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Let $X$ be a topological space and let $\mathcal{B}(X)$ be its Borel $\sigma$-algebra. That is, $\mathcal{B}(X)$ is the smallest $\sigma$-algebra on $X$ containing all the open sets. Let $\mu, \eta : \mathcal{B}(X) \to [0,\infty]$ be two Borel measures.

Question: If $\mu(U) = \eta(U)$ for all open sets $U \subset X$, does it necessarily follow that $\mu = \eta$?

I suspect that the answer is "no". Obviously it would suffice to prove

$\{ S : \mu(S)=\eta(S)\}$ is a $\sigma$-algebra,

but I don't see why this should hold. In general, the sets where two measures agree does not seem to be a $\sigma$-algebra. For example, consider two trivial measures on $2^X$, one which assigns zero measure to all sets, one which assigns infinite measure to all nonempty sets. They agree only on the empty set which is not a $\sigma$-algebra.

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  • $\begingroup$ I am pretty sure the standard proof uses an exhausting sequence and $\sigma$-finite-ness to show this, but I cant think of a counterexample quickly. $\endgroup$ Jul 21, 2014 at 21:21

3 Answers 3

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Your guess is correct : the answer is "No" in general;

For example, let $\mu_1$ be the counting measure on $\mathbb R$, and let $\mu_2$ be the measure defined by $\mu_2(\emptyset)=0$ and $\mu_2(A)=\infty$ if $A\neq\emptyset$.

On the other hand, if the space $X$ is the union of an increasing sequence of open sets on which the two measures are finite, the the answer is "Yes". This follows from the monotone class theorem.

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  • $\begingroup$ Perfect! This works since every (standard) open set has infinitely many points. $\endgroup$
    – Mike F
    Jul 21, 2014 at 21:36
  • $\begingroup$ @Etienne: For the case in which $X$ is the union of an increasing sequence of open sets on which the two measures are finite, how does the monotone class theorem imply the result? I believe we would define our class to be $\mathcal M = \{A : \mu_1(A) = \mu_2(A)\}$, our $\sigma$-algebra to be the Borel $\sigma$-algebra, but how would we define the algebra $A_0$ that is contained in these two? And how do we show $\mathcal M$ is in fact a class? The hard part being $A_k \searrow A$ implies $A \in \mathcal M$ (which I'm assuming would use the $\sigma$-finite property?) $\endgroup$ Sep 13, 2014 at 20:08
  • $\begingroup$ @Robert Assume first that the measures are finite, and apply the monotone class theorem to your family $\mathcal M$. (btw, I don't understand why you want an algebra contained in something). Then, apply the "finite" result in the obvious way to get what you want in the more general situation. $\endgroup$
    – Etienne
    Sep 14, 2014 at 18:19
  • $\begingroup$ @Etienne, the definition I have for Monotone Class Theorem is as follows: $\mathcal A_0$ an algebra, $\mathcal A$ is the smallest $\sigma$-algebra containing $\mathcal A_0$, $\mathcal M$ is the smallest monotone class containing $\mathcal A_0$, then $\mathcal M = \mathcal A$. (Chapter 2 of Richard Bass' book) $\endgroup$ Feb 6, 2015 at 4:32
  • $\begingroup$ @RobertCardona I guess maybe Etienne was referring to the $\pi - \lambda$-theorem. $\endgroup$
    – Anguepa
    Jul 9, 2017 at 7:09
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Here is an alternative, perhaps slightly easier proof using Dynkin's π−λπ−λ Theorem: https://math.stackexchange.com/a/813414/283164 Although it is sketched for $\mathbb R$, it works more generally.

BTW, Lemma 7.1.2. (p. 68) of Measure Theory, volume 1, Vladimir I. Bogachev:
If two finite signed Borel measures on any topological space coincide on all open sets, they coincide on all Borel sets.

Its simple proof uses:
Lemma 1.9.4. If two probability measures on a measurable space $(X,A)$ coincide on some class $E\subset A$ that is closed with respect to finite intersections, then they coincide on the $\sigma$-algebra generated by $E$.

Link to Lemma 7.1.2

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Hint: what's