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If the value of a measure on any subset in a generator of a sigma algebra is known, will the measure for the sigma algebra also be uniquely determined? Thanks!

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Thank you! What is ]a,b[? –  steveO Dec 11 '11 at 16:55
    
I don't understand if your counterexample works. The Lebesgue measure for (a,b) may not be infinite as long as a and b are not infinite –  steveO Dec 11 '11 at 16:59

6 Answers 6

up vote 7 down vote accepted

Consider the Borel $\sigma$-algebra $\mathcal B(\mathbb R)$, and the class $\mathcal C:=\left\{(a,+\infty),a\in\mathbb R\right\}$. Then the class $\mathcal C$ generates $\mathcal B(\mathbb R)$. Consider the counting measure $\mu$ over $\mathcal B(\mathbb R)$, that is $\mu(B)=\begin{cases} \operatorname{card} A&\mbox{ if }A \mbox{ is finite, }\\\ +\infty&\mbox{ otherwise,} \end{cases}$ and the Lebesgue measure. These measures have the same value over the elements of $\mathcal C$, but since for example $\{0\}\in\mathcal B(\mathbb R)$, and $\mu(\{0\})=1\neq \lambda(\{0\})=0$, these measure can't be the same.

However, we can show the following result:

Let $\mu$ a $\sigma$-finite measure on a measurable space $(X,\mathcal S)$, $\mathcal A$ an algebra which generates $\mathcal S$ and $\mu_1,\mu_2$ two measures on $\mathcal S$ such that for each $A\in\mathcal A$, $\mu_1(A)=\mu_2(A)=\mu(A)$.

Then $\mu_1(B)=\mu_2(B)$ for each $B\in\mathcal S$.

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@DavideGiraudo: Thanks! Let us talk about another example. The product measure is defined to be the unique measure that is the product of component measures on the product rectangles. Since the product sigma algebra is generated by the product rectangles, why can the product measure be uniquely determined by being specified on all the product rectangles? I think the result in your last paragraph does not apply to this case? –  steveO Dec 11 '11 at 20:15
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The finite disjoint union of product rectangle is an algebra, so putting $\mu(\bigsqcup_{i=1}^n A_i\times B_i):=\sum_{i=1}^n\mu_1(A_i)\mu_2(B_i)$ determines uniquely the measure if $\mu_1$ and $\mu_2$ are $\sigma$-finite. If it's not the case, look at this topic math.stackexchange.com/questions/70888/… –  Davide Giraudo Dec 11 '11 at 20:29

There is a famous example: A compact separable metric space, two different finite Borel measures on the space, but the two measures agree with each other on all of the balls for the metric.

R. O. Davies, "Measures not approximable or not specifiable by means of balls." Mathematika 18 (1971) 157--160

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Consider flipping two coins. Let $A$ be the event that the first coin is heads, and $B$ the event that the second coin is heads. $A$ and $B$ together generate the $\sigma$-algebra of all possible events. Suppose we know that $P(A) = P(B) = 1/2$ (i.e. each coin is unbiased). This is not enough information to determine whether the two coins are independent, so $P$ is not completely determined.

This is the same counterexample that I gave in this answer to another question. In its notation, $P$ and $Q$ agree on the events in $\mathcal{L}$, and $\sigma(\mathcal{L}) = \mathcal{F}$, but $P \ne Q$.

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Consider the $\sigma$-algebra given by all subsets of $\{a,b,c\}$. It is generated by $A = \{a,b\}$ and $C = \{b,c\}$. Let $\mu(A) = \mu(C) = 1$.

It could be that $\mu(\{a\}) = \mu(\{b\}) = \mu(\{c\}) = \frac{1}{2}$. It could also be that $\mu(\{a\}) = \mu(\{c\}) = \frac{1}{3}$ and $\mu(\{b\}) = \frac{2}{3}$.

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This is a strengthening of the sufficient condition presented by Davide Giraudo.

Let $(X,\Sigma)$ be a measurable space and $\mathcal{F}$ be a family of measurable sets such that

  1. $\Sigma=\sigma(\mathcal{F})$ and
  2. $\mathcal{F}$ is closed under pairwise intersections.

Then any $\sigma$-finite measure whose restriction to $\mathcal{F}$ is finite is completely determined by it's restriction to $\mathcal{F}$.

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Counterexample: Take $X$ to be the real line, $\Sigma$ the Borel $\sigma$-algebra, and ${\mathcal F}$ the class of half-lines $\{(-\infty,b], b\in X\}$. Then Lebesgue measure and twice Lebesgue measure agree on ${\mathcal F}$, but not on $\Sigma$. –  John Dawkins Jan 14 '12 at 17:58
    
That's true. I'll correct the statement. –  Michael Greinecker Jan 14 '12 at 18:01
    
Upvoted a long time ago. Nice reply! (1) I wonder if the description of $\mathcal{F}$ is actually the definition of a $\pi$ system? (2) Is your reply related to this result? –  Tim Mar 6 '12 at 11:59
    
Yes for both questions. –  Michael Greinecker Mar 6 '12 at 12:10
    
Thanks! Does the linked result generalize your reply, or do they not imply each other? –  Tim Mar 6 '12 at 12:15

Let consider the above mentioned assertion due to Davide Giraudo.

