Uniqueness of a continuation – Serlo

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We derive conditions under which a measure on a $\sigma$-algebra is uniquely determined by the values on a generator. On our way, we will learn about Dynkin systems and their relation to $\sigma$-algebras. Finally, we will prove the uniqueness theorem for measure continuation.

Problem

Continuation of functions on sets to measures are usually done in a way that the as many subsets of a basic set $\Omega$ enter the $\sigma$-algebra, as possible (while sustaining some nice properties). Often, one requires additional conditions, for instance when constructing a geometric volume, that cuboids in $\mathbb {R} ^{n}$ get assigned their geometric volume. In general, it is not clear whether such a measure with the desired properties even exists. If it does, we might have several $\sigma$-algebras where it can be defined on.

The general construction of a measure is as follows: we start with a small set system ${\mathcal {C}}$, such that the function on sets restricted to it fulfills the desired properties. For example, for defining a geometric volume, we choose ${\mathcal {C}}$ as the set system of cuboids and $\mu$ as the set function that assigns to each cuboid its geometric volume.

Using the existence of a continuation - theorem we know when a function $\mu \colon {\mathcal {C}}\to [0,\infty ]$ can be continued to a measure on the $\sigma$-algebra $\sigma ({\mathcal {C}})$ generated by $\sigma$. For the proof of the continuation theorem, one possible continuation has been explicitly stated via outer measures. But can there also be other ways to continue $\mu$ to a measure on $\sigma ({\mathcal {C}})$? In other words, we are interested in whether a measure $\mu$ on the $\sigma$-algebra $\sigma ({\mathcal {C}})$ is already uniquely determined by its values on the smaller set system ${\mathcal {C}}$.

To make them easier to check, uniqueness statements are often re-formulated in mathematics: Suppose, $\mu$ and $\nu$ are measures on the $\sigma$-algebra $\sigma ({\mathcal {C}})$ generated by a set system ${\mathcal {C}}$. Further, let $\mu$ and $\nu$ coincide on ${\mathcal {C}}$, which means $\mu (A)=\nu (A)$ for all $A\in {\mathcal {C}}$. Then uniqueness just means $\mu =\nu$ on the whole $\sigma$ algebra $\sigma ({\mathcal {C}})$.

In the following, we derive conditions for when this is the case.

The principle of good sets

We will proceed step by step to find conditions on the generator ${\mathcal {C}}$ and the two measures $\mu$ and $\nu$ under which uniqueness holds. For this we consider the system of "good sets"

${\mathcal {G}}:=\{A\in \sigma ({\mathcal {C}})\colon \mu (A)=\nu (A)\}.$

It contains all sets from $\sigma ({\mathcal {C}})$ on which $\mu$ and $\nu$ coincide. Uniqueness would mean that all sets in $\sigma ({\mathcal {C}})$ are good, i.e. ${\mathcal {G}}=\sigma ({\mathcal {C}})$.

Actually, this is equivalent to saying that ${\mathcal {G}}$ is a $\sigma$-algebra: Since by assumption, $\mu (C)=\nu (C)$ is satisfied for all $C\in {\mathcal {C}}$, we have ${\mathcal {C}}\subseteq {\mathcal {G}}$ holds. But ${\mathcal {C}}$ already generates $\sigma ({\mathcal {C}})$, i.e. there is no smaller $\sigma$-algebra which contains ${\mathcal {C}}$. So if ${\mathcal {G}}$ is a $\sigma$-algebra, then ${\mathcal {G}}$ (which was contained in $\sigma ({\mathcal {C}})$) must be the entire $\sigma$-algebra $\sigma ({\mathcal {C}})$ (see: monotonicity and idempotence of the $\sigma$-operator).

This type of approach is often used to show that a given property is satisfied for all sets of a set system ${\mathcal {M}}$ (like a $\sigma$-algebra). It is called the "principle of good sets" and it works like this:

Suppose one can only make statements about a generator of ${\mathcal {M}}$, perhaps because ${\mathcal {M}}$ can only be characterized in terms of the generator. An example is the Borel $\sigma$-algebra, which is extremely large and can only be written down by means of generators and relations. Then it might be smart to proceed indirectly when showing a property of all sets in ${\mathcal {M}}$.

We define the set system ${\mathcal {G}}=\{A\in {\mathcal {M}}\mid A{\text{ has the given property}}\}$ of "good sets". Then we show:

• ${\mathcal {G}}$ is a $\sigma$-algebra.
• ${\mathcal {G}}$ contains a generator ${\mathcal {C}}$ of ${\mathcal {M}}$.

It follows that ${\mathcal {M}}={\mathcal {G}}$, i.e. all sets in ${\mathcal {M}}$ are "good". Hence, we gained control over the extremely tedious set ${\mathcal {M}}$ by only using properties of the simple sets in ${\mathcal {C}}$.

Theorem (The principle of good sets)

Let ${\mathcal {M}}$ be a $\sigma$-algebra and let ${\mathcal {G}}$ be the system of all sets from ${\mathcal {M}}$ for which a given property holds. Assume that

• ${\mathcal {G}}$ is a $\sigma$-algebra.
• ${\mathcal {G}}$ contains a generator ${\mathcal {C}}$ of ${\mathcal {M}}$.

Then the property holds for all sets from ${\mathcal {M}}$, i.e., we have that ${\mathcal {M}}={\mathcal {G}}$.

Hint

The principle of good sets is very powerful. It can also be applied to other kinds of set systems (not only for $\sigma$-algebras). For instance, it can also be applied to the ring or $\sigma$-ring generated by a set system.

In our case, the sets of good sets ${\mathcal {G}}=\{A\in \sigma ({\mathcal {C}})\colon \mu (A)=\nu (A)\}$ contains all the sets $A\in \sigma ({\mathcal {C}})$ on which the measures $\mu$ and $\nu$ coincide. The equality of the measures on the generator ${\mathcal {C}}$ is known, so ${\mathcal {C}}\subseteq {\mathcal {G}}$ holds. Now we are still looking for conditions so that ${\mathcal {G}}$ becomes a $\sigma$-algebra. So we want to find conditions on the generator ${\mathcal {C}}$ and the two measures $\mu ,\nu$ such that:

• The basic set $\Omega$ is contained in ${\mathcal {G}}$.
• ${\mathcal {G}}$ is complement stable.
• ${\mathcal {G}}$ is union stable with respect to countable unions.

Existence of an inner approximation

Every $\sigma$-algebra contains the basic set $\Omega$, so $\Omega$ should be in the set of good sets ${\mathcal {G}}$. That is, it should hold that $\mu (\Omega )=\nu (\Omega )$ for both measures $\mu$ and $\nu$. In general, this need not be the case even if the two measures agree on ${\mathcal {C}}$:

Example (Measure of the basic set is not uniquely determined)

Let ${\mathcal {C}}$ be the set system of all one-element subsets of $\mathbb {R}$:

${\mathcal {C}}=\{\{x\}\mid x\in \mathbb {R} \}$

The $\sigma$-algebra generated by this is

$\sigma ({\mathcal {C}})=\{A\subseteq \mathbb {R} :A{\text{ countable or }}A^{\complement }{\text{ countable}}\}$

(This is proved in the article on generated $\sigma$-algebras). We define on $\sigma ({\mathcal {C}})$ the two measures $\mu =0$ and $\nu$ via

$\nu (A):={\begin{cases}0,{\text{ if }}A{\text{ countable}}\\\infty {\text{ else.}}\end{cases}}$

Then $\mu$ and $\nu$ are equal on the generator, but not on all of $\sigma ({\mathcal {C}})$. Thus, they are not yet uniquely determined by specifying the values on ${\mathcal {C}}$. In particular $\mu (\mathbb {R} )=0\neq \infty =\nu (\mathbb {R} )$.

