Uniqueness of a continuation – "Math for Non-Geeks"

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 $\mathrm{\Omega }$ 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 ${ℝ}^n$ 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 $𝒞$, such that the function on sets restricted to it fulfills the desired properties. For example, for defining a geometric volume, we choose $𝒞$ 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 $\mu :𝒞\to [0,\mathrm{\infty }]$ can be continued to a measure on the $\sigma$-algebra $\sigma (𝒞)$ 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 $\sigma (𝒞)$? In other words, we are interested in whether a measure $\mu$ on the $\sigma$-algebra $\sigma (𝒞)$ is already uniquely determined by its values on the smaller set system $𝒞$.

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

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 $𝒞$ and the two measures $\mu$ and $\nu$ under which uniqueness holds. For this we consider the system of "good sets"

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

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

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

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

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

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

Theorem (The principle of good sets)

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

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

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

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

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

Existence of an inner approximation

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

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

Let $𝒞$ be the set system of all one-element subsets of $ℝ$:

$𝒞=\{\{x\}\mid x\in ℝ\}$

The $\sigma$-algebra generated by this is

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

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

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

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

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

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

Idea and definition of an inner approximation

We know that the measures $\mu$ and $\nu$ coincide on the sets in $𝒞$. The idea is to cover the basic set with at most countably many sets from $E_1,E_2,\dots \in 𝒞$, i.e. their union should be $\bigcup _{n=1}^{\mathrm{\infty }}{E}_n=\mathrm{\Omega }$.

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

• 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 $𝒞$), that $\mu (\mathrm{\Omega })=\nu (\mathrm{\Omega })$ 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: $\mu (\mathrm{\Omega })=\mu \left(\underset{i=1}{\overset{\mathrm{\infty }}{⨄}}{E}_i\right)=\sum _{n=1}^{\mathrm{\infty }}\mu (E_i)=\sum _{n=1}^{\mathrm{\infty }}\nu (E_i)=\nu \left(\underset{i=1}{\overset{\mathrm{\infty }}{⨄}}{E}_i\right)=\nu (\mathrm{\Omega })$
• If they are contained in each other in ascending order (i.e. $E_1\subseteq E_2\subseteq \dots$), they form a monotonic set sequence with limit $\bigcup _{i=1}^{\mathrm{\infty }}{E}_i=\mathrm{\Omega }$. Then we can use the continuity of the measures $\mu$ and $\nu$: $\mu (\mathrm{\Omega })=\mu (\underset{n\to \mathrm{\infty }}{lim}{E}_n)=\underset{n\to \mathrm{\infty }}{lim}\mu (E_i)=\underset{n\to \mathrm{\infty }}{lim}\nu (E_i)=\nu (\underset{n\to \mathrm{\infty }}{lim}{E}_n)=\nu (\mathrm{\Omega })$

If neither is the case, then we can not necessarily conclude $\mu (\mathrm{\Omega })=\nu (\mathrm{\Omega })$ 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 $\mathrm{\Omega }$ be a set and let $E_n\subseteq \mathrm{\Omega }$ be subsets for each $n\in \mathbb{N}$ with $E_n\subseteq E_{n+1}$. If $\mathrm{\Omega }=\bigcup _{n\in \mathbb{N}}{E}_n$, then the sequence $(E_n{)}_{n\in \mathbb{N}}$ is called an exhaustion of $\mathrm{\Omega }$.

Example (Exhaustion)

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

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

Intermediate result

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

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

Hint

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

The inner approximating sets have finite measure

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

Unions

Let $A$ and $B$ be two sets from the set of good sets $𝒢$. 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 $𝒢$. 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 $𝒢$.

