Continuity of volumes on rings – Serlo

In this article we derive the definition of continuity of volumes. We investigate how the notions of continuity from below and from above are related and learn about $\sigma$-rings as a domain of definition of continuous volumes.

Motivation for continuity of volumes

In the last chapter, we learned about volumes defined on rings defined as an abstract form of measuring (extensive) quantities. When measuring such quantities, we expect that small changes in the measured object will result in only small changes in the measurement result. Examples are

• Ingredients for cooking: if you add only a little more, weight and taste will only change slightly.
• Counting of objects: If one determines the number of an at most countably infinite quantity of objects, then this changes only little, if only few objects are added or taken away.
• The area/circumference of a circle: If the radius is changed only little, area and circumference also only change little.

A similar behaviour occurs for functins and is called continuity, there. So we would also like to define a continuity for sets - more precisely, sequences of sets

In fact, it is difficult to find extensive quantities in nature which are not continuous. It is also a natural and important question whether content functions are continuous (this should often be the case). Moreover, continuity has a very useful consequence: it allows an approximation of quantities to be measured. If small differences between sets cause only small differences between the measured values, then the error of the approximation can be controlled by the accuracy of the approximating sets. Thus, even "complicated" sets can be measured by approximating them with easier sets. An example is the approximation of the area of a circle (area complicated to describe) by rectangular figures (area easy to describe).

Exhaustion of an ellipse by rectangles

Continuity therefore seems to be a desirable property of a volume. Before we examine this notion in more detail: Is there any (intuitively) discontinuous volume at all? If yes, we know what to exclude from the definition.

Example (A discontinuous volume)

We consider the basic set $\mathbb {N}$ and the volume $\mu :{\mathcal {P}}(\mathbb {N} )\to [0,\infty ]$, which assigns any subset of natural numbers either finite amount or infinity:

$\mu (A)={\begin{cases}0,{\text{ if }}|A|<\infty ,\\\infty {\text{ else.}}\end{cases}}$

In the article volumes on rings we saw that $\mu$ is really a volume defined on the ring ${\mathcal {P}}(\mathbb {N} )$. Now consider for $n\in \mathbb {N}$ the sets $A_{n}=\{1,2,\dots ,n\}$. For $n\to \infty$ these obviously approximate the basic set $\mathbb {N}$. But since every $A_{n}$ is finite, $\mu (A_{n})=0$ for every $n$. So the function value $\mu (\mathbb {N} )=\infty$ is not approximated in the limit $n\to \infty$. So the volume is intuitively "discontinuous".

So definitely not every volumes is continuous,and it makes sense to define a precise notion of continuity. How can we formalize the continuity of a content mathematically? Following the sequence definition of continuity for real functions, we try the following definition:

Definition (Continuous volume, first attempt)

A volume $\mu :{\mathcal {R}}\to [0,\infty ]$ on a ring ${\mathcal {R}}$ is called continuous, if for a sequence $(A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {R}}$ with $A_{n}{\overset {n\to \infty }{\longrightarrow }}A\in {\mathcal {R}}$ we have that $\lim _{n\to \infty }\mu (A_{n})=\mu (A)$.

But here we have to be careful: What is meant by $A_{n}{\overset {n\to \infty }{\longrightarrow }}A$ if the $A_{n}$ are sets? We first need a notion of convergence for sequences of sets.

Sequences of sets

Imagine a sequence of intervals of length 1, which constantly move to the right by step length 1, i.e., $(A_{n})_{n}\subseteq {\mathcal {P}}(\mathbb {R} )$, $A_{n}=[n,n+1]$ . It is hard to determine a limit set, to which this sequence of sets converges. In contrast, consider the sequence of sets $A_{n}=[0,1+1/n]$. Those seem to shrink to a set $[0,1]$: Since the $A_{n}$ are contained within each other, one can take the set sequence to be an approximation of the set $[0,1]$ "from outside/above". Similarly, one can take increasing sequences of sets to be approximations of a set "from inside/below" (for example, the sequence of $A_{n}=[0,1-1/n]$ exhausting $[0,1)$). It is then meaningful to set the intersection or union of $A_{n}$ as the limit of the sequence.

• 
  The set sequence $A_{n}=[n,n+1]$ "escapes to the right" and hence does not converge.
• 
  The set sequence $A_{n}=[0,1+1/n]$ shrinks down to $[0,1]$ (blue region) and hence converges.

Definition (Monotonic sequence of sets)

Let $\Omega$ be a set and $(A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {P}}(\Omega )$ a sequence of subsets.