We can still generalize this example: Let $\Omega \neq \emptyset$ be a set and let ${\mathcal {C}}=\{\emptyset \}$. The $\sigma$-algebra generated by ${\mathcal {C}}$ is $\sigma ({\mathcal {C}})=\{\emptyset ,\Omega \}$. Let $\mu =0$ be the zero measure on it, and for a $r>0$ let the measure $\nu$ be defined by $\nu (\emptyset )=0,\nu (\Omega )=r$. Then, we have $\mu (\emptyset )=\nu (\emptyset )$, so $\mu$ and $\nu$ agree on ${\mathcal {C}}$. However, $\mu (\Omega )=0\neq r=\nu (\Omega )$.

So with which conditions on the generator ${\mathcal {C}}$ or the two measures $\mu ,\nu$ can we enforce $\mu (\Omega )=\nu (\Omega )$?

Idea and definition of an inner approximation

We know that the measures $\mu$ and $\nu$ coincide on the sets in ${\mathcal {C}}$. The idea is to cover the basic set with at most countably many sets from $E_{1},E_{2},\ldots \in {\mathcal {C}}$, i.e. their union should be $\bigcup _{n=1}^{\infty }E_{n}=\Omega$.

Example (Coverings of $\Omega =[0,\infty )$)

• The sets $E_{n}=[n,n+2]$ are neither pairwise disjoint nor contained in each other.
• The sets $E_{n}=[n,n+1)$ are pairwise disjoint.
• The sets $E_{n}=[0,n)$ are contained in each other in ascending order.

Now we want to infer from $\mu (E_{n})=\nu (E_{n})$ for all $n\in \mathbb {N}$ (which holds since these sets are from ${\mathcal {C}}$), that $\mu (\Omega )=\nu (\Omega )$ holds as well. For this the $E_{n}$ should be either pairwise disjoint or contained in each other in ascending order:

• If the $E_{n}$ are pairwise disjoint, we can use the $\sigma$-additivity of measures: $\mu (\Omega )=\mu \left(\biguplus _{i=1}^{\infty }E_{i}\right)=\sum _{n=1}^{\infty }\mu (E_{i})=\sum _{n=1}^{\infty }\nu (E_{i})=\nu \left(\biguplus _{i=1}^{\infty }E_{i}\right)=\nu (\Omega )$
• If they are contained in each other in ascending order (i.e. $E_{1}\subseteq E_{2}\subseteq \ldots$), they form a monotonic set sequence with limit $\bigcup _{i=1}^{\infty }E_{i}=\Omega$. Then we can use the continuity of the measures $\mu$ and $\nu$: $\mu (\Omega )=\mu (\lim _{n\to \infty }E_{n})=\lim _{n\to \infty }\mu (E_{i})=\lim _{n\to \infty }\nu (E_{i})=\nu (\lim _{n\to \infty }E_{n})=\nu (\Omega )$

If neither is the case, then we can not necessarily conclude $\mu (\Omega )=\nu (\Omega )$ from $\mu (E_{n})=\nu (E_{n})$ for all $n$. We consider the case where the $E_{n}$ are contained in each other in ascending order and define the notion of exhaustion:

Definition (Exhaustion)

Let $\Omega$ be a set and let $E_{n}\subseteq \Omega$ be subsets for each $n\in \mathbb {N}$ with $E_{n}\subseteq E_{n+1}$. If $\Omega =\bigcup _{n\in \mathbb {N} }E_{n}$, then the sequence $(E_{n})_{n\in \mathbb {N} }$ is called an exhaustion of $\Omega$.

Example (Exhaustion)

For $\Omega :=\mathbb {R}$ and ${\cal {{C}:=\{(a,b]:a,b\in \mathbb {R} \}}}$ there exists an exhaustion $(E_{n})_{n}\subseteq {\mathcal {C}}$ with $E_{n}:=(-n,n]$.

Since measures are continuous, íf ${\mathcal {C}}$ contains an exhaustion $(E_{n})_{n\in \mathbb {N} }$ of $\Omega$, then we have $\mu (\Omega )=\lim _{n\to \infty }\mu (E_{n})=\lim _{n\to \infty }\nu (E_{n})=\nu (\Omega )$.

Intermediate result

To conclude, w have found the following first condition on the set system ${\mathcal {C}}$:

To ensure that the basic set is in the "set of good sets" ${\mathcal {G}}$, that is, that $\mu (\Omega )=\nu (\Omega )$ holds, we require that ${\mathcal {C}}$ contains an exhaustion of $\Omega$.

Hint

The condition is automatically satisfied if $\Omega \in {\mathcal {C}}$ holds, e.g. if $\mu$ and $\nu$ are probability measures: Then one can assume a priori $\Omega \in {\mathcal {C}}$.

The inner approximating sets have finite measure

By requiring that $\Omega$ has an exhaustion by sets from ${\mathcal {C}}$, we ensured that $\Omega$ lies in the "set of good sets" ${\mathcal {G}}=\{A\in \sigma ({\mathcal {C}})\colon \mu (A)=\nu (A)\}$. It remains to investigate under which conditions ${\mathcal {G}}$ is closed under formation of differences and countable unions. For this purpose let us first examine under which operations ${\mathcal {G}}$ is already closed with our previous assumptions:

Unions

Let $A$ and $B$ be two sets from the set of good sets ${\mathcal {G}}$. Thus we have $\mu (A)=\nu (A)$ and the same for $B$. Now, we take their union. Things look good if $B\subseteq A$: in this case $A\cup B=A$ , so $\mu (A\cup B)=\mu (A)=\nu (A)=\nu (A\cup B)$ is definitely also in the set of good sets ${\mathcal {G}}$. The same happens for $A\subseteq B$

Similarly, the measure value of the union of $A$ and $B$ is uniquely determined if the two sets are disjoint: In that case, from additivity of the measures $\mu$ and $\nu$ it follows that

$\mu (A\uplus B)=\mu (A)+\mu (B)=\nu (A)+\nu (B)=\nu (A\uplus B)$

So the disjoint union is also uniquely measurable and lies again in ${\mathcal {G}}$.

If we additionally exploit the $\sigma$-additivity of measures, then we even know the measure of countably infinite unions of disjoint sets. Given a sequence of pairwise disjoint sets $A_{1},A_{2},\dots \in {\mathcal {G}}$, the $\sigma$-additivity of measures $\mu$ and $\nu$ implies

$\mu \left(\biguplus _{i=1}^{\infty }A_{i}\right)=\sum _{i=1}^{\infty }\mu (A_{i})=\sum _{i=1}^{\infty }\nu (A_{i})=\nu \left(\biguplus _{i=1}^{\infty }A_{i}\right).$

Set differences

Let again $A$ and $B$ be two sets from ${\mathcal {G}}$, i.e. let $\mu (A)=\nu (A)$, $\mu (B)=\nu (B)$ hold. As in the case of the union, the difference of the two sets is again in ${\mathcal {G}}$ if $A$ and $B$ are disjoint. In that case $A\setminus B=A$ and we have that $\mu (A\setminus B)=\mu (A)=\nu (A)=\nu (A\setminus B)$. (Likewise with the roles on $A$ and $B$ exchanged).