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 $A_1,A_2,\cdots \in 𝒢$, the $\sigma$-additivity of measures $\mu$ and $\nu$ implies

$\mu \left(\underset{i=1}{\overset{\mathrm{\infty }}{⨄}}{A}_i\right)=\sum _{i=1}^{\mathrm{\infty }}\mu (A_i)=\sum _{i=1}^{\mathrm{\infty }}\nu (A_i)=\nu \left(\underset{i=1}{\overset{\mathrm{\infty }}{⨄}}{A}_i\right).$

Set differences

Let again $A$ and $B$ be two sets from $𝒢$, 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 $𝒢$ 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 $A\setminus B=\mathrm{\emptyset }$. Since $\mu (\mathrm{\emptyset })=0=\nu (\mathrm{\emptyset })$ , it lies again in $𝒢$. 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 "$\mathrm{\infty }-\mathrm{\infty }$", which cannot be sensibly defined:

Example ($\mathrm{\infty }-\mathrm{\infty }$ cannot be defined)

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

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

This shows that in general we cannot make sense of the expression $\mathrm{\infty }-\mathrm{\infty }$, 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 $\mathrm{\Omega }$ with sets from the set system $𝒞$, i.e., a monotonically growing set sequence in $𝒞$ with limit $\mathrm{\Omega }$. (This was to guarantee $\mu (\mathrm{\Omega })=\nu (\mathrm{\Omega })$.)

Then the sets $A_n:=A\cap E_n$ also form an increasing set sequence with limit

$\bigcup _{n=1}^{\mathrm{\infty }}{A}_n=\bigcup _{n=1}^{\mathrm{\infty }}(A\cap E_n)=A\cap \bigcup _{n=1}^{\mathrm{\infty }}{E}_n=A\cap \mathrm{\Omega }=A$.

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 $n\to \mathrm{\infty }$. 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 $C\in 𝒞$ with the $E_n$ are said to lie in $𝒢$.
• Each set of the exhaustion has finite measure, i.e., $\mu (E_n)<\mathrm{\infty }$ for all $n\in \mathbb{N}$. Only then, because of monotonicity, we can be sure that $\mu (A_n)=\mu (A\cap E_n)\le \mu (E_n)<\mathrm{\infty }$ and $\mu (B_n)=\mu (B\cap E_n)\le \mu (E_n)<\mathrm{\infty }$ 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 $\mathrm{\infty }$ to every set except $\mathrm{\emptyset }$.

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

$\mu (B\setminus A)=\underset{n\to \mathrm{\infty }}{lim}\mu (B_n\setminus A_n)=\underset{n\to \mathrm{\infty }}{lim}(\mu (B_n)-\mu (A_n))=\underset{n\to \mathrm{\infty }}{lim}(\nu (B_n)-\nu (A_n))=\underset{n\to \mathrm{\infty }}{lim}\nu (B_n\setminus A_n)=\nu (B\setminus A).$

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 show 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 $ℝ$

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

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

$\mu (A)=\{\begin{array}{c}0,\text{ if }A=\mathrm{\emptyset },\\ \mathrm{\infty },\text{ else.}\end{array}$

and

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

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

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

Intermediate result

We conclude what has been found out:

• The "set of good sets" $𝒢$ 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 $𝒢$, that is, $\mu (\mathrm{\Omega })=\nu (\mathrm{\Omega })$, we require that $𝒞$ contains an exhaustion $(E_n{)}_{n\in \mathbb{N}}$ of $\mathrm{\Omega }$.
• In order for differences of sets of infinite measure, which are subsets of each other, are in $𝒢$, we require, that the exhaustion sets $(E_n{)}_{n\in \mathbb{N}}$ all have finite measure and that for all $A\in 𝒢$ the intersections $A\cap E_n$ lie again in $𝒢$.

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

The last condition is somewhat unsatisfactory, because it involves $𝒢$ (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 $𝒞$, respectively. We will still work on this and weaken the condition later.

Hint

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

Dynkin systems

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

Thus, we can already characterize the set system $𝒢$ 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 $𝒟\subseteq 𝒫(\mathrm{\Omega })$ is called a Dynkin system if it holds that:

1. $\mathrm{\Omega }\in 𝒟$
2. for every two sets $A,B\in 𝒟$ with $A\subseteq B$ we also have $B\setminus A\in 𝒟$.
3. for each countably many pairwise disjoint sets $A_1,A_2,\cdots \in 𝒟$ we also have $\underset{n\in \mathbb{N}}{⨄}{A}_n\in 𝒟$.