The sequence $(A_{n})_{n\in \mathbb {N} }$ is called increasing or monotonically increasing if $A_{n}\subseteq A_{n+1}$ for all $n\in \mathbb {N}$. In this case we set

$\lim _{n\to \infty }A_{n}=\bigcup _{n=1}^{\infty }A_{n}=:A$

and write $A_{n}\uparrow A$.

The sequence $(A_{n})_{n\in \mathbb {N} }$ is called descending or monotonically descending if $A_{n}\supseteq A_{n+1}$ for all $n\in \mathbb {N}$ . In this case we set

$\lim _{n\to \infty }A_{n}=\bigcap _{n=1}^{\infty }A_{n}=:A$

and write $A_{n}\downarrow A$.

Hint

With our definition we can speak of the limit of a set sequence only if it is monotonic. This is enough for us at this point. But one can define $\liminf$ and $\limsup$ for arbitrary set sequences, which always exist as for real number sequences. Then, we are able to define even more converging sequences of sets.

Let's have a look at some examples:

Example (Monotonic sequence of sets)

The sequence $A_{n}=[0,1+1/n]\subseteq \mathbb {R}$ is monotonically decreasing with limit

$\lim _{n\to \infty }A_{n}=\bigcap _{n=1}^{\infty }[0,1+1/n]=[0,1].$

The inclusion "$\supseteq$" holds since $[0,1]\subseteq A_{n}$ for all $n\in \mathbb {N}$. On the other hand, the limiting set cannot contain more elements, since none of the $A_{n}$ contains negative numbers and every number strictly greater than one is no longer in $A_{n}$ for an index $n$, so it is not in the intersection either.

Another example for a monotonically decreasing set sequence is $A_{n}=\{n,n+1,\dots \}\subseteq \mathbb {N}$. One shows with similar arguments that for this sequence $\lim _{n\to \infty }A_{n}=\emptyset$ is the limit.

The sequence of $A_{n}=[0,1-1/n]\subseteq \mathbb {R}$ is an example of a monotonically increasing set sequence. Its limit is

$\lim _{n\to \infty }A_{n}=\bigcup _{n=1}^{\infty }[0,1-1/n]=[0,1).$

The inclusion "$\supseteq$" holds since every $x\in [0,1)$ lies in the set $A_{n}$ for some index $n$. On the other hand, the limit cannot be a larger interval since none of the $A_{n}$ contains negative numbers or one.

Another monotonically increasing sequence of sets is $A_{n}=\{1,2,\dots ,n\}\subseteq \mathbb {N}$. The limit is $\lim _{n\to \infty }A_{n}=\mathbb {N}$, as can be seen in a similar way.

How do limits of monotonic set sequences get along with set rings? Is the limit in the ring? Let's look at the ring ${\mathcal {R}}$ of cuboids in $\mathbb {R} ^{2}$ (i.e., rectangles) and two examples of monotonic growing set sequences in this ring. In the left picture a rectangle is approximated by a sequence of smaller rectangles. The limit is itself a rectangles and again lies in ${\mathcal {R}}$. On the right we see how a circle is approximated by rectangles. But the limiting set is no longer a rectangle and hence not in ${\mathcal {R}}$.

Obviously, the limit of a monotonic sequence of sets from a ring does not necessarily have to lie in the ring again. Our reasoning for this was rather intuitive with the rectangles (= 2D-cuboids), but we can also give a very concrete (not only intuitive) and short example:

Example (Limit is no element of the ring)

Consider the ring ${\mathcal {R}}\subseteq {\mathcal {P}}(\mathbb {N} )$ of all finite subsets of $\mathbb {N}$ (we learned about this ring here). The sequence $(A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {R}}$ with $A_{n}=\{1,2,\dots ,n\}$ is monotonically increasing and lies in ${\mathcal {R}}$. But its limit is all of $\mathbb {N}$ and thus no longer lies in the ring of all finite subsets of $\mathbb {N}$.

Continuity from below and above

Definition of continuity

Equipped with this definition of limits of set sequences we can now make another attempt to define the continuity of a volume $\mu$ on a ring ${\mathcal {R}}$. We have just seen that for a monotonic set sequence on a ring ${\mathcal {R}}$ its limit is not necessarily again in the ring. Therefore, we must impose the restrictive condition that the limit of the set sequence lies again in the ring for the definition to make sense.

Definition (Continuous volume, second attempt)

A volume $\mu :{\mathcal {R}}\to [0,\infty ]$ on a ring ${\mathcal {R}}$ is called continuous from below (from above), if for every ascending (descending) set sequence $(A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {R}}$ with limit $A\in {\mathcal {R}}$, it holds that

$\lim _{n\to \infty }\mu (A_{n})=\mu (A)$

A volume is just called continuous if it is continuous of below and of above.

Let's take an example to see if this improved definition already describes the concept of continuity to our satisfaction.