Fact 1. Let $μ$ a $\sigma$-finite measure on a measurable space $(X,S)$ , $\cal{A}$ an algebra which generates $S$ and $μ_1 ,μ_2$ two measures on $S$ such that for each $A\in\cal{A}$ , $μ_1 (A)=μ_2 (A)=μ(A)$. Then $μ_1 (B)=μ_2 (B)$ for each $B\in S$ .

Remark 1. Fact 1 , in general, is not true if $X$ is not covered by a countable family of elements of $\cal{A}$ which have finite $\mu$-measures.

Example 1. Let $N$ denotes a set of all natural munbers. Let $ {\bf R}^N $ be the topological vector space of all real-valued sequences equipped with the Tykhonoff topology. Let us denote by $ B({\bf R}^N) $ the $\sigma$-algebra of all Borel subsets in $ {\bf R}^N $.

Let $ (a_i)_{i \in N} $ and $ (b_i)_{i \in N} $ be sequences of real numbers such that $$ ( \forall i )( i \in N \rightarrow a_i < b_i ). $$

We put $$ A_n={\bf R}_0 \times \cdots \times {\bf R}_n \times (\prod \limits_{i > n}\Delta_i)~, $$ for $n \in N$, where $$ (\forall i)( i \in N \rightarrow {\bf R}_i={\bf R}~ \& ~ \Delta_i=[a_i;b_i[). $$ We put also $$ \Delta=\prod_{i \in N}\Delta_i. $$

For an arbitrary natural number $i \in N$, consider the Lebesgue measure $ \mu_i $ defined on the space ${\bf R}_i$ and satisfying the condition $\mu_i(\Delta_i)=1$. Let us denote by $\lambda_i$ the normed Lebesgue measure defined on the interval $\Delta_i$.

For an arbitrary $n \in N$, let us denote by $\nu_n$ the measure defined by $$ \nu_n= \prod \limits_{1 \le i \le n} \mu_i \times \prod\limits_{i > n} \lambda_i, $$ and by ${\overline{\nu}}_n$ the Borel measure in the space ${\bf R}^N$ defined by $$ ( \forall X)(X \in B({\bf R}^N) \rightarrow {\overline{\nu}}_n(X)= \nu_n(X \cap A_n)). $$

Following [G.Pantsulaia , Invariant and quasiinvariant measures in infinite-dimensional topological vector spaces. Nova Science Publishers, Inc., New York, 2007. xii+234 pp.](see Lemma 5.1, p. 93), for an arbitrary Borel set $X \subseteq {\bf R}^N $ there exists a limit $$ {\nu}_{\Delta}(X)= \lim \limits_{n \rightarrow \infty} \overline{\nu}_n(X). $$ Moreover, the functional ${\nu}_{\Delta}$ is a nontrivial $\sigma$-finite measure defined on the Borel $\sigma$-algebra $B({\bf R}^N)$.

Let $\Delta^{(1)}=[0,1[^{N}$ and $\Delta^{(2)}=[2,3[^{N}$. Let consider $\cal{A}$-a class of subsets of ${\bf R}^N$ defined by $$ {\cal{A}}=\{X \times R^{N \setminus \{1,\cdots,n\}}:X \in \cal{B}({\bf R}^n)~\&~n \in N\}. $$ Obviously, $\cal{A}$ is an algebra of subsets of ${\bf R}^N$ which generates the Borel $\sigma$-algebra $B({\bf R}^N)$.

On the one hand, the measures $\mu_1:={\nu}_{\Delta^{(1)}}$ and $\mu_2:={\nu}_{\Delta^{(2)}}$ are agree on $\cal{A}$. In particular, $\mu_1(X \times R^{N \setminus \{1,\cdots,n\}})=\mu_2(X \times R^{N \setminus \{1,\cdots,n\}})=+\infty$ if $n$-dimensional Lebesgue measure of $X$ is positive and $=0$ if $n$-dimensional Lebesgue measure of $X$ is zero.

On the other hand, we have that $\mu_1(\Delta^{(2)})=0$ and $\mu_2(\Delta^{(2)})=1$.

Example 2(Simple example) Let $X=[0,1[$ and $(x_k)_{k \in N}$ and $(y_k)_{k \in N}$ be two different everywere dense in $[0,1[$ sequences. Let $\cal{A}$ be a class of all subsets of $[0,1[$ every element of which is presented as a union of a finite family of left closed and right open subintervals of $[0,1[$. Obviously, $\cal{A}$ is the algebra of subset of $[0,1[$ which generates the Borel $\sigma$-algebra $\cal{B}([0,1[)$. Let define two measures $\mu_1$ and $\mu_2$ as follows: $\mu_1(X)=\# (\{x_k:k \in N\} \cap X)$ and $\mu_2(X)=\# (\{y_k:k \in N\} \cap X)$ for $X \in \cal{B}([0,1[)$, where $\#$ denotes the counting measure. Then $\mu_1$ and $\mu_2$ are agree on $\cal{A}$ but they are dfferent because $(x_k)_{k \in N}$ and $(y_k)_{k \in N}$ are different.

The following assertion is valid.

Fact 2. Let $μ$ a $\sigma$-finite measure on a measurable space $(X,S)$ , $\cal{A}$ an algebra which generates $S$ such that $X$ is covered by a countable family of elements of $\cal{A}$ which have finite $\mu$-measures. If $μ_1 ,μ_2$ are two measures on $S$ such that for each $A\in\cal{A}$ , $μ_1 (A)=μ_2 (A)=μ(A)$ then $\mu_1(B)=\mu_2(B)$ for all $B \in S$.

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