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

1. $\mathrm{\Omega }\in 𝒟$
2. for every countably many $A\in 𝒟$, we also have $A^{\complement }\in 𝒟$.
3. for each countably many pairwise disjoint sets $A_1,A_2,\cdots \in 𝒟$ we also have $\underset{n\in \mathbb{N}}{⨄}{A}_n\in 𝒟$.

Proof

Let $𝒟$ 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 $A\in 𝒟$ be arbitrary. We have that $\mathrm{\Omega }\in 𝒟$ and $A\subseteq \mathrm{\Omega }$, so it follows from property 2. of the definition that also $A^{\complement }=\mathrm{\Omega }\setminus A\in 𝒟$.

Conversely, let the three properties from the theorem be satisfied. We show that then $𝒟$ is a Dynkin system in the sense of the definition above. Again, we only need to show point 2. So let $A,B\in 𝒟$ 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 $𝒟$.

So with the preconditions from the intermediate result of the previous section, $𝒢$ 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 $𝒞$, we have that $𝒞\subseteq 𝒢$ holds. So the Dynkin system generated by $𝒞$ also lies in the set of good sets $𝒢$, which is defined analogously to the generated $\sigma$-algebra:

Definition (Generated Dynkin system)

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

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

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

As with the definition of the generated $\sigma$-algebra, we need to convince ourselves that $\delta (𝒞)$ 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 $𝒞$:

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

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 $\mathrm{\Omega }$ be a set with exactly $n$ elements. The set system

$𝒟:=\{A\subseteq \mathrm{\Omega }\mid \text{number of elements in }A\text{ is even}\}$ is a Dynkin system:

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

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

Since $\mathrm{\Omega }$ is finite, we need only consider unions of finitely many disjoint sets for the third condition. Let $A_1,\dots ,A_k\in 𝒟$ be pairwise disjoint sets, each containing an even number of elements. Then, because of disjointness, the union $\underset{i=1}{\overset{k}{⨄}}{A}_i$ also contains an even number of elements and so lies in $𝒟$.

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

Motivation for cut-stability

We considered conditions under which the set system $𝒢=\{A\in \sigma (𝒞)\mid \mu (A)=\nu (A)\}$ 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 $𝒞$, $𝒞\subseteq 𝒢$ holds. Since $𝒢$ is itself a Dynkin system, therefore the Dynkin system $\delta (𝒞)$ generated by $𝒞$ is also contained in $𝒢$.

We would like to have $\sigma (𝒞)\subseteq 𝒢$ , such that $\mu$ and $\nu$ agree on their whole domain of definition. So $𝒢$ 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 $A,B\in 𝒢$ 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 $\mu (A\cup B)\ge \mu (A)$ and $\mu (A\cup B)\ge \mu (B)$.

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 $\mu (A\cup B)=\mathrm{\infty }=\nu (A\cup B)$ 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 $ℳ$ is called cut stable, if

$A,B\in ℳ⟹A\cap B\in ℳ$

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 $𝒢$ in addition to our previous conditions already enough for it to be an $\sigma$-algebra? Suppose $𝒢$ is cut-stable. The previous reasoning for two sets can be extended by induction to any finite union of good sets: Let $A_1,A_2,\dots ,A_n\in \sigma (𝒞)$ 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

$\begin{array}{cc}\mu \left(\bigcup _{i=1}^n{A}_i\right)=\mu (A_n\cup \bigcup _{i=1}^{n-1}{A}_i)& =\mu \left(A_n\right)+\mu \left(\bigcup _{i=1}^{n-1}{A}_i\right)-\mu (A_n\cap \bigcup _{i=1}^{n-1}{A}_i)\\ & \stackrel{\text{i.a.}}{=}\nu \left(A_n\right)+\nu \left(\bigcup _{i=1}^{n-1}{A}_i\right)-\mu \left(\bigcup _{i=1}^{n-1}\underset{\in 𝒢}{\underbrace{(A_n\cap A_i)}}\right)\\ & \stackrel{\text{i.a.}}{=}\nu \left(A_n\right)+\nu \left(\bigcup _{i=1}^{n-1}{A}_i\right)-\nu (\bigcup _{i=1}^{n-1}(A_n\cap A_i))=\nu \left(\bigcup _{i=1}^n{A}_i\right).\end{array}$