Example (A problem with the definition)

In the introduction we learned about counting the elements contained in a set as an example of an (intuitively) continuous volume. Let us look at the set system

${\mathcal {R}}=\{A\subseteq \mathbb {N} :A{\text{ finite }}\}$

of all finite subsets of $\mathbb {N}$. In the article volumes on rings we saw that counting the elements of finite subsets of $\mathbb {N}$ really defines a volume:

$\mu :{\mathcal {R}}\to [0,\infty ],\quad \mu (A)=|A|$

We now expect $\mu$ to be continuous from below and from above in the sense of the new definition. Let then $(A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {R}}$ be a monotonically growing sequence whose limit $A$ is also in ${\mathcal {R}}$. Since the $A_{n}$ converge to $A$, the number of elements in $A$ must be the limit of the number of elements of $A_{n}$. In other words, we have that

$\lim _{n\to \infty }\mu (A_{n})=\lim _{n\to \infty }|A_{n}|=|A|=\mu (A).$

Thus $\mu$ is continuous of below in the sense of our definition. Let now $(A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {R}}$ be a monotonically decreasing sequence whose limit $A$ is also in ${\mathcal {R}}$. Again, for the same reasons, the continuity condition must be satisfied. Thus $\mu$ is continuous on the ring ${\mathcal {R}}$ in the sense of the definition.

One can also consider the volume $\mu$ on the whole power set ${\mathcal {P}}(\mathbb {N} )$. Intuitively, counting of elements should be just as continuous on arbitrary subsets of $\mathbb {N}$ as on finite subsets. In the argumentation, however, we should make a distinction between finite and infinite to be on the safe side:

Let $(A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {P}}(\mathbb {N} )$ again be a monotonically increasing sequence of subsets with limit $A\subseteq \mathbb {N}$. If $A$ contains only finitely many elements, then because of monotonicity, $A_{n}\subseteq A$ and we can argue as above. But since $A$ is the limit of $A_{n}$, the argument works even if $A$ contains infinitely many elements: If $\lim _{n\to \infty }|A_{n}|\neq \infty$, then the number of elements in the $A_{n}$ would not grow beyond a maximal index. Thus the sequence of $A_{n}$ would be constant from an index and in particular could not converge to an infinite set. So $\lim _{n\to \infty }|A_{n}|=\infty$ must hold and also for infinite $A$, we have that

$\lim _{n\to \infty }\mu (A_{n})=\lim _{n\to \infty }|A_{n}|=\infty =|A|=\mu (A).$

If the $A_{n}$ now form a monotonically decreasing sequence and all sets of the sequence are finite, then because of the monotonicity, also the limit $A$ is a finite set and we can argue as above. It works the same way if only finitely many of the $A_{n}$ are infinite: We can omit all infinite sequences elements without restriction, since omitting finitely many elements of a sequence does not change its limit, and get again a sequence of finite sets only. The situation is different if all $A_{n}$ of the sequence are infinite sets. To satisfy the continuity condition, we would need

$\mu (A)=\lim _{n\to \infty }\mu (A_{n})=\infty$

meaning that the limit $A$ of the sequence contains infinitely many elements. But since the sequence is monotonically decreasing, this is not guaranteed. We have not excluded that a descending sequence of infinite sets converges to a finite set. Try to find such a sequence before reading on!

Consider the sequence $(A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {P}}(\mathbb {N} )$ with $A_{n}=\{n,n+1,\dots \}$. The $A_{n}$ converge decreasingly to the empty set. But at the same time we have that $\mu (A_{n})=\infty$ for all $n$. So

$\lim _{n\to \infty }\mu (A_{n})=\infty \neq 0=\mu (\emptyset ).$

I.e., the volume $\mu$, which intuitively should be continuous, is not continuous in the sense of our definition.

So there are volumes which are intuitively continuous, but discontinuous in the sense of our definition. Obviously there is a problem with the definition of the continuity of above. Note: With the second definition of continuity, such "unreasonable" cases can also occur. For instance, the monotonic sequence of $\mu (A_{n})$ can be bounded, while $\mu (A)=\infty$ holds for the limit of the sequence of sets. However, calling such a volume discontinuous does not contradict our intuition. We saw an example of this at the beginning of the article.

Now what is the problem with the above definition of continuity? It is apparently due to the occurrence of the value infinity: while every $A_{n}$ is slightly "less" infinite than its predecessors, nevertheless $\mu (A_{n})=\infty$. We have to exclude this case in the definition of continuity. For this we require finiteness of the decreasing sequence, at least after passing a certain minimal index (like a sequence, which decreases starting from some minimal index).

Continuous volumes are a special kind of volumes.