In the case $A\subseteq B$, the difference of $A$ and $B$ is equal to $A\setminus B=\emptyset$. Since $\mu (\emptyset )=0=\nu (\emptyset )$ , it lies again in ${\mathcal {G}}$. Moreover, because of the additivity of the measure $\mu$, we have that

$\mu (B)=\mu ((B\setminus A)\uplus A)=\mu (B\setminus A)+\mu (A).$

In the first equation, we used that $A$ is a subset of $B$. The same is true for the measure $\nu$ instead of $\mu$. Rearranging the above formula together with $\mu (A)=\nu (A)$ and $\mu (B)=\nu (B)$ yields

$\mu (B\setminus A)=\mu (B)-\mu (A)=\nu (B)-\nu (A)=\nu (B\setminus A).$

This equation is dangerous! If $A$ and $B$ have infinite measure, we get the ill-defined expression "$\infty -\infty$", which cannot be sensibly defined:

Example ($\infty -\infty$ cannot be defined)

We consider the measure $\lambda$ defined by the geometric length in $\mathbb {R}$. (We know that such a measure exists by the continuation theorem.) For the two sets $A=\mathbb {R}$ and $B=[0,\infty )$, we have that $A\setminus B=(-\infty ,0)$. So it follows $\lambda (A\setminus B)=\lambda ((-\infty ,0))=\infty$, and in this case, "$\infty -\infty =\infty$".

By contrast, for the sets $A=B=\mathbb {R}$, we have that $A\setminus B=\emptyset$. So $\lambda (A\setminus B)=\lambda (\emptyset )=0$ and we get "$\infty -\infty =0$".

This shows that in general we cannot make sense of the expression $\infty -\infty$, without further assumptions.

Differences of sets with infinite measure

A way out of this problem is to approximate $A$ and $B$ by ascending sequences $(A_{n})_{n\in \mathbb {N} },(B_{n})_{n\in \mathbb {N} }$ of sets of finite measure and take a limit. For this, the sets of the set sequence should also be good sets. We can then calculate the measure of the differences $A_{n}\setminus B_{n}$ as above, since $\mu (A_{n})$ and $\mu (B_{n})$ are both finite. But we have to be careful: For this to work, the subset relation $A_{n}\subseteq B_{n}$ must also hold for the set sequences $(A_{n})_{n\in \mathbb {N} },(B_{n})_{n\in \mathbb {N} }$ for all $n\in \mathbb {N}$. So we cannot just choose the sequences $(A_{n})_{n\in \mathbb {N} }$ and $(B_{n})_{n\in \mathbb {N} }$ in an arbitrary manner. They need to grow "equally fast" and at the same time approximate $A$ and $B$ equally fast.

Recall: we assumed that there is an exhaustion $(E_{n})_{n\in \mathbb {N} }$ of the basic set $\Omega$ with sets from the set system ${\mathcal {C}}$, i.e., a monotonically growing set sequence in ${\mathcal {C}}$ with limit $\Omega$. (This was to guarantee $\mu (\Omega )=\nu (\Omega )$.)

Then the sets $A_{n}:=A\cap E_{n}$ also form an increasing set sequence with limit

$\bigcup _{n=1}^{\infty }A_{n}=\bigcup _{n=1}^{\infty }(A\cap E_{n})=A\cap \bigcup _{n=1}^{\infty }E_{n}=A\cap \Omega =A$.

The same is true for the sequence $B_{n}:=B\cap E_{n}$. Moreover, because $B\subseteq A$, we also have $B_{n}=B\cap E_{n}\subseteq A\cap E_{n}=A_{n}$, so the subset relation is satisfied for every member of the sequence. Because of $B_{n}\setminus A_{n}=(B\cap E_{n})\setminus (A\cap A_{n})=(B\setminus A)\cap E_{n}$ the sequence of $(B_{n}\setminus A_{n})_{n\in \mathbb {N} }$ is also monotonically increasing.

Let us now use the same calculation as for the differences of sets with finite measure

$\mu (B_{n}\setminus A_{n})=\mu (B_{n})-\mu (A_{n})=\nu (B_{n})-\nu (A_{n})=\nu (B_{n}\setminus A_{n})$

and then turn to the limit $n\to \infty$. For this we need:

• $\mu (A_{n})=\nu (A_{n})$ and $\mu (B_{n})=\nu (B_{n})$ for all $n$. That is, intersections $C\cap E_{n}$ of sets $C\in {\mathcal {C}}$ with the $E_{n}$ are said to lie in ${\mathcal {G}}$.
• Each set of the exhaustion has finite measure, i.e., $\mu (E_{n})<\infty$ for all $n\in \mathbb {N}$. Only then, because of monotonicity, we can be sure that $\mu (A_{n})=\mu (A\cap E_{n})\leq \mu (E_{n})<\infty$ and $\mu (B_{n})=\mu (B\cap E_{n})\leq \mu (E_{n})<\infty$ also holds, ´which is our goal.

Hint

Note that the "finiteness" of an exhaustion depends on the considered measure. For example, the exhaustion $((-n,n])_{n\in \mathbb {N} }$ is finite with respect to $\mu \equiv 0$, but not with respect to the measure ${\tilde {\mu }}$ that assigns the value $\infty$ to every set except $\emptyset$.

If $(A_{n})_{n\in \mathbb {N} }$ and $(B_{n})_{n\in \mathbb {N} }$ satisfy these conditions, we can calculate the original difference $B\setminus A$ ($A\subseteq B$):

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

Note that we were only able to swap difference and measure because the sets $A_{n},B_{n}$ had finite measure. We could do that for the sets $A,B$ with infinite measure. For this we had to construct $(A_{n})_{n\in \mathbb {N} }=(A\cap E_{n})_{n\in \mathbb {N} },(B_{n})_{n\in \mathbb {N} }=(B\cap E_{n})_{n\in \mathbb {N} }$.

The following examples shows that the sets $E_{n}$ of the exhaustion indeed need to have finite measure in order to get uniqueness.

Example

We consider the set system of half-open intervals in $\mathbb {R}$

${\mathcal {C}}=\{(a,b]:-\infty \leq a\leq b<\infty \}.$

It generates a $\sigma$-algebra $\sigma ({\mathcal {C}})$ (the so-called Borel $\sigma$-algebra), which also contains all one-element subsets of $\mathbb {R}$. Now consider the two measures $\sigma ({\mathcal {C}})$ defined on $\mu$ and $\nu$ by

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

and

$\nu (A)=\#(\mathbb {Q} \cap A)$

for all $A\in \sigma ({\mathcal {C}})$. ($\mu$ is thus the trivial measure that takes only the values $0$ and $\infty$, while $\nu$ counts the number of rational numbers contained in $A$). Because $\mathbb {Q}$ is dense in $\mathbb {R}$, the measures $\mu$ and $\nu$ coincide on the set system ${\mathcal {C}}$. Furthermore, the exhaustion $(E_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {C}},E_{n}:=(-n,n]$ of $\mathbb {R}$ exists with $\mu (E_{n})=\infty =\nu (E_{n})$ for all $n\in \mathbb {N}$.

