(x)\right)\\[0.3em]\leq \ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}(f(x)+g(x))\\[0.3em]=\ &U({\tilde {\Delta }},f+g)\\[0.3em]\leq \ &I_{-}(f+g,[a,b])\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e2a70b2d90a5fcdd1a1437aa9ec5110927c550bd)
![{\displaystyle {\begin{aligned}&I_{+}(f,[a,b])+I_{+}(g,[a,b])\\[0.3em]=\ &\inf _{\Delta _{1}{\text{ Zerlegung}}}O(\Delta _{1},f)+\inf _{\Delta _{2}{\text{ Zerlegung}}}O(\Delta _{2},g)\\[0.3em]=\ &\inf _{\Delta _{1},\Delta _{2}{\text{ Zerlegungen}}}\left(O(\Delta _{1},f)+O(\Delta _{2},g)\right)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ O(\Delta _{1},f)+O(\Delta _{2},g)\geq I_{+}(f+g,[a,b]){\text{ für alle Zerlegungen }}\Delta _{1},\Delta _{2}\right.}\\[0.3em]\geq \ &I_{+}(f+g,[a,b])\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ed3e4354ba336df8d64f867d029af243ebde5789)
![{\displaystyle I_{-}(f+g,[a,b])\geq I_{-}(f,[a,b])+I_{-}(g,[a,b])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1764714381ca332403e05706747f0453b52bf814)
![{\displaystyle {\begin{aligned}&I_{-}(f,[a,b])+I_{-}(g,[a,b])\\[0.3em]=\ &\sup _{\Delta _{1}{\text{ Zerlegung}}}U(\Delta _{1},f)+\sup _{\Delta _{2}{\text{ Zerlegung}}}U(\Delta _{2},g)\\[0.3em]=\ &\sup _{\Delta _{1},\Delta _{2}{\text{ Zerlegungen}}}\left(U(\Delta _{1},f)+U(\Delta _{2},g)\right)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ U(\Delta _{1},f)+U(\Delta _{2},g)\leq I_{-}(f+g,[a,b]){\text{ für alle Zerlegungen }}\Delta _{1},\Delta _{2}\right.}\\[0.3em]\leq \ &I_{-}(f+g,[a,b])\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/abfd97d35fd1ab5658672a3683ab314f9e15cf15)
![{\displaystyle {\begin{aligned}I_{+}(f+g,[a,b])&\leq I_{+}(f,[a,b])+I_{+}(g,[a,b])\\[0.3em]I_{-}(f+g,[a,b])&\geq I_{-}(f,[a,b])+I_{-}(g,[a,b])\\[0.3em]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe383dab3e150da8b757909ce39ad63d5ce087b0)
![{\displaystyle {\begin{aligned}&I_{+}(f+g,[a,b])\\[0.3em]\leq \ &I_{+}(f,[a,b])+I_{+}(g,[a,b])\\[0.3em]&{\color {OliveGreen}\left\downarrow \ I_{+}(f,[a,b])=I_{-}(f,[a,b]){\text{ und }}I_{+}(g,[a,b])=I_{-}(g,[a,b])\right.}\\[0.3em]=\ &I_{-}(f,[a,b])+I_{-}(g,[a,b])\\[0.3em]\leq \ &I_{-}(f+g,[a,b])\\[0.3em]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5b7fef2c96e4354d88cfb98e35d302ca488d76c8)
![{\displaystyle I_{+}(f+g,[a,b])\geq I_{-}(f+g,[a,b])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2bbcf24605bb6036bed3407f094812a0f2046229)
![{\displaystyle I_{+}(f+g,[a,b])=I_{-}(f+g,[a,b])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d8b8faf85f5a50f2f8a53b6aeb23224926cc1c35)
![{\displaystyle \int _{a}^{b}\left(f(x)+g(x)\right)\mathrm {d} x=I_{+}(f+g,[a,b])=I_{+}(f,[a,b])+I_{+}(g,[a,b])=\int _{a}^{b}f(x)\mathrm {d} x+\int _{a}^{b}g(x)\mathrm {d} x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/27a51b1b13099350ef24303557eeca9feadbe83b)
![{\displaystyle (\lambda f):[a,b]\to \mathbb {R} ,x\mapsto \lambda \cdot f(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c831a49643034ab3b0941fee7dfa91d1ff730922)