Definition (Continuous volume (final definition))

A volume $\mu :{\mathcal {R}}\to [0,\infty ]$ on a ring ${\mathcal {R}}$ is called continuous from below if for every increasing set sequence $(A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {R}}$ with limit $A\in {\mathcal {R}}$, it holds that

$\lim _{n\to \infty }\mu (A_{n})=\mu (A)$

It is called continuous from above if for every decreasing set sequence $(A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {R}}$ with $\mu (A_{n})<\infty$ for one (and thus for all further) $n\in \mathbb {N}$ and with limit $A\in {\mathcal {R}}$, it holds that

$\lim _{n\to \infty }\mu (A_{n})=\mu (A)$

A volume is just called continuous if it is continuous from below and above.

Hint

The continuity of volumes or more generally of functions on sets is sometimes also called $\sigma$-continuity ("sigma-continuity"). The prefix $\sigma$ is meant to remind of "sum" and means in this context something like "countable": $\sigma$-continuous volumes behave continuously at the transition from finite to countable unions or cuts.

Note: It is not necessary to require finiteness starting from an index within the continuity from below: If $\mu (A_{N})=\infty$ holds for an index $N\in \mathbb {N}$, then we have that $\mu (A_{n})=\infty$ also for all subsequent $n\geq N$ because of the monotonicity of the volume $\mu$. For the same reason, infiniteness then also holds for the limit $\bigcup _{n\in \mathbb {N} }A_{n}$.

Continuity from below implies continuity from above

continuity from above and continuity from below do not seem to be quite equivalent: At least, with the continuity from above one has to make restrictions which are not necessary with the continuity from below. How exactly are the two notions related? Does one imply the other? To get an answer, we try to find an analogy to sequences of real numbers.

Let $(x_{n})_{n\in \mathbb {N} }\subseteq \mathbb {R}$ be a monotonically decreasing sequence of non-negative real numbers. You can make it a monotonically increasing sequence of non-negative real numbers by considering the sequence $(x_{1}-x_{n})_{n\in \mathbb {N} }$. If this sequence converges to a value $y\in \mathbb {R}$, we can conclude that the original sequence $(x_{n})_{n\in \mathbb {N} }$ converges to $x_{1}-y$.

Suppose we now know that a volume is continuous from below. Then we can infer the continuity from above by turning in the same way from a monotonically decreasing set sequence $(A_{n})_{n\in \mathbb {N} }$ into a monotonically increasing set sequence $(A_{1}\setminus A_{n})_{n\in \mathbb {N} }$. After that transformation, we can then exploit the continuity from below. Crucially, we must only need to consider decreasing sequences of sets with finite volume, so we don't get any problems with subtraction.

Theorem (continuity from below implies continuity from above)

Let $\mu :{\mathcal {R}}\to [0,\infty ]$ be a volume on a ring ${\mathcal {R}}$. If $\mu$ is continuous from below, then $\mu$ is also continuous from above. In particular, continuity of a content from below is equivalent to just continuity.

Proof (continuity from below implies continuity from above)

Let $(A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {R}}$ be a monotonically decreasing set sequence with $\mu (A_{n})<\infty$ starting at an index $N\in \mathbb {N}$. Since omitting finitely many sequences elements does not change the limit, we can assume $N=1$ without restriction. The limit $A=\bigcap _{n\in \mathbb {N} }A_{n}$ lies in ${\mathcal {R}}$, as well. We want to show $\lim _{n\to \infty }\mu (A_{n})=\mu (A)$.

The sequence $(A_{1}\setminus A_{n})_{n\in \mathbb {N} }$ is monotonically increasing and lies in the ring ${\mathcal {R}}$, as well. Also the limit $A_{1}\setminus A$ lies in ${\mathcal {R}}$. Furthermore, due to monotonicity, all these sets have finite content. From the continuity from below we know that $\lim _{n\to \infty }\mu (A_{1}\setminus A_{n})=\mu (A_{1}\setminus A)$ holds. Since moreover $A_{n}\subseteq A_{1}$ for all $n\in \mathbb {N}$ , we have that $A_{1}=A_{n}\uplus (A_{1}\setminus A_{n})$. With the additivity of the volume $\mu$ it follows that

$\mu (A_{1})=\mu (A_{n}\uplus (A_{1}\setminus A_{n}))=\mu (A_{n})+\mu (A_{1}\setminus A_{n}),$

so $\mu (A_{n})=\mu (A_{1})-\mu (A_{1}\setminus A_{n})$. (Alternatively, one can argue directly with the subtractivity of volumes on rings). Likewise we have that $\mu (A)=\mu (A_{1})-\mu (A_{1}\setminus A)$. Since the contents of all involved volumes are finite, these differences make sense and by continuity from below,