The two measures $\mu$ and $\nu$ are not equal on the $\sigma$-algebra generated by ${\mathcal {C}}$: For all one-element sets $\{x\}\subseteq \mathbb {R}$, we have that $\mu (\{x\})=\infty$ holds, but $\nu (\{x\})=1$ or $=0$. So the measures $\mu$ or $\nu$ are not yet uniquely determined by their values on ${\mathcal {C}}$.

Intermediate result

We conclude what has been found out:

• The "set of good sets" ${\mathcal {G}}$ is already closed under (at most countably infinitely many) disjoint unions; unions of sets which are subsets of each other; differences of disjoint sets; and differences of sets which are subsets of each other and have finite measure.
• To ensure that the basic set is in ${\mathcal {G}}$, that is, $\mu (\Omega )=\nu (\Omega )$, we require that ${\mathcal {C}}$ contains an exhaustion $(E_{n})_{n\in \mathbb {N} }$ of $\Omega$.
• In order for differences of sets of infinite measure, which are subsets of each other, are in ${\mathcal {G}}$, we require, that the exhaustion sets $(E_{n})_{n\in \mathbb {N} }$ all have finite measure and that for all $A\in {\mathcal {G}}$ the intersections $A\cap E_{n}$ lie again in ${\mathcal {G}}$.

Apart from these conditions and $\mu |_{\mathcal {C}}=\nu |_{\mathcal {C}}$, we make no requirements on $\mu$, $\nu$, and ${\mathcal {C}}$.

The last condition is somewhat unsatisfactory, because it involves ${\mathcal {G}}$ (which may include complicated sets). But we want to find conditions that refer only to the a priori given measures $\mu ,\nu$ and the generator ${\mathcal {C}}$, respectively. We will still work on this and weaken the condition later.

Hint

In particular, the exhaustion requirements are directly satisfied if $\mu (\Omega )=\nu (\Omega )<\infty$ holds This is always the case if $\mu$ and $\nu$ are probability measures. Then $(\Omega )_{n\in \mathbb {N} }$ is a monotonically increasing sequence of sets with finite measure, which obviously converges to $\Omega$.

Dynkin systems

As before, let ${\mathcal {G}}=\{A\in \sigma ({\mathcal {C}})\mid \mu (A)=\nu (A)\}$ be the system of good sets on which the two measures $\mu$ and $\nu$ coincide. Assuming that the conditions from the previous intermediate result are satisfied, we now know that the two measures are equal on the following sets:

• The basic set $\Omega$: this is guaranteed by the exhaustion $(E_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {C}}$.
• Unions of finitely or countably infinitely many pairwise disjoint sets in ${\mathcal {G}}$: this holds because of $\sigma$-additivity and continuity of the measures $\mu$ and $\nu$.
• Differences of sets from ${\mathcal {G}}$, where one is contained in the other: this is guaranteed by finiteness of the exhaustion and by the condition that intersections of sets from ${\mathcal {G}}$ with sets of the exhaustion are again in ${\mathcal {C}}$.

Thus we can already characterize the set system ${\mathcal {G}}$ of "good sets" more precisely. It contains the basic set and is closed under the operations "disjoint union" and "difference of sets contained in each other".

A set system with these properties is called a "Dynkin system".

Definition

Relationship between different set systems.

Definition (Dynkin system)

A set system ${\mathcal {D}}\subseteq {\mathcal {P}}(\Omega )$ is called a Dynkin system if it holds that:

1. $\Omega \in {\mathcal {D}}$
2. for every two sets $A,B\in {\mathcal {D}}$ with $A\subseteq B$ we also have $B\setminus A\in {\mathcal {D}}$.
3. for each countably many pairwise disjoint sets $A_{1},A_{2},\dots \in {\mathcal {D}}$ we also have $\biguplus _{n\in \mathbb {N} }A_{n}\in {\mathcal {D}}$.

Hint

It follows directly from the definition that every $\sigma$-algebra is a Dynkin system. The converse is not true: A Dynkin system is not always a $\sigma$-algebra, because it lacks closure under non-disjoint countable unions.

An equivalent characterization of a Dynkin system is the following.

Theorem

A set system ${\mathcal {D}}\subseteq {\mathcal {P}}(\Omega )$ is a Dynkin system if and only if:

1. $\Omega \in {\mathcal {D}}$
2. for every countably many $A\in {\mathcal {D}}$, we also have $A^{\complement }\in {\mathcal {D}}$.
3. for each countably many pairwise disjoint sets $A_{1},A_{2},\dots \in {\mathcal {D}}$ we also have $\biguplus _{n\in \mathbb {N} }A_{n}\in {\mathcal {D}}$.

Proof

Let ${\mathcal {D}}$ be a Dynkin system in the sense of the definition above. Points 1. and 3. in the theorem are identical to the definition. We only need to show point 2. So let $A\in {\mathcal {D}}$ be arbitrary. We have that $\Omega \in {\mathcal {D}}$ and $A\subseteq \Omega$, so it follows from property 2. of the definition that also $A^{\complement }=\Omega \setminus A\in {\mathcal {D}}$.

Conversely, let the three properties from the theorem be satisfied. We show that then ${\mathcal {D}}$ is a Dynkin system in the sense of the definition above. Again, we only need to show point 2. So let $A,B\in {\mathcal {D}}$ be arbitrary with $A\subseteq B$. Let $B\setminus A=B\cap A^{\complement }=(B^{\complement }\uplus A)^{\complement }$. The union on the right hand side is disjoint because of $A\subseteq B$. By assumptions 2. and 3. the set $(B^{\complement }\uplus A)^{\complement }=B\setminus A$ thus indeed lies in ${\mathcal {D}}$.

So with the preconditions from the intermediate result of the previous section, ${\mathcal {G}}$ is already a Dynkin system. All that is still missing for a $\sigma$ algebra is closure under arbitrary countable unions.

Moreover, since the two measures $\mu$ and $\nu$ agree on the generator ${\mathcal {C}}$, we have that ${\mathcal {C}}\subseteq {\mathcal {G}}$ holds. So the Dynkin system generated by ${\mathcal {C}}$ also lies in the set of good sets ${\mathcal {G}}$, which is defined analogously to the generated $\sigma$-algebra:

Definition (Generated Dynkin system)

Let $\Omega$ be a set and ${\mathcal {C}}\subseteq {\mathcal {P}}(\Omega )$ be a set system. The Dynkin system

$\delta ({\mathcal {C}}):=\bigcap \{{\mathcal {D}}:{\mathcal {D}}{\text{ is a Dynkin system with }}{\mathcal {C}}\subseteq {\mathcal {D}}\}$

is then called the Dynkin system generated by ${\mathcal {C}}$. In other words, $\delta ({\mathcal {C}})$ is the smallest Dynkin system containing ${\mathcal {C}}$.

As with the definition of the generated $\sigma$-algebra, we need to convince ourselves that $\delta ({\mathcal {C}})$ is well-defined. This can be done completely analogously to the proof we already gave for generated $\sigma$-algebras.

Theorem

The intersection in the above definition is not empty. Furthermore, the intersection of any number of Dynkin systems is again a Dynkin system.