![{\displaystyle {\begin{aligned}&O({\tilde {\Delta }},\lambda f)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}\lambda \cdot f(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ \lambda \geq 0\right.}\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\cdot \lambda \cdot \sup _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]=\ &\lambda \cdot \sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]=\ &\lambda \cdot O({\tilde {\Delta }},f)\\[0.3em]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dc43e1b8a281ccfc11ab54747240857e955b78a4)
![{\displaystyle {\begin{aligned}&U({\tilde {\Delta }},\lambda f)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}\lambda \cdot f(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ \lambda \geq 0\right.}\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\cdot \lambda \cdot \inf _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]=\ &\lambda \cdot \sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]=\ &\lambda \cdot U({\tilde {\Delta }},f)\\[0.3em]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/877c0ef89869763ed24cd80ab814f5198c4fc79b)
![{\displaystyle {\begin{aligned}&I_{+}(\lambda f,[a,b])\\[0.3em]=\ &\inf _{\Delta {\text{ Zerlegung}}}O(\Delta ,\lambda f)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ O({\tilde {\Delta }},\lambda f)=\lambda O({\tilde {\Delta }},f){\text{ für alle Zerlegungen }}{\tilde {\Delta }}\right.}\\[0.3em]=\ &\inf _{\Delta {\text{ Zerlegung}}}\lambda O(\Delta ,f)\\[0.3em]=\ &\lambda \inf _{\Delta {\text{ Zerlegung}}}O(\Delta ,f)\\[0.3em]=\ &\lambda \cdot I_{+}(f,[a,b])\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/db27fc0c09ee6eaf0c86d15e271dec3c7d8d979d)
![{\displaystyle {\begin{aligned}&I_{-}(\lambda f,[a,b])\\[0.3em]=\ &\sup _{\Delta {\text{ Zerlegung}}}U(\Delta ,\lambda f)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ U({\tilde {\Delta }},\lambda f)=\lambda U({\tilde {\Delta }},f){\text{ für alle Zerlegungen }}{\tilde {\Delta }}\right.}\\[0.3em]=\ &\sup _{\Delta {\text{ Zerlegung}}}\lambda U(\Delta ,f)\\[0.3em]=\ &\lambda \sup _{\Delta {\text{ Zerlegung}}}U(\Delta ,f)\\[0.3em]=\ &\lambda \cdot I_{-}(f,[a,b])\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4f35d095257a007237698654d524ae723cd9db84)
![{\displaystyle {\begin{aligned}&I_{+}(\lambda f,[a,b])\\[0.3em]=\ &\lambda \cdot I_{+}(f,[a,b])\\[0.3em]&{\color {OliveGreen}\left\downarrow \ f{\text{ ist riemannintegrierbar, d.h. }}I_{+}(f,[a,b])=I_{-}(f,[a,b])\right.}\\[0.3em]=\ &\lambda \cdot I_{-}(f,[a,b])\\[0.3em]=\ &I_{-}(\lambda f,[a,b])\\[0.3em]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/73dadda619d4a14ece559b8359e2c2d368a65fb0)
![{\displaystyle {\begin{aligned}&\int _{a}^{b}\lambda f(x)\mathrm {d} x\\[0.3em]=\ &I_{+}(\lambda f,[a,b])\\[0.3em]=\ &\lambda \cdot I_{+}(f,[a,b])\\[0.3em]=\ &\lambda \int _{a}^{b}f(x)\mathrm {d} x\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c2aea73fd49eb7f7480f12abdf191a1e8a827cd8)