$\lim _{n\to \infty }\mu (A_{n})=\mu (A_{1})-\lim _{n\to \infty }\mu (A_{1}\setminus A_{n})=\mu (A_{1})-\mu (A_{1}\setminus A)=\mu (A).$

Continuity from above does not imply continuity from below

In the same way it should work if we know that a volume is continuous from above: we turn an increasing sequence into a descending one and exploit the continuity from above. But here we have to be careful! The continuity from above only holds for sequences which have finite content starting from an index. This condition is not guaranteed to be satisfied if we construct a descending sequence starting from an arbitrary increasing one. This problem can also occur with real-valued sequences: From a monotonically increasing sequence $(x_{n})_{n\in \mathbb {N} }$ convergent to some $x\in \mathbb {R}$, we can construct the monotonically decreasing sequence $(x-x_{n})_{n\in \mathbb {N} }$. But if $(x_{n})_{n\in \mathbb {N} }$ is not upper bounded but tends to infinity, this is no longer possible.

For set sequences, on one hand, for every monotonically growing sequence $(A_{n})_{n\to \infty }$ with limit $A$ the sequence $(A\setminus A_{n})_{n\in \mathbb {N} }$ is monotonically decreasing. But on the other hand, it can occur that $\mu (A\setminus A_{n})=\infty$ for all $n$. With such a sequence, we cannot use the continuity from above. So we cannot expect that the continuity from above always follows from the continuity from below. This is illustrated by the following example, which we already got to know in the first section as an example of a discontinuous volume:

Example (Continuity from above does not imply continuity from below)

The set sequence $(A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {P}}(\mathbb {N} )$ with $A_{n}=\{1,2,\dots ,n\}$ is incraesing with $A_{n}\uparrow \mathbb {N}$. Thus the sequence of $B_{n}=\mathbb {N} \setminus A_{n}=\{n+1,n+2,\dots \}$ is a descending sequence with $B_{n}\downarrow \emptyset$. While all $A_{n}$ are finite sets, the $B_{n}$ are all infinite. Let us now consider the volume that determines whether a subset of natural numbers is finite or infinite:

$\mu :{\mathcal {P}}(\mathbb {N} )\to [0,\infty ],\quad \mu (A)={\begin{cases}0,{\text{ if }}|A|<\infty ,\\\infty {\text{ else.}}\end{cases}}$

Obviously $\mu$ is continuous from above: The only descending set sequences that satisfy the finiteness condition are finite from an index, and then, their content is constantly $0$.

But now the sequence of $B_{n}$ does not satisfy the finiteness condition from an index, so we cannot use it to infer the continuity from below of $\mu$ for the sequence of $A_{n}$. And indeed,

$\lim _{n\to \infty }\mu (A_{n})=\lim _{n\to \infty }0=0\neq \infty =\mu (\lim _{n\to \infty }A_{n})=\mu (\mathbb {N} ),$

So $\mu$ is not continuous from below. In particular, from the continuity of a content from above, we generally not get the continuity from below!

We already got to know this volume in the introduction as an example for a (intuitively) discontinuous volume. The example shows that $\mu$ is not continuous also in the sense of our definition.

We capture this observation:

Warning

From the continuity of a content from above, we can generally not imply the continuity from below.

For finite rings, continuity from above and below are equivalent

Intuitively, the continuity from above is weaker, because one has to impose the condition of finiteness of the volume starting from an index. So one "loses" some sequences if the volume does not take only finite values. In fact, "from above" and "from below" are equivalent for finite volumes:

Theorem (For finite rings, continuity from above and below are equivalent)

Let $\mu$ be a finite volume on a ring ${\mathcal {R}}$, i.e.. $\mu (A)<\infty$ for all $A\in {\mathcal {R}}$. Then, we have

$\mu {\text{ continuous from above}}\iff \mu {\text{ continuous from below.}}$

Proof (For finite rings, continuity from above and below are equivalent)

We have already shown that continuity from below always implies continuity from above. Consequently, it is now sufficient for the finite case to show the other direction.

So let $\mu$ be continuous from above and $(A_{n})_{n\in \mathbb {N} }$ be a sequence converging from below to $A\in {\mathcal {R}}$ with sets from ${\mathcal {R}}$, i.e. $\bigcup _{n\in \mathbb {N} }A_{n}=A$.

We note that $(A\setminus A_{n})$ is always back in the ring ${\mathcal {R}}$. (stability under taking differences).

We construct a decreasing set sequence $(B_{n})_{n\in \mathbb {N} }$ from the increasing one to exploit the continuity given above.

For this we define $B_{n}=A\setminus A_{n}$ for all $n\in \mathbb {N}$. Since $(A_{n})_{n\in \mathbb {N} }$ was increasing, $(B_{n})_{n\in \mathbb {N} }$ is decreasing.