(with "i.a." meaning "induction assumption") Making use of the cut-stability, we have that $𝒢$ 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

$B_n:=A_n\setminus (A_1\cup \cdots \cup A_{n-1})=A_n\cap (A_1\cup \cdots \cup A_{n-1}{)}^{\complement },$

Then $\bigcup _{n\in \mathbb{N}}{A}_n=\underset{n\in \mathbb{N}}{⨄}{B}_n$ is a disjoint union of sets. If the $A_n$ are from $𝒢$, then so are the $B_n$, provided that $𝒢$ is cut-stable: According to the preceding, the finite union $A_1\cup \cdots \cup A_{n-1}$ lies in $𝒢$, and so does the complement of this set (since $𝒢$ is a Dynkin system). And by cut-stability also the intersection of it with $A_n$ lies in $𝒢$. Now, $𝒢$ is closed under countable disjoint unions (Dynkin system), so also $\underset{n\in \mathbb{N}}{⨄}{B}_n=\bigcup _{n\in \mathbb{N}}{A}_n$ is a good set.

These considerations show: If the Dynkin system $𝒢$ 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 $𝒟\subseteq 𝒫(\mathrm{\Omega })$ be a cut stable Dynkin system. Then $\mathrm{\Omega }\in 𝒟$ 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 $𝒟$.

Now let $(A_n{)}_{n\in \mathbb{N}}$ be any sequence of sets in $𝒟$. 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

$\bigcup _{i=1}^n{B}_i=\bigcup _{i=1}^{n-1}{B}_i\cup (A_n\setminus \left(\bigcup _{i=1}^{n-1}{A}_i\right))=\bigcup _{i=1}^{n-1}{B}_i\cup A_n.$

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, $B_1=A_1\in 𝒟$. Suppose now that $B_1,\dots ,B_n\in 𝒟$ holds (induction assumption). Then with stability with respect to disjoint unions it follows that $\bigcup _{i=1}^n{B}_i\in 𝒟$. With difference stability, we further have $B_{n+1}=A_{n+1}\setminus \left(\bigcup _{i=1}^n{B}_i\right)\in 𝒟$. Then, by induction, it also holds that $B_i\in 𝒟$ for all $i\in \mathbb{N}$.

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

From this we finally get

$\bigcup _{n\in \mathbb{N}}{A}_n=\bigcup _{n\in \mathbb{N}}\bigcup _{i=1}^n{A}_i=\bigcup _{n\in \mathbb{N}}\bigcup _{i=1}^n{B}_i=\bigcup _{n\in \mathbb{N}}{B}_n\in 𝒟.$

Intermediate result

We conclude our so far obtained results:

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

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

Hint

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

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

Cut-stability of the generator

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

So under what conditions is the Dynkin system generated by $𝒞$ cut-stable? This apparently depends only on the properties of the set system $𝒞$, not on the measures $\mu$ or $\nu$. In fact, it is sufficient if $𝒞$ 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 $𝒞$ is itself cut-stable.

Proof

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

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

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

So let us now show the necessary statements.

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

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

Let $A\in 𝒟(B)$ be arbitrary. Then $A\cap B\in \delta (𝒞)$ holds. Furthermore, because of $B\in \delta (𝒞)$, of course $B^{\complement }\in \delta (𝒞)$. 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 $\delta (𝒞)$, is also an element of $\delta (𝒞)$ and so $A^{\complement }\in 𝒟(B)$, i.e. $𝒟(B)$ is complement stable.