![{\displaystyle {\begin{aligned}&O({\tilde {\Delta }},\lambda f)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}\lambda \cdot f(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ \lambda <0\right.}\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\cdot \lambda \cdot \inf _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]=\ &\lambda \cdot \sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]=\ &\lambda \cdot U({\tilde {\Delta }},f)\\[0.3em]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c6728ec07ba12e4b040915160769db8dd4937df)
![{\displaystyle {\begin{aligned}&U({\tilde {\Delta }},\lambda f)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}\lambda \cdot f(x)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ \lambda <0\right.}\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\cdot \lambda \cdot \sup _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]=\ &\lambda \cdot \sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]=\ &\lambda \cdot O({\tilde {\Delta }},f)\\[0.3em]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b5fecfd03c2e934236d519f3f8a4313c433418c2)
![{\displaystyle {\begin{aligned}&I_{+}(\lambda f,[a,b])\\[0.3em]=\ &\inf _{\Delta {\text{ Zerlegung}}}O(\Delta ,\lambda f)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ O({\tilde {\Delta }},\lambda f)=\lambda U({\tilde {\Delta }},f){\text{ für alle Zerlegungen }}{\tilde {\Delta }}\right.}\\[0.3em]=\ &\inf _{\Delta {\text{ Zerlegung}}}\lambda U(\Delta ,f)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ \lambda <0\right.}\\[0.3em]=\ &\lambda \sup _{\Delta {\text{ Zerlegung}}}U(\Delta ,f)\\[0.3em]=\ &\lambda \cdot I_{-}(f,[a,b])\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d9f47ad2416f7466e2f0484df2d1e63edfed2789)
![{\displaystyle {\begin{aligned}&I_{-}(\lambda f,[a,b])\\[0.3em]=\ &\sup _{\Delta {\text{ Zerlegung}}}U(\Delta ,\lambda f)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ U({\tilde {\Delta }},\lambda f)=\lambda O({\tilde {\Delta }},f){\text{ für alle Zerlegungen }}{\tilde {\Delta }}\right.}\\[0.3em]=\ &\sup _{\Delta {\text{ Zerlegung}}}\lambda O(\Delta ,f)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ \lambda <0\right.}\\[0.3em]=\ &\lambda \inf _{\Delta {\text{ Zerlegung}}}O(\Delta ,f)\\[0.3em]=\ &\lambda \cdot I_{+}(f,[a,b])\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48e67be0b29f50e44bc5ed269113ae413c9c9602)
![{\displaystyle {\begin{aligned}&I_{+}(\lambda f,[a,b])\\[0.3em]=\ &\lambda \cdot I_{-}(f,[a,b])\\[0.3em]&{\color {OliveGreen}\left\downarrow \ f{\text{ ist riemannintegrierbar, d.h. }}I_{+}(f,[a,b])=I_{-}(f,[a,b])\right.}\\[0.3em]=\ &\lambda \cdot I_{+}(f,[a,b])\\[0.3em]=\ &I_{-}(\lambda f,[a,b])\\[0.3em]\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/920fb6e15400cf1d41230077062f3f2d9a72791e)
![{\displaystyle {\begin{aligned}&\int _{a}^{b}\lambda f(x)\mathrm {d} x\\[0.3em]=\ &I_{+}(\lambda f,[a,b])\\[0.3em]=\ &\lambda \cdot I_{-}(f,[a,b])\\[0.3em]=\ &\lambda \int _{a}^{b}f(x)\mathrm {d} x\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/099712c9f117da2294c0e428093bcb8858f25ff5)
![{\displaystyle I_{+}(f,[a,c])+I_{+}(f,[c,b])\geq I_{+}(f,[a,b])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/de237770bc2f56c98a16e69214c16fa64de887ff)
![{\displaystyle I_{-}(f,[a,c])+I_{-}(f,[c,b])\leq I_{-}(f,[a,b])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7347e5f47c5fb0cbabafa382775e1cac83c376af)