Hint

Just as for generated $\sigma$-algebras, the following properties hold for the Dynkin system generated by a set system ${\mathcal {C}}$:

1. Extensivity: ${\mathcal {C}}\subseteq \delta ({\mathcal {C}})$
2. Minimality: $\delta ({\mathcal {C}})$ is the smallest Dynkin system containing ${\mathcal {C}}$. If ${\mathcal {C}}$ is a Dynkin system, then $\delta ({\mathcal {C}})={\mathcal {C}}$.
3. Idempotence: $\delta ({\mathcal {C}}))=\delta ({\mathcal {C}})$.
4. Monotonicity: ${\mathcal {C}}\subseteq {\mathcal {E}}\implies \delta ({\mathcal {C}})\subseteq \delta ({\mathcal {E}})$

Examples

Here are a few examples of Dynkin systems:

Example ($\sigma$-algebras)

It follows directly from the definition that any $\sigma$-algebra is also a Dynkin system.

However, not every Dynkin system is a $\sigma$-algebra:

Example (Subsets with even cardinality)

Let $n>2$ be an even number and $\Omega$ be a set with exactly $n$ elements. The set system

${\mathcal {D}}:=\{A\subseteq \Omega \mid {\text{number of elements in }}A{\text{ is even}}\}$

is a Dynkin system:

Since $|\Omega |=n$ is even, we have that $\Omega \in {\mathcal {D}}$.

If $A,B\in {\mathcal {D}}$ are even with $A\subseteq B$, then $|B\setminus A|=|B|-|A|$ is even since $|A|,|B|$ are even.

Since $\Omega$ is finite, we need only consider unions of finitely many disjoint sets for the third condition. Let $A_{1},\dots ,A_{k}\in {\mathcal {D}}$ be pairwise disjoint sets, each containing an even number of elements. Then, because of disjointness, the union $\biguplus _{i=1}^{k}A_{i}$ also contains an even number of elements and so lies in ${\mathcal {D}}$.

However, ${\mathcal {D}}$ is not a $\sigma$-algebra: Let $x,y,z\in \Omega$ be three distinct elements (these exist, since by assumption $|\Omega |=n>2$). Then $\{x,y\},\{y,z\}\in {\mathcal {D}}$ holds, but their union $\{x,y,z\}$ contains an odd number of elements, so it is not contained in ${\mathcal {D}}$.

Motivation for cut-stability

We considered conditions under which the set system ${\mathcal {G}}=\{A\in \sigma ({\mathcal {C}})\mid \mu (A)=\nu (A)\}$ of "good" sets is a Dynkin system, i.e.

• it contains the basic set,
• it is closed under taking complements,
• it is closed under taking (countable) disjoint unions.

Since we assume that the measures $\mu$ and $\nu$ coincide on the generator ${\mathcal {C}}$, ${\mathcal {C}}\subseteq {\mathcal {G}}$ holds. Since ${\mathcal {G}}$ is itself a Dynkin system, therefore the Dynkin system $\delta ({\mathcal {C}})$ generated by ${\mathcal {C}}$ is also contained in ${\mathcal {G}}$.

We would like to have $\sigma ({\mathcal {C}})\subseteq {\mathcal {G}}$ , such that $\mu$ and $\nu$ agree on their whole domain of definition. So ${\mathcal {G}}$ should not just a Dynkin system, but a $\sigma$-algebra. So what about closedness under non-disjoint finite/countably infinite unions? Let us first look at finite unions. Let $A,B\in {\mathcal {G}}$ be good sets (i.e., $\mu (A)=\nu (A)$ and $\mu (B)=\nu (B)$), non-disjoint, and neither $A\subseteq B$ nor $B\subseteq A$.

Initially, we have $\mu (A)=\nu (A)$ and $\mu (B)=\nu (B)$, but it is not yet clear whether also $\mu (A\cup B)=\nu (A\cup B)$ . Actually every value may appear on the lift-hand side, as long as we do not violate the monotonicity of $\mu$, i.e. as long as $\mu (A\cup B)\geq \mu (A)$ and $\mu (A\cup B)\geq \mu (B)$.

What other conditions must be satisfied for $\mu (A\cup B)=\nu (A\cup B)$ to hold? It suffices if the intersection $A\cap B$ is again a good set, i.e. $\mu (A\cap B)=\nu (A\cap B)$: If $A$ and $B$ both have finite measure, then we have

$\mu (A\cup B)=\mu (A)+\mu (B)-\mu (A\cap B)=\nu (A)+\nu (B)-\nu (A\cap B)=\nu (A\cup B).$

If either set has infinite measure, the equality $\mu (A\cup B)=\infty =\nu (A\cup B)$ holds anyway. Intersections of good sets should therefore be good sets again.

A set system, which is not left, when taking arbitrary intersections between its sets is called "cut-stable":

Definition (Cut stability of a set system)

A set system ${\mathcal {M}}$ is called cut stable, if

$A,B\in {\mathcal {M}}\implies A\cap B\in {\mathcal {M}}$

Hint

By induction, the set system is then closed under intersection of any finite number of sets (not only 2), See below.

Is the cut-stability of the system of good sets ${\mathcal {G}}$ in addition to our previous conditions already enough for it to be an $\sigma$-algebra? Suppose ${\mathcal {G}}$ is cut-stable. The previous reasoning for two sets can be extended by induction to any finite union of good sets: Let $A_{1},A_{2},\dots ,A_{n}\in \sigma ({\mathcal {C}})$ be good sets, i.e., $\mu (A_{i})=\nu (A_{i})$ for all $i=1,\dots ,n$. We also assume again that all $A_{i}$ have finite measure (otherwise the equality holds anyway). Then

${\begin{aligned}\mu \left(\bigcup _{i=1}^{n}A_{i}\right)=\mu \left(A_{n}\cup \bigcup _{i=1}^{n-1}A_{i}\right)&=\mu \left(A_{n}\right)+\mu \left(\bigcup _{i=1}^{n-1}A_{i}\right)-\mu \left(A_{n}\cap \bigcup _{i=1}^{n-1}A_{i}\right)\\&{\overset {\text{i.a.}}{=}}\nu \left(A_{n}\right)+\nu \left(\bigcup _{i=1}^{n-1}A_{i}\right)-\mu \left(\bigcup _{i=1}^{n-1}\underbrace {(A_{n}\cap A_{i})} _{\in {\mathcal {G}}}\right)\\&{\overset {\text{i.a.}}{=}}\nu \left(A_{n}\right)+\nu \left(\bigcup _{i=1}^{n-1}A_{i}\right)-\nu \left(\bigcup _{i=1}^{n-1}(A_{n}\cap A_{i})\right)=\nu \left(\bigcup _{i=1}^{n}A_{i}\right).\end{aligned}}$

(with "i.a." meaning "induction assumption") Making use of the cut-stability, we have that ${\mathcal {G}}$ is closed under arbitrary finite unions.

Countably infinite unions $\bigcup _{n\in \mathbb {N} }A_{n}$ can be made "artificially" disjoint by cutting the preceding ones out of each set: Define

$B_{n}:=A_{n}\setminus (A_{1}\cup \dots \cup A_{n-1})=A_{n}\cap (A_{1}\cup \dots \cup A_{n-1})^{\complement },$

Then $\bigcup _{n\in \mathbb {N} }A_{n}=\biguplus _{n\in \mathbb {N} }B_{n}$ is a disjoint union of sets. If the $A_{n}$ are from ${\mathcal {G}}$, then so are the $B_{n}$, provided that ${\mathcal {G}}$ is cut-stable: According to the preceding, the finite union $A_{1}\cup \dots \cup A_{n-1}$ lies in ${\mathcal {G}}$, and so does the complement of this set (since ${\mathcal {G}}$ is a Dynkin system). And by cut-stability also the intersection of it with $A_{n}$ lies in ${\mathcal {G}}$. Now, ${\mathcal {G}}$ is closed under countable disjoint unions (Dynkin system), so also $\biguplus _{n\in \mathbb {N} }B_{n}=\bigcup _{n\in \mathbb {N} }A_{n}$ is a good set.