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

$\left(\underset{k\in \mathbb{N}}{⨄}{A}_k\right)\cap B=\underset{k\in \mathbb{N}}{⨄}(A_k\cap B)\in \delta (𝒞).$

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

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

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

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

Theorem

If $𝒞$ is a cut-stable set system, then $\delta (𝒞)=\sigma (𝒞)$ 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 $\sigma (𝒞)=𝒢$, that is, equality of $\mu$ and $\nu$ on all of $\sigma (𝒞)$:

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

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

Example

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

$\mu :=\frac{1}{2}({\delta }_1+{\delta }_2),\nu :=\frac{1}{2}({\delta }_2+{\delta }_4).$

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

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

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

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 $𝒜$ over the basic set $\mathrm{\Omega }$. Let there be a generator $𝒞$ of $𝒜$ with the following properties:

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

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

Proof (Uniqueness of measure continuations)

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

$𝒜=\sigma (𝒞)=\delta (𝒞)\subseteq \delta (𝒢)=𝒢$

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 $𝒢$ in two steps: First assuming that $\mu ,\nu$ are finite measures (i.e. $\mu (A)<\mathrm{\infty },\nu (A)<\mathrm{\infty }$ for all $A\in 𝒜$), then generalizing to the infinite measure case.

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

We have $\mathrm{\Omega }\in 𝒢$: Let $(E_n{)}_{n\in \mathbb{N}}\subseteq 𝒞$ be the exhaustion of $\mathrm{\Omega }$ 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 $\mu (\mathrm{\Omega })=\underset{n\to \mathrm{\infty }}{lim}\mu (E_n)=\underset{n\to \mathrm{\infty }}{lim}\nu (E_n)=\nu (\mathrm{\Omega })$.

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

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

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

$\mu \left(\underset{n=1}{\overset{\mathrm{\infty }}{⨄}}{A}_n\right)=\sum _{n=1}^{\mathrm{\infty }}\mu (A_n)=\sum _{n=1}^{\mathrm{\infty }}\nu (A_n)=\nu \left(\underset{n=1}{\overset{\mathrm{\infty }}{⨄}}{A}_n\right).$

Proof step: Proof in the general case

Now, we turn to non-finite measures.

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

Since by assumption $𝒞$ is cut-stable and $\mu =\nu$ holds on $𝒞$, the measures ${\mu }_n$ and ${\nu }_n$ also coincide on $𝒞$. Furthermore, because of ${\mu }_n(A)=\mu (A\cap E_n)\le \mu (E_n)<\mathrm{\infty }$ 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 $𝒜$ for all $n\in \mathbb{N}$. The limit transition $n\to \mathrm{\infty }$ gives that also $\mu$ and $\nu$ on all $𝒜$ 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 $𝒞$ the sequence of sets $(E_n{)}_n$ exhausting $\mathrm{\Omega }$ need not be monotonically increasing. It is sufficient to require that there exists a sequence $(A_n{)}_n\subseteq 𝒞$ such that $\mathrm{\Omega }=\bigcup _{n=1}^{\mathrm{\infty }}{A}_n$ and $\mu (A_n)=\nu (A_n)<\mathrm{\infty }$ 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 $\mathrm{\Omega }$. Moreover, these sets also satisfy $\mu (E_n)=\nu (E_n)<\mathrm{\infty }$, 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 $𝒜$ over the basic set $\mathrm{\Omega }$. Let there be a generator $𝒞$ of $𝒜$ with the following properties:

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

Then $\mu =\nu$ holds on all of $𝒜$.

If $\mu$ and $\nu$ are probability measures, then the second condition is always automatically satisfied: Because of $\mu (\mathrm{\Omega })=\nu (\mathrm{\Omega })=1<\mathrm{\infty }$ one can assume without restriction $\mathrm{\Omega }\in 𝒞$ and choose the constant sequence $(\mathrm{\Omega }{)}_{n\in \mathbb{N}}$. 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 $𝒜$ over the basic set $\mathrm{\Omega }$. Let there be a cut-stable generator $𝒞$ of $𝒜$ on which $\mu$ and $\nu$ coincide. Then $\mu =\nu$ holds on all of $𝒜$.