![{\displaystyle {\begin{aligned}&O(\Delta _{1},f)+O(\Delta _{2},f)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}f(x)+\sum _{k=0}^{l-1}(w_{k+1}-w_{k})\sup _{x\in [w_{k},w_{k+1}]}f(x)\\[0.3em]=\ &O(\Delta ,f)\\[0.3em]\geq \ &I_{+}(f,[a,b])\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/754a5e05e7f572559289079f2792f8754038ed83)
![{\displaystyle {\begin{aligned}&U(\Delta _{1},f)+U(\Delta _{2},f)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}f(x)+\sum _{k=0}^{l-1}(w_{k+1}-w_{k})\inf _{x\in [w_{k},w_{k+1}]}f(x)\\[0.3em]=\ &U(\Delta ,f)\\[0.3em]\leq \ &I_{-}(f,[a,b])\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/06c8c6e78217e8e5374b3e933e194851ad1a8ff4)
![{\displaystyle {\begin{aligned}&I_{+}(f,[a,c])+I_{+}(f,[c,b])\\[0.3em]=\ &\left(\inf _{\Delta _{1}{\text{ Zerlegung}}}O(\Delta _{1},f)\right)+\left(\inf _{\Delta _{2}{\text{ Zerlegung}}}O(\Delta _{2},f)\right)\\[0.3em]=\ &\inf _{\Delta _{1},\Delta _{2}{\text{ Zerlegungen}}}\left(O(\Delta _{1},f)+O(\Delta _{2},f)\right)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ O(\Delta _{1},f)+O(\Delta _{2},f)\geq I_{+}(f,[a,b]){\text{ für alle Zerlegungen }}\Delta _{1},\Delta _{2}{\text{ von }}[a,c]{\text{ bzw. }}[c,b]\right.}\\[0.3em]\geq \ &I_{+}(f,[a,b])\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a99b26b4260f2bc55204321a11d8ea633b6d2819)
![{\displaystyle {\begin{aligned}&I_{-}(f,[a,c])+I_{-}(f,[c,b])\\[0.3em]=\ &\left(\sup _{\Delta _{1}{\text{ Zerlegung}}}U(\Delta _{1},f)\right)+\left(\sup _{\Delta _{2}{\text{ Zerlegung}}}U(\Delta _{2},f)\right)\\[0.3em]=\ &\sup _{\Delta _{1},\Delta _{2}{\text{ Zerlegungen}}}\left(U(\Delta _{1},f)+U(\Delta _{2},f)\right)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ U(\Delta _{1},f)+U(\Delta _{2},f)\leq I_{-}(f,[a,b]){\text{ für alle Zerlegungen }}\Delta _{1},\Delta _{2}{\text{ von }}[a,c]{\text{ bzw. }}[c,b]\right.}\\[0.3em]\leq \ &I_{-}(f,[a,b])\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ea9fa9e99454a1a86f70f24f015e0a926b099fcd)
![{\displaystyle I_{+}(f,[a,b])\geq I_{+}(f,[a,c])+I_{+}(f,[c,b])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a9f512e4fea36eda8366bc7aab8bb847121056af)
![{\displaystyle I_{-}(f,[a,b])\leq I_{-}(f,[a,c])+I_{-}(f,[c,b])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2538292af2a41aa6af7e1aabc12228d62cff1621)
![{\displaystyle {\begin{aligned}&O(\Delta ,f)\\[0.3em]\geq \ &O({\tilde {\Delta }},f)\\[0.3em]=\ &\sum _{k=0}^{m-1}(y_{k+1}-y_{k})\sup _{x\in [y_{k},y_{k+1}]}f(x)\\[0.3em]=\ &\sum _{k=0}^{i-1}(y_{k+1}-y_{k})\sup _{x\in [y_{k},y_{k+1}]}f(x)+\sum _{k=i}^{m-1}(y_{k+1}-y_{k})\sup _{x\in [y_{k},y_{k+1}]}f(x)\\[0.3em]=\ &O(\Delta _{1},f)+O(\Delta _{2},f)\\[0.3em]\geq \ &I_{+}(f,[a,c])+I_{+}(f,[c,b])\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f768f2c9d493d408610cb2d487dcbab05bcf644a)
![{\displaystyle {\begin{aligned}&U(\Delta ,f)\\[0.3em]\leq \ &U({\tilde {\Delta }},f)\\[0.3em]=\ &\sum _{k=0}^{m-1}(y_{k+1}-y_{k})\inf _{x\in [y_{k},y_{k+1}]}f(x)\\[0.3em]=\ &\sum _{k=0}^{i-1}(y_{k+1}-y_{k})\inf _{x\in [y_{k},y_{k+1}]}f(x)+\sum _{k=i}^{m-1}(y_{k+1}-y_{k})\inf _{x\in [y_{k},y_{k+1}]}f(x)\\[0.3em]=\ &U(\Delta _{1},f)+U(\Delta _{2},f)\\[0.3em]\leq \ &I_{-}(f,[a,c])+I_{-}(f,[c,b])\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b9a25273f916b5d498f248e36b0fe29704072338)