It follows that $\bigcap _{n\in \mathbb {N} }B_{n}=\bigcap _{n\in \mathbb {N} }(A\setminus A_{n})=A\setminus \bigcup _{n\in \mathbb {N} }(A_{n})=A\setminus A=\emptyset$.

Thus $(B_{n})_{n\in \mathbb {N} }$ is a set sequence converging to $\emptyset$ in ${\mathcal {R}}$. Because of the assumed continuity of above, we have

$\lim _{n\to \infty }\mu (B_{n})=\mu (A\setminus A)=\mu (\emptyset ).$

Here we also used the finiteness of $\mu$ in order to get continuity from above ($\mu (B_{n})$ is always finite).

For every $n\mu$, due to additivity, $\mu (A)=\mu (A_{n}\uplus B_{n})=\mu (A_{n})+\mu (B_{n})$.

Thus $\lim _{n\to \infty }\mu (A_{n})=\lim _{n\to \infty }\mu (A)-\mu (B_{n})=\mu (A)-\mu (\emptyset )=\mu (A).$

Equivalent characterization of continuity from above

Finally, we give a simpler characterization of the continuity from above. It can be useful to prove the continuity of finite contents:

Theorem (Continuity from above and in $\emptyset$)

For a volume $\mu$ on a ring ${\mathcal {R}}$ , the following statements are equivalent:

1. $\mu$ is continuous from above.
2. $\mu$ is continuous in $\emptyset$, i.e. for all monotonically decreasing sequences $(A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {R}}$ with $\mu (A_{n})<\infty$ starting from some $n$ and $A_{n}\downarrow \emptyset$ we have that $\lim _{n\to \infty }\mu (A_{n})=0$.

Proof (Continuity from above and in $\emptyset$)

$1\implies 2$: This is simply the definition of "continuity of above" applied to $\emptyset$.

$2\implies 1$: Let now $\mu$ be continuous in $\emptyset$, and let $(A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {R}}$ with $\mu (A_{n})<\infty$ starting from some $n$ and $A_{n}\downarrow A\in {\mathcal {R}}$.

Then $(A_{n}\setminus A)_{n\in \mathbb {N} }\subseteq {\mathcal {R}}$ is monotonically decreasing, and with the monotonicity property of volumes we have that $\mu (A_{n}\setminus A)\leq \mu (A_{n})<\infty$ starting from some $n$.

Since moreover we have that $\bigcap _{n\in \mathbb {N} }(A_{n}\setminus A)=\left(\bigcap _{n\in \mathbb {N} }A_{n}\right)\setminus A=A\setminus A=\emptyset$, it follows that $(A_{n}\setminus A)_{n\in \mathbb {N} }$ converges to $\emptyset$ from above.

By our assumption then $\lim _{n\to \infty }\mu (A_{n}\setminus A)=0$.

Since $\mu (A_{n})$ is finite for sufficiently large $n$, and since always $A\subseteq A_{n}$ holds, it follows that $\mu (A_{n}\setminus A)=\mu (A_{n})-\mu (A)$.

So we can conclude:

$\lim _{n\to \infty }\mu (A_{n})-\mu (A)=\lim _{n\to \infty }\mu (A_{n}\setminus A)=0.$

Examples for continuous volumes

Now that we know so much about continuous volumes, we can look at some concrete examples.

Example

Let ${\mathcal {R}}$ be a Ring. a volume of the form

$\delta _{x}(A)={\begin{cases}1,{\text{ if }}x\in A,\\0{\text{ else.}}\end{cases}}$

is always continuous. If $(A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {R}}$ from below against a $A\in {\mathcal {R}}$, then there are two possibilities:

$x\notin A$: Then, we have because of $A_{n}\subseteq A$ that $x\notin A_{n}$ for all $n\in \mathbb {N}$. In addition, we have that $\delta _{x}(A_{n})=0{\overset {n\to \infty }{\longrightarrow }}0=\delta _{x}(A)$.

$x\in A$: Then, we have since $\bigcup _{n\in \mathbb {N} }A_{n}=A$ that there is an $n_{0}\in \mathbb {N}$ with $x\in A_{n_{0}}$. Hence, from the monotonicity of the set sequence we get $x\in A_{n}$ for all $n>n_{0}$. Hence, $\delta _{x}(A_{n}){\overset {n\to \infty }{\longrightarrow }}1=\delta (A)$.

Thus $\delta _{x}$ is continuous from below and therefore also continuous.

Example (Volumes on rings of finite cardinality)

If $\mu$ is a volume on ${\mathcal {R}}$, with $|{\mathcal {R}}|<\infty$, then the sequences, which converge in ${\mathcal {R}}$ are exactly the set sequences which are constant starting from some index. The continuity of $\mu$ then follows directly.

