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Let $\mathscr{S}$ be a nonempty Set. Let $\mathscr{E}\subset \mathscr{P}( \mathscr{S})$ be the generator of the $\sigma$-algebra $\sigma (\mathscr{E}) =:\mathscr{A}$. Let $\mu$ and $\mu'$ be measures on $\mathscr{A}$ and $\mu(\mathscr{S})=\mu'(\mathscr{S})$ and if necessary $\mu(\mathscr{S})<\infty$

Is the following statement true: $\Big(\forall A\in\mathscr{E}:\mu(A)=\mu'(A)\Big)\Rightarrow \Big(\forall A \in \mathscr{A}:\mu(A)=\mu'(A)\Big)$

Thanks in Advance

Just dropped in
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Börge
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1 Answers1

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No in general. There are some counterexamples in the duplicates.

However there are some interesting results. All the them can be proved using what seem to be the two standard tools to prove these kinds of results: Dynkin's $\pi-\lambda$ Theorem and the Monotone Class Theorem. Both are not very different, I might have read that the first one is more used in probability theory and the latest in measure theory for some reason.

$1)$ If two measures $\mu_1$ and $\mu_2$ agree on $\mathcal{A}\subseteq \mathcal{P}(X)$, where $\mathcal{A}$ is closed under finite intersections (a $\pi$-system) and satisfy that $\mu_1(X)=\mu_2(X)<\infty$, then $\mu_1=\mu_2$ on the $\sigma$-algebra generated by $\mathcal{A}$ (see a proof sketch here, under the Theorem; basically it is enough to show that $\{A\in \sigma(\mathcal{A}): \mu_1(A)=\mu_2(A)\}$ is a $\lambda$-system). For example, if two finite Borel measures on any topological space coincide on all open sets, then they coincide on all Borel sets.

$2)$ [Generalization of $1$] Suppose two measures $\mu_1$ and $\mu_2$ agree on $\mathcal{A}\subseteq \mathcal{P}(X)$, where $\mathcal{A}$ is closed under finite intersections (a $\pi$-system). Moreover suppose there exists a countable nested subfamily of sets $\mathcal{B}\subseteq \mathcal{A}$ that cover $X$ and have finite $\mu_1$-(and $\mu_2$-) measure. Then $\mu_1=\mu_2$ in the $\sigma$-algebra generated by $\mathcal{A}$ (proof scheme similar to $1$, essentially a straightforward generalization of the answer here). So for example the Lebesgue measure in all Borel sets in $\mathbb{R}^n$ is uniquely determined by its values on boxes/intervals.

$3)$ Up to a multiplicative constant, Lebesgue measure is the only translation-invariant measure on the Borel sets that puts finite measure on the unit interval. This can be generalised to higher dimensions considering the unit box $[0,1]^n$ instead of the unit interval. You can find proofs here using the $\pi - \lambda$ Theorem. A proof using the Monotone Class Theorem also seems quite straightforward.

Anguepa
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    Two results on equality of measures can be found in Proposition 3.7 and Exercise 3.18, both in Erhan Çınlar's "Probability and Stochastics". – user5280911 Sep 26 '20 at 21:37
  • Would you please provide the counter example? I am searching for a counter example but unable to find... – Badshah Khan Oct 19 '20 at 20:48
  • @MohammadShoaib Sure you can find one in the accepted answer here – Anguepa Oct 19 '20 at 22:22
  • But what if I need u(X)=v(X)< ♾️ also..? there is not any such condition mentioned in answers to that question... – Badshah Khan Oct 19 '20 at 22:55
  • @MohammadShoaib I suppose you can construct an easy one. For example on the set ${1,2,3,4}$ with $\mathscr{E} ={ {1,2}, {2,3}}$. Pick $\mu$ with $\mu({1})=\mu({2})=\mu({3})=\mu({4})=1$ and $\mu'$ with $\mu'({1})=\mu'({3})=0$, $\mu'({2})=2$, $\mu'({4})=2$. – Anguepa Oct 22 '20 at 00:36