![{\displaystyle {\begin{aligned}&I_{+}(f,[a,b])\\[0.3em]=\ &\inf _{\Delta {\text{ Zerlegung}}}O(\Delta ,f)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ O(\Delta ,f)\geq I_{+}(f,[a,c])+I_{+}(f,[c,b]){\text{ für alle Zerlegungen }}\Delta {\text{ von }}[a,b]\right.}\\[0.3em]\geq \ &I_{+}(f,[a,c])+I_{+}(f,[c,b])\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c35e3c2be32b286cf7bca2efd6b00a708d1e0f74)
![{\displaystyle {\begin{aligned}&I_{-}(f,[a,b])\\[0.3em]=\ &\sup _{\Delta {\text{ Zerlegung}}}U(\Delta ,f)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ U(\Delta ,f)\leq I_{-}(f,[a,c])+I_{-}(f,[c,b]){\text{ für alle Zerlegungen }}\Delta {\text{ von }}[a,b]\right.}\\[0.3em]\leq \ &I_{-}(f,[a,c])+I_{-}(f,[c,b])\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f305536815abca026e6a52eccd4c126866385f42)
![{\displaystyle I_{+}(f,[a,b])=I_{+}(f,[a,c])+I_{+}(f,[c,b])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0afbabf944b6074ac3e21a0b48c8dfa1e558261a)
![{\displaystyle I_{-}(f,[a,b])=I_{-}(f,[a,c])+I_{-}(f,[c,b])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dcc2688fe51bd02fa534deb7bc356a1662648c79)
![{\displaystyle I_{+}(f,[a,c])=I_{-}(f,[a,c])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/50ad86b39c64eb02a0dfe68a926caa8f6799468a)
![{\displaystyle I_{+}(f,[c,b])=I_{-}(f,[c,b])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aaf0122a5f7c5d5cf7efc086cdf609cf51aef269)
![{\displaystyle I_{+}(f,[a,b])=I_{-}(f,[a,b])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/178d85d3ef1f01c0877295a3d65fb7bdb1ac7abe)
![{\displaystyle I_{+}(f,[a,c])>I_{-}(f,[a,c])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6d9f0742423b75daa025f261984bab742e4fd77c)
![{\displaystyle I_{+}(f,[c,b])>I_{-}(f,[c,b])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f020b0c1a7bb64e6a49695b3f45f46d520ec95f6)
![{\displaystyle I_{+}(f,[a,c])\geq I_{-}(f,[a,c])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ef54ae4d8ee778c21466c80ac1a79f6a9119c47)
![{\displaystyle I_{+}(f,[c,b])\geq I_{-}(f,[c,b])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d465231af5963abab915087caf39313261a77a3)
![{\displaystyle I_{+}(f,[a,b])>I_{-}(f,[a,b])}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e2a2eb34416fefa55af809b1223d5cef3861000)
![{\displaystyle \int _{a}^{b}f(x)\mathrm {d} x=I_{+}(f,[a,b])=I_{+}(f,[a,c])+I_{+}(f,[c,b])=\int _{a}^{c}f(x)\mathrm {d} x+\int _{c}^{b}f(x)\mathrm {d} x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/af060c6fdf6d478511e405ae41ce6b41b7cfeb23)
![{\displaystyle \sup _{x\in [c,d]}|f(x)|-\inf _{x\in [c,d]}|f(x)|\leq \sup _{x\in [c,d]}f(x)-\inf _{x\in [c,d]}f(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/05a2cfcaac01f7165e1e308191ded82fc7ae906e)
![{\displaystyle [c,d]\subseteq [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/32c4b90060cfad2522da60caeab608de43226f6e)