These considerations show: If the Dynkin system ${\mathcal {G}}$ is additionally cut stable, it is also closed under arbitrary, at most countable unions, i.e., it is a sigma-algebra. We summarize this in a theorem:

Theorem

Cut stable Dynkin systems are $\sigma$-algebras.

Proof

Let ${\mathcal {D}}\subseteq {\mathcal {P}}(\Omega )$ be a cut stable Dynkin system. Then $\Omega \in {\mathcal {D}}$ and complement stability is given, as we have a Dynkin system. The remaining $\sigma$-algebra property is the union stability with respect to countable unions. So far, this is only given for pairwise disjoint unions. We first note that because $A\setminus B=A\cap B^{\complement }$, difference stability follows from complement stability and intersection stability of ${\mathcal {D}}$.

Now let $(A_{n})_{n\in \mathbb {N} }$ be any sequence of sets in ${\mathcal {D}}$. We define $(B_{n})_{n\in \mathbb {N} }$ by $B_{1}=A_{1}$, $B_{n+1}=A_{n+1}\setminus \bigcup _{i=1}^{n}A_{i}$ for all $n\in \mathbb {N}$. Since $A_{i}\subseteq B_{i}$ for all $i\in \mathbb {N}$ we have that

$\bigcup _{i=1}^{n}B_{i}=\bigcup _{i=1}^{n-1}B_{i}\cup \left(A_{n}\setminus \left(\bigcup _{i=1}^{n-1}A_{i}\right)\right)=\bigcup _{i=1}^{n-1}B_{i}\cup A_{n}.$

With a simple induction argument it follows that $\bigcup _{i=1}^{n}B_{i}=\bigcup _{i=1}^{n}A_{i}$ and finally $B_{n+1}=A_{n+1}\setminus \left(\bigcup _{i=1}^{n}B_{i}\right)$.

In particular, it follows from the construction of the sequence $(B_{n})_{n\in \mathbb {N} }$ that elements of this sequence are pairwise disjoint. By definition, $B_{1}=A_{1}\in {\mathcal {D}}$. Suppose now that $B_{1},\dots ,B_{n}\in {\mathcal {D}}$ holds (induction assumption). Then with stability with respect to disjoint unions it follows that $\bigcup _{i=1}^{n}B_{i}\in {\mathcal {D}}$. With difference stability, we further have $B_{n+1}=A_{n+1}\setminus \left(\bigcup _{i=1}^{n}B_{i}\right)\in {\mathcal {D}}$. Then, by induction, it also holds that $B_{i}\in {\mathcal {D}}$ for all $i\in \mathbb {N}$.

But since the $B_{i}$ were additionally pairwise disjoint, it follows from the disjoint countable union stability that $\bigcup _{n\in \mathbb {N} }B_{n}\in {\mathcal {D}}$.

From this we finally get

$\bigcup _{n\in \mathbb {N} }A_{n}=\bigcup _{n\in \mathbb {N} }\bigcup _{i=1}^{n}A_{i}=\bigcup _{n\in \mathbb {N} }\bigcup _{i=1}^{n}B_{i}=\bigcup _{n\in \mathbb {N} }B_{n}\in {\mathcal {D}}.$

Intermediate result

We conclude our so far obtained results:

• To ensure that the basic set lies in the set of good sets ${\mathcal {G}}$, that is $\mu (\Omega )=\nu (\Omega )$, we require that ${\mathcal {C}}$ contains an exhaustion $(E_{n})_{n\in \mathbb {N} }$ of $\Omega$.
• To ensure that differences of sets of infinite measure which are subsets of each other, are again in ${\mathcal {G}}$, we require, that the exhaustion sets $(E_{n})_{n\in \mathbb {N} }$ all have finite measure and that for all $A\in {\mathcal {G}}$ cuts $A\cap E_{n}$ lie again in ${\mathcal {G}}$.
• With these two conditions, ${\mathcal {G}}$ is already a Dynkin system. For it to be a $\sigma$-algebra, ${\mathcal {G}}$ should be cut stable, that is, cuts of good sets should be good again.

Except for these conditions and $\mu |_{\mathcal {C}}=\nu |_{\mathcal {C}}$, we make no further requirements on $\mu$, $\nu$, and ${\mathcal {C}}$.

Hint

While we had previously required in the second point that for all $A\in {\mathcal {G}}$ cuts $A\cap E_{n}$ must also be in ${\mathcal {G}}$, this is no longer necessary because of the third condition.

Next, we will answer the question, which additional conditions on $\mu$, $\nu$, or ${\mathcal {C}}$ will make ${\mathcal {G}}$ cut-stable.

Cut-stability of the generator

We have found conditions on the measures $\mu ,\nu$ and the generator ${\mathcal {C}}$ by which the set of good sets ${\mathcal {G}}=\{A\in \sigma ({\mathcal {C}})\mid \mu (A)=\nu (A)\}$ is a Dynkin system. In particular, ${\mathcal {G}}$ thus contains the Dynkin system $\delta ({\mathcal {C}})$ generated by ${\mathcal {C}}$, since ${\mathcal {C}}\subseteq {\mathcal {G}}$. We want $\sigma ({\mathcal {C}})\subseteq {\mathcal {G}}$ to hold as well. For this it suffices to find additional conditions under which $\delta ({\mathcal {C}})$ is cut-stable: Since every cut-stable Dynkin system is a $\sigma$-algebra, it then follows that $\sigma ({\mathcal {C}})=\delta ({\mathcal {C}})\subseteq {\mathcal {G}}\subseteq \sigma ({\mathcal {C}})$, and we are done.

So under what conditions is the Dynkin system generated by ${\mathcal {C}}$ cut-stable? This apparently depends only on the properties of the set system ${\mathcal {C}}$, not on the measures $\mu$ or $\nu$. In fact, it is sufficient if ${\mathcal {C}}$ is cut-stable. This has to do with the fact that the cut operation is compatible with the union and complement operations of a Dynkin system, and thus the cut-stability is inherited from the generator to the generated Dynkin system. We show this in the following theorem.

Theorem

Any Dynkin system generated by a cut-stable set system ${\mathcal {C}}$ is itself cut-stable.

Proof

For $B\in \delta ({\mathcal {C}})$ we define ${\mathcal {D}}(B)=\{D\in \delta ({\mathcal {C}})|D\cap B\in \delta ({\mathcal {C}})\}$, i.e., the set of sets in our Dynkin system that are cut-stable "with respect to $B$". By definition, ${\mathcal {D}}(B)\subseteq \delta ({\mathcal {C}})$ holds for any $B\in \delta ({\mathcal {C}})$. Now we show two things:

1. ${\mathcal {D}}(B)$ is a Dynkin system for every $B\in \delta ({\mathcal {C}})$.
2. ${\mathcal {C}}\subseteq {\mathcal {D}}(B)$ for all $B\in \delta ({\mathcal {C}})$.