Example (The geometric volume)

We already know the ring of cuboids : this is the set of all finite unions of axis-parallel cuboids in $\mathbb {R} ^{n}$. (An axis-parallel cuboid in $\mathbb {R} ^{n}$ is a product $I_{1}\times \dots \times I_{n}$ of intervals (each open, half-open, or closed).)

We defined the geometric volume $\lambda$ on this ring as follows:

For a single cuboid $Q=I_{1}\times \dots I_{n}$, $\lambda (Q):=|I_{1}|\dots |I_{n}|$ is the product of the side lengths.

For a cuboid $A=\biguplus _{i=1}^{k}Q_{i}$ (where the $Q_{i}$ are axis-parallel cuboids, without restriction assumed to be pairwise disjoint) define $\lambda (A):=\sum _{i=1}^{k}\lambda (Q_{i})$.

One can show that the geometric volume $\lambda$ is continuous.

To-Do:

Show that \lambda is continuous (from below) / sketch the proof. this goes just as one also shows that \lambda is a premeasure, with some epsilontics...

Furthermore, one can easily show that finite linear combinations of continuous volumes are continuous again.

Sigma-rings

• 
  A ring (of sets) does not necessarily have to include all limiting sets.
• 
  However, a $\sigma$-Ring must also include all limiting sets.

We now know what continuity of a volume on a ring means. In the introduction we stated that continuity allows to measure sets by approximation, since small deviations of the sets induce only small deviations of the measured values. Thus, if a volume on a ring ${\mathcal {R}}$ is continuous, it should be possible to measure not only the sets from ${\mathcal {R}}$, but also all with sets from ${\mathcal {R}}$ that can be approximated. The approximable sets are even the limits of monotonically increasing or decreasing sequences of sets. So we have that for continuous volumes it makes sense to use a ring as domain of definition, which also contains the limit values of such sequences:

A $\sigma$-ring is a special ring. The arrow means "is included in"

Definition ($\sigma$-ring)

A ring ${\mathcal {R}}\subseteq {\mathcal {P}}(\Omega )$ over some basic set $\Omega$ is called $\sigma$-Ring, if:

• limits of increasing sets sequences in ${\mathcal {R}}$ lie in ${\mathcal {R}}$,
• limits of decreasing sets sequences in ${\mathcal {R}}$ lie in ${\mathcal {R}}$.

For the continuity of a volume the continuity of below was sufficient, because one can construct an increasing set sequence out of every decreasing one. In the same way it is sufficient to formulate the closedness of ${\mathcal {R}}$ only for limits of increasing set sequences: If $(A_{n})_{n\in \mathbb {N} }$ is a monotonically decreasing set sequence with limit $A$, then the sequence of $B_{n}=A_{1}\setminus A_{n}$ is a monotonically increasing sequence with limit $A_{1}\setminus A$. Since ${\mathcal {R}}$ is a ring, so in particular stable under differences, we have that

$A_{n}=A_{1}\setminus (A_{1}\setminus A_{n})\in {\mathcal {R}}\iff A_{1}\setminus A_{n}\in {\mathcal {R}}$

for all $n$ and

$A=A_{1}\setminus (A_{1}\setminus A)\in {\mathcal {R}}\iff A_{1}\setminus A\in {\mathcal {R}}.$

For the equalities on the left hand side we have exploited $A_{n}\subseteq A_{1}$ and $A\subseteq A_{1}$ respectively. So it is enough to require that limit values of monotonically increasing set sequences are again in the set system ${\mathcal {R}}$, and we have the equivalent definition:

Definition ($\sigma$-ring (equivalent definition))

A ring ${\mathcal {R}}\subseteq {\mathcal {P}}(\Omega )$ over a basic set $\Omega$ is called a $\sigma$-ring if for every increasing set sequence in ${\mathcal {R}}$ its limit is also in ${\mathcal {R}}$.

In the literature $\sigma$-rings are often defined differently. We prove that the following is an equivalent characterization.