![{\displaystyle A:=\inf _{x\in [c,d]}f(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39a01c302f0ad6158cac62f032954a30a0d1da28)
![{\displaystyle B:=\sup _{x\in [c,d]}f(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/525f991b81f168da16506bd1f6f8db0627a48c7e)
![{\displaystyle C:=\inf _{x\in [c,d]}|f(x)|}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a83edfa46168d642499a7fe6bb3ac5af7dd026d8)
![{\displaystyle D:=\sup _{x\in [c,d]}|f(x)|}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6bed5edf60cbfe9823cbe258941b343e3819175)
![{\displaystyle x\in [c,d]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/aef10ac7d23aac26f4cb71a2ad9745f574ca5af1)
![{\displaystyle f(x)\geq \inf _{x\in [c,d]}f(x)=A\geq 0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0894ec8cd643b185ba1e3120243c25f25ae6317a)
![{\displaystyle f(x)\leq \sup _{x\in [c,d]}f(x)=B<0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c98b845293436265875bf9edfa5a26b7fc6bf638)
![{\displaystyle [x_{k},x_{k+1}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ae285fbca56a685135b31cf996cdd519f8e21e0e)
![{\displaystyle {\begin{aligned}&O({\tilde {\Delta }},|f|)-U({\tilde {\Delta }},|f|)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}|f(x)|-\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}|f(x)|\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\left(\sup _{x\in [x_{k},x_{k+1}]}|f(x)|-\inf _{x\in [x_{k},x_{k+1}]}|f(x)|\right)\\[0.3em]\leq \ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\left(\sup _{x\in [x_{k},x_{k+1}]}f(x)-\inf _{x\in [x_{k},x_{k+1}]}f(x)\right)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}f(x)-\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]=\ &O({\tilde {\Delta }},f)-U({\tilde {\Delta }},f)\\[0.3em]<\ &\epsilon \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ac2d04ba0889534cb2f06da0bee4acb8b7ef4c69)
![{\displaystyle -f:[a,b]\to \mathbb {R} ,x\mapsto -f(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5b6ecc3470c04b52bcaf32e00d2d947571f77e71)
![{\displaystyle f:[0,1]\to \mathbb {R} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/2de6d0d4c98d4ca7ad937c772dc3e3e914b062f5)
![{\displaystyle x\in [0,1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/64a15936df283add394ab909aa7a5e24e7fb6bb2)
![{\displaystyle [c,d]\subseteq [0,1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1a0633c5008c3a58fb86b7f190ed0b21e31e9671)
![[0,1]](https://wikimedia.org/api/rest_v1/media/math/render/svg/738f7d23bb2d9642bab520020873cccbef49768d)
![{\displaystyle \sup _{x\in [c,d]}f(x)g(x)-\inf _{x\in [c,d]}f(x)g(x)\leq S\left(\sup _{x\in [c,d]}g(x)-\inf _{x\in [c,d]}g(x)\right)+T\left(\sup _{x\in [c,d]}f(x)-\inf _{x\in [c,d]}f(x)\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3bd8e0c9b6082d74466d0ce6c84ae1c4a5fb705d)
![{\displaystyle {\begin{aligned}&\sup _{x\in [c,d]}f(x)g(x)-\inf _{x\in [c,d]}f(x)g(x)\\[0.3em]=\ &\sup _{x,y\in [c,d]}\left(f(x)g(x)-f(y)g(y)\right)\\[0.3em]=\ &\sup _{x,y\in [c,d]}\left(f(x)(g(x)-g(y))+g(y)(f(x)-f(y))\right)\\[0.3em]\leq \ &\sup _{x,y\in [c,d]}\left(f(x)(g(x)-g(y))\right)+\sup _{x,y\in [c,d]}\left(g(y)(f(x)-f(y))\right)\\[0.3em]\leq \ &S\sup _{x,y\in [c,d]}\left(g(x)-g(y)\right)+T\sup _{x,y\in [c,d]}\left(f(x)-f(y)\right)\\[0.3em]=\ &S\left(\sup _{x\in [c,d]}g(x)-\inf _{x\in [c,d]}g(x)\right)+T\left(\sup _{x\in [c,d]}f(x)-\inf _{x\in [c,d]}f(x)\right)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/59a61e6720de1115816f2d2e36d7f7dda7c51242)