For once we have shown these statements, it follows that ${\mathcal {D}}(B)\subseteq \delta ({\mathcal {C}})\subseteq \delta ({\mathcal {D}}(B))={\mathcal {D}}(B)$, so ${\mathcal {D}}(B)=\delta ({\mathcal {C}})$. Here we used monotonicity in the second step and idempotence of the $\delta$ operator in the third step. Now if $A,B\in \delta ({\mathcal {C}})$ are arbitrary, then because of ${\mathcal {D}}(B)=\delta ({\mathcal {C}})$ we know in particular that $A\in {\mathcal {D}}(B)$ holds, so $A\cap B\in \delta ({\mathcal {C}})$. This shows the desired cut-stability.

So let us now show the necessary statements.

Proof step: ${\mathcal {D}}(B)$ is a Dynkin system for all $B\in \delta ({\mathcal {C}})$.

Let $B\in \delta ({\mathcal {C}})$ be arbitrary. Obviously $\Omega \cap B=B\in \delta ({\mathcal {C}})$, so $\Omega \in {\mathcal {D}}(B)$.

Let $A\in {\mathcal {D}}(B)$ be arbitrary. Then $A\cap B\in \delta ({\mathcal {C}})$ holds. Furthermore, because of $B\in \delta ({\mathcal {C}})$, of course $B^{\complement }\in \delta ({\mathcal {C}})$. From this we conclude

$A^{\complement }\cap B=(A^{\complement }\cup B^{\complement })\cap B=(A\cap B)^{\complement }\cap B=((A\cap B)\uplus B^{\complement })^{\complement }.$

This, as the complement of a disjoint union of two elements of $\delta ({\mathcal {C}})$, is also an element of $\delta ({\mathcal {C}})$ and so $A^{\complement }\in {\mathcal {D}}(B)$, i.e. ${\mathcal {D}}(B)$ is complement stable.

Now let $(A_{k})_{k\in \mathbb {N} }$ be a sequence of pairwise disjoint sets in ${\mathcal {B}}$. Then for all $k\in \mathbb {N}$, it holds that $A_{k}\cap B\in \delta ({\mathcal {C}})$. The sequence $(A_{k}\cap B)_{k\in \mathbb {N} }$ is also a sequence of pairwise disjoint sets in $\delta ({\mathcal {C}})$. From the stability of $\delta ({\mathcal {C}})$ under countable disjoint unions, it follows that

$\left(\biguplus _{k\in \mathbb {N} }A_{k}\right)\cap B=\biguplus _{k\in \mathbb {N} }(A_{k}\cap B)\in \delta ({\mathcal {C}}).$

So $\bigcup _{k\in \mathbb {N} }A_{k}\in {\mathcal {D}}(B)$. Thus the three properties of a Dynkin system are satisfied and we are done.

Proof step: ${\mathcal {C}}\subseteq {\mathcal {D}}(B)$ for all $B\in \delta ({\mathcal {C}})$.

Let $B\in \delta ({\mathcal {C}})$ and $E\in {\mathcal {C}}$ be arbitrary. Then for all $D\in {\mathcal {C}}$, due to the cut-stability of ${\mathcal {C}}$, it holds that $E\cap D\in {\mathcal {C}}\subseteq \delta ({\mathcal {C}})$. So, in particular, $D\in {\mathcal {D}}(E)$. Since $D\in {\mathcal {C}}$ was arbitrary, we also have ${\mathcal {C}}\subseteq {\mathcal {D}}(E)$. As ${\mathcal {D}}(E)$ is a Dynkin system (as shown before), we get $\delta ({\mathcal {C}})\subseteq \delta ({\mathcal {D}}(E))={\mathcal {D}}(E)$. Further, $B\in \delta ({\mathcal {C}})\subseteq {\mathcal {D}}(E)$ holds. That is, $B\cap E\in \delta ({\mathcal {C}})$ and so $E\in {\mathcal {D}}(B)$. Since $E\in {\mathcal {C}}$ was chosen arbitrary, it finally follows that ${\mathcal {C}}\subseteq {\mathcal {D}}(B)$.

Since every average-stable Dynkin system is a $\sigma$ algebra, it follows directly:

Theorem

If ${\mathcal {C}}$ is a cut-stable set system, then $\delta ({\mathcal {C}})=\sigma ({\mathcal {C}})$ holds.

This relation between Dynkin systems and $\sigma$-algebras is very useful and simplifies many proofs about measures. This is because for Dynkin systems one can exploit the $\sigma$-additivity of the measure, since only disjoint unions need to be considered. In the proof of the uniqueness theorem we will see in a moment a first example where this enables to perform a proof.

Intermediate result

We summarize the conditions we found to get $\sigma ({\mathcal {C}})={\mathcal {G}}$, that is, equality of $\mu$ and $\nu$ on all of $\sigma ({\mathcal {C}})$:

• To ensure that the basic set is in the set of good sets ${\mathcal {G}}$, that is, $\mu (\Omega )=\nu (\Omega )$, we require that ${\mathcal {C}}$ contains an exhaustion $(E_{n})_{n\in \mathbb {N} }$ of $\Omega$.
• To ensure that differences of sets of infinite measure, which are subsets of each other, are again in ${\mathcal {G}}$, we require that the exhaustion sets $(E_{n})_{n\in \mathbb {N} }$ all have finite measure.
• To ensure that the Dynkin system $\delta ({\mathcal {C}})$ generated by ${\mathcal {C}}$ is cut-stable, i.e., it is a $\sigma$-algebra, we require that the generator ${\mathcal {C}}$ must be cut-stable.

In general, one cannot drop the cut- stability of ${\mathcal {C}}$ if the measures $\mu$ and $\nu$ shall also agree on $\sigma ({\mathcal {C}})$, as the following example shows.

Example

Consider the (rather small as it is finite) basic set $\Omega =\{1,2,3,4\}$ and the set system ${\mathcal {C}}=\{\emptyset ,\{1,2\},\{2,3\},\Omega \}$. Then ${\mathcal {C}}$ is not cut-stable, because $\{2\}=\{1,2\}\cap \{2,3\}\notin {\mathcal {C}}$. Consider the two (probability) measures

$\mu :={\frac {1}{2}}(\delta _{1}+\delta _{2}),\quad \nu :={\frac {1}{2}}(\delta _{2}+\delta _{4}).$

Here $\delta _{x}$ denotes the Dirac measure, which is defined by.

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

Then $\mu$ and $\nu$ agree on ${\mathcal {C}}$. Moreover, because of $\Omega \in {\mathcal {C}}$, there is a finite exhaustion of the basic set. Nevertheless, $\mu$ and $\nu$ are not equal on the $\sigma$-algebra generated by ${\mathcal {C}}$ : This $\sigma$-algebra contains the set $\{1,2,3\}=\{1,2\}\cup \{2,3\}$, and we have $\mu (\{1,2,3\})=2\neq 1=\nu (\{1,2,3\})$.

Uniqueness of measure continuations

We can now formulate and prove the uniqueness theorem.

Theorem (Uniqueness of measure continuations)

Let $\mu$ and $\nu$ be two measures on the $\sigma$-algebra ${\mathcal {A}}$ over the basic set $\Omega$. Let there be a generator ${\mathcal {C}}$ of ${\mathcal {A}}$ with the following properties:

• $\mu$ and $\nu$ coincide on ${\mathcal {C}}$, i.e. $\mu (C)=\nu (C)$ for all $C\in {\mathcal {C}}$,
• There exists in ${\mathcal {C}}$ an exhaustion of $\Omega$ with sets of finite measure: a monotonically increasing sequence $(E_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {C}}$ with limit $\Omega$ and $\mu (E_{n})=\nu (E_{n})<\infty$,
• ${\mathcal {C}}$ is cut-stable, i.e. $A,B\in {\mathcal {C}}\implies A\cap B\in {\mathcal {C}}$.