Theorem (Alternative characterization of $\sigma$-rings)

A set system ${\mathcal {R}}\subseteq {\mathcal {P}}(\Omega )$ over a basic set $\Omega$ is a $\sigma$-ring exactly if:

1. ${\mathcal {R}}\neq \emptyset$
2. $A_{1},A_{2},\dots \in {\mathcal {R}}\implies \bigcup _{i=1}^{\infty }A_{i}\in {\mathcal {R}}$
3. $A,B\in {\mathcal {R}}\implies A\setminus B\in {\mathcal {R}}$

Proof (Alternative characterization of $\sigma$-rings)

Let ${\mathcal {R}}$ be a $\sigma$-ring. Since ${\mathcal {R}}$ is a ring, properties 1 and 3 of this theorem are satisfied. Now let $(A_{n})_{n\in \mathbb {N} }$ be a sequence in ${\mathcal {R}}$. As ${\mathcal {R}}$ is a ring, finite unions of the form $B_{n}=\bigcup _{k=1}^{n}A_{n}\in {\mathcal {R}}$. Since ${\mathcal {R}}$ is a $\sigma$-ring, and because $(B_{n})_{n\in \mathbb {N} }$ is an increasing set sequence in ${\mathcal {R}}$, it follows that $\bigcup _{n\in \mathbb {N} }A_{n}=\bigcup _{n\in \mathbb {N} }B_{n}=\lim _{n\to \infty }B_{n}\in {\mathcal {R}}$. So property 2 also holds.

Let now ${\mathcal {R}}$ be a set system with properties 1, 2 and 3. Because of 1 and 3, $\emptyset \in {\mathcal {R}}$ holds: we have $A\in {\mathcal {R}}$ from which follows $\emptyset =A\setminus A\in {\mathcal {R}}$. Thus, from the union stability with respect to countable unions, we also get the union stability with respect to finite unions. Thus, ${\mathcal {R}}$ is a ring. To show that ${\mathcal {R}}$ is a $\sigma$-ring, let $(B_{n})_{n\in \mathbb {N} }$ be an increasing set sequence in ${\mathcal {R}}$. Then, because of property 2, we have $\lim _{n\to \infty }B_{n}=\bigcup _{n\in \mathbb {N} }B_{n}\in {\mathcal {R}}$. That is, ${\mathcal {R}}$ is closed with respect to taking limits of increasing set sequences and hence ${\mathcal {R}}$ is a $\sigma$-ring.

Examples for sigma-rings

Example (Power set and finite rings)

For every basic set $\Omega$ , the power set ${\mathcal {P}}(\Omega )$ is a $\sigma$-ring.

Every finite ring ${\mathcal {R}}$ (i.e. contains only finitely many sets) is a $\sigma$-ring: The second property in the definition is trivially satisfied, since there are only finitely many sets which can be joined, and ${\mathcal {R}}$ is closed as a ring under finite unions.

Example (Countable subsets)

We have already met the ring ${\mathcal {R}}=\{A\subseteq \mathbb {R} \mid A{\text{ countable}}\}$ of countable subsets of $\mathbb {R}$. This is a $\sigma$ ring:

1. Since the empty set is countable, ${\mathcal {R}}$ is non-empty.
2. If $A_{1},A_{2},\dots \subseteq \mathbb {R}$ is a sequence of countable sets, then their union is also countable, so we have that $\bigcup _{i=1}^{\infty }A_{i}\in {\mathcal {R}}$.
3. If $A,B\subseteq \mathbb {R}$ are countable, then because $A\setminus B\subseteq A$, their difference is also countable and so $A\setminus B\in {\mathcal {R}}$.

Example (Finite subsets)

By contrast, the set system ${\mathcal {R}}=\{A\subseteq \mathbb {N} \mid A{\text{ finite}}\}$ of finite subsets of $\mathbb {N}$ is indeed a ring, but not a $\sigma$-ring: For the sets $A_{n}=\{1,\dots ,n\}\in {\mathcal {R}}$ we have that

$\bigcup _{n=1}^{\infty }A_{n}=\mathbb {N} \notin {\mathcal {R}}.$

Example ($\sigma$-ring of cuboids)

Consider the set system

${\mathcal {Q}}=\left\{\prod _{k=1}^{n}I_{k}\mid I_{k}\subseteq \mathbb {R} {\text{ interval}}\right\}\subseteq \mathbb {R} ^{n}$

of the axis-parallel cuboids in $\mathbb {R} ^{n}$. (The intervals in the product may be open, half-open, or closed.) We already know the ring of cuboids in $\mathbb {R} ^{n}$. It is defined as the ring generated by ${\mathcal {Q}}$, that is, the smallest ring containing ${\mathcal {Q}}$. In the same way one can consider the $\sigma$-ring ${\mathcal {R}}$ generated by ${\mathcal {Q}}$. Like the ring of cuboids, ${\mathcal {R}}$ can be defined as the intersection of all $\sigma$-rings over $\mathbb {R} ^{n}$ containing the cuboids ${\mathcal {Q}}$. (Compare also the article on generated $\sigma$-algebras).

A union of cuboids $\mathbb {R} ^{2}$

While the ring generated by the cuboids contains only cuboid unions, the $\sigma$-ring ${\mathcal {R}}$ additionally contains all sets which can be approximated by cuboid figures, such as the circle.

Exhaustion of a circle with rectangles