![{\displaystyle {\begin{aligned}&O({\tilde {\Delta }},fg)-U({\tilde {\Delta }},fg)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}f(x)g(x)-\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}f(x)g(x)\\[0.3em]=\ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\left(\sup _{x\in [x_{k},x_{k+1}]}f(x)g(x)-\inf _{x\in [x_{k},x_{k+1}]}f(x)g(x)\right)\\[0.3em]\leq \ &\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\left(S\left(\sup _{x\in [c,d]}g(x)-\inf _{x\in [c,d]}g(x)\right)+T\left(\sup _{x\in [c,d]}f(x)-\inf _{x\in [c,d]}f(x)\right)\right)\\[0.3em]=\ &S\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\left(\sup _{x\in [x_{k},x_{k+1}]}g(x)-\inf _{x\in [x_{k},x_{k+1}]}g(x)\right)+T\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\left(\sup _{x\in [c,d]}f(x)-\inf _{x\in [c,d]}f(x)\right)\\[0.3em]=\ &S\left(\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}g(x)-\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}g(x)\right)\\&\qquad {}+T\left(\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}f(x)-\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}f(x)\right)\\[0.3em]=\ &S\left(O({\tilde {\Delta }},g)-U({\tilde {\Delta }},g)\right)+T\left(O({\tilde {\Delta }},f)-U({\tilde {\Delta }},f)\right)\\[0.3em]&{\color {OliveGreen}\left\downarrow \ {\tilde {\Delta }}{\text{ ist Verfeinerung von }}\Delta _{1}{\text{ und }}\Delta _{2}\right.}\\[0.3em]\leq \ &S\left(O(\Delta _{2},g)-U(\Delta _{2},g)\right)+T\left(O(\Delta _{1},f)-U(\Delta _{1},f)\right)\\[0.3em]<\ &S\cdot {\tfrac {\epsilon }{2S}}+T\cdot {\tfrac {\epsilon }{2T}}\\[0.3em]=\ &\epsilon \end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c4a918ffbef145adad881201455c0683f3e1a29a)
![{\displaystyle \inf _{x\in [c,d]}f(x)=f(c)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f2023cd6fbcb147485ddc45148827b8fd692df64)
![{\displaystyle \sup _{x\in [c,d]}f(x)=f(d)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ecd40351b69deb2aa3d482f50ad658c155cb1b5)
![{\displaystyle {\begin{aligned}O(\Delta _{n},f)&=\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\sup _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]&=\sum _{k=0}^{n-1}{\tfrac {b-a}{n}}f(x_{k+1})\\[0.3em]&={\tfrac {b-a}{n}}\left(f(x_{1})+\ldots +f(x_{n-1})+f(b)\right)\\[0.3em]U(\Delta _{n},f)&=\sum _{k=0}^{n-1}(x_{k+1}-x_{k})\inf _{x\in [x_{k},x_{k+1}]}f(x)\\[0.3em]&=\sum _{k=0}^{n-1}{\tfrac {b-a}{n}}f(x_{k})\\[0.3em]&={\tfrac {b-a}{n}}\left(f(a)+f(x_{1})+\ldots +f(x_{n-1})\right)\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2446c9b635817663dde015c9d84fbeed43d31db2)
![{\displaystyle [a,x_{1}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ae2c07161cd155ac8fb5c82f6469286c13d767f0)
![{\displaystyle [x_{1},b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/964ceef28eee0ca6ed47d83ddd6c966ea7956b24)