Then $\mu =\nu$ holds on all of ${\mathcal {A}}$. So in particular a measure $\mu$ is then already uniquely determined by the values on ${\mathcal {C}}$.

Proof (Uniqueness of measure continuations)

We perform the proof with the "principle of good sets" and define the set system ${\mathcal {G}}=\{A\in {\mathcal {A}}\mid \mu (A)=\nu (A)\}\subseteq {\mathcal {A}}$ as containing "good sets". It contains those sets from ${\mathcal {A}}$ on which $\mu$ and $\nu$ coincide. By assumption, ${\mathcal {C}}\subseteq {\mathcal {G}}$ holds. We still have to show that ${\mathcal {G}}$ is a $\sigma$-algebra. It is enough to show that ${\mathcal {C}}$ is a Dynkin system since its generator is cut-stability . Now

${\mathcal {A}}=\sigma ({\mathcal {C}})=\delta ({\mathcal {C}})\subseteq \delta ({\mathcal {G}})={\mathcal {G}}$

where we have exploited in the second equality that the Dynkin system generated by a cut-stable set system is already a $\sigma$-algebra. We now establish the properties of a Dynkin system for ${\mathcal {G}}$ in two steps: First assuming that $\mu ,\nu$ are finite measures (i.e. $\mu (A)<\infty ,\nu (A)<\infty$ for all $A\in {\mathcal {A}}$), then generalizing to the infinite measure case.

Proof step: Proof for finite $\mu ,\nu$

We have $\Omega \in {\mathcal {G}}$: Let $(E_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {C}}$ be the exhaustion of $\Omega$ from the assumption. For this exhaustion, we have equality $\mu (E_{n})=\nu (E_{n})$ for all $n\in \mathbb {N}$. Then from the continuity of the two measures $\mu ,\nu$ we get $\mu (\Omega )=\lim _{n\to \infty }\mu (E_{n})=\lim _{n\to \infty }\nu (E_{n})=\nu (\Omega )$.

${\mathcal {G}}$ is complement stable: let $A\in {\mathcal {G}}$. Because of $\mu (\Omega )=\nu (\Omega )<\infty$ for finite measures, we can exploit the subtractivity and obtain

$\mu (A^{\complement })=\mu (\Omega \setminus A)=\mu (\Omega )-\mu (A)=\nu (\Omega )-\nu (A)=\nu (\Omega \setminus A)=\nu (A^{\complement }).$

${\mathcal {G}}$ is also stable under disjoint unions: Let $(A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {G}}$ be a sequence of pairwise disjoint sets in ${\mathcal {G}}$. Taking advantage of the $\sigma$-additivity of $\mu$ and $\nu$ we obtain

$\mu \left(\biguplus _{n=1}^{\infty }A_{n}\right)=\sum _{n=1}^{\infty }\mu (A_{n})=\sum _{n=1}^{\infty }\nu (A_{n})=\nu \left(\biguplus _{n=1}^{\infty }A_{n}\right).$

Proof step: Proof in the general case

Now, we turn to non-finite measures.

We define for $n\in \mathbb {N}$ the dimensions $\mu _{n},\nu _{n}\colon {\mathcal {A}}\to [0,\infty ]$ with $\mu _{n}(A):=\mu (A\cap E_{n})$ and $\nu _{n}(A):=\nu (A\cap E_{n})$, where $E_{n}\in {\mathcal {C}}$ are the exhaustion sets from the assumption.

Since by assumption ${\mathcal {C}}$ is cut-stable and $\mu =\nu$ holds on ${\mathcal {C}}$, the measures $\mu _{n}$ and $\nu _{n}$ also coincide on ${\mathcal {C}}$. Furthermore, because of $\mu _{n}(A)=\mu (A\cap E_{n})\leq \mu (E_{n})<\infty$ and an analogous statement for $\nu _{n}$, the two measures $\mu _{n}$ , $\nu _{n}$ are finite. So we can apply the statement already proved for the finite case and obtain that $\mu _{n}=\nu _{n}$ holds on all ${\mathcal {A}}$ for all $n\in \mathbb {N}$. The limit transition $n\to \infty$ gives that also $\mu$ and $\nu$ on all ${\mathcal {A}}$ are equal.

Hint

This theorem is a good example of the usefulness of Dynkin systems: Because of the disjointness of all unions, we could conveniently exploit the $\sigma$-additivity of the measures $\mu$ and $\nu$ in the proof.

Because of the cut-stability of ${\mathcal {C}}$ the sequence of sets $(E_{n})_{n}$ exhausting $\Omega$ need not be monotonically increasing. It is sufficient to require that there exists a sequence $(A_{n})_{n}\subseteq {\mathcal {C}}$ such that $\Omega =\bigcup _{n=1}^{\infty }A_{n}$ and $\mu (A_{n})=\nu (A_{n})<\infty$ holds: If there exists such a sequence, we can define $E_{n}=\bigcup _{i=1}^{n}A_{i}$ and obtain a monotonically increasing sequence with limit $\Omega$. Moreover, these sets also satisfy $\mu (E_{n})=\nu (E_{n})<\infty$, as we saw in the section on cut-stability: Cut-stability ensures that $\mu$ and $\nu$ coincide even on finite (possibly non-disjoint) unions.

One therefore sometimes finds the following formulation of the uniqueness theorem:

Theorem (Uniqueness theorem (alternative version))

Let $\mu$ and $\nu$ be two measures on the $\sigma$-algebra ${\mathcal {A}}$ over the basic set $\Omega$. Let there be a generator ${\mathcal {C}}$ of ${\mathcal {A}}$ with the following properties:

• $\mu$ and $\nu$ coincide on ${\mathcal {C}}$, i.e. $\mu (C)=\nu (C)$ for all $C\in {\mathcal {C}}$,
• There is a sequence $(A_{n})_{n\in \mathbb {N} }\subseteq {\mathcal {C}}$ with $\Omega =\bigcup _{n=1}^{\infty }A_{n}$ and $\mu (A_{n})=\nu (A_{n})<\infty$,
• ${\mathcal {C}}$ is cut-stable, i.e. $A,B\in {\mathcal {C}}\implies A\cap B\in {\mathcal {C}}$.

Then $\mu =\nu$ holds on all of ${\mathcal {A}}$.

If $\mu$ and $\nu$ are probability measures, then the second condition is always automatically satisfied: Because of $\mu (\Omega )=\nu (\Omega )=1<\infty$ one can assume without restriction $\Omega \in {\mathcal {C}}$ and choose the constant sequence $(\Omega )_{n\in \mathbb {N} }$. In probability theory, therefore, one often finds the following version of the uniqueness theorem:

Theorem (Uniqueness theorem for probability measures)

Let $\mu$ and $\nu$ be two probability measures on the $\sigma$-algebra ${\mathcal {A}}$ over the basic set $\Omega$. Let there be a cut-stable generator ${\mathcal {C}}$ of ${\mathcal {A}}$ on which $\mu$ and $\nu$ coincide. Then $\mu =\nu$ holds on all of ${\mathcal {A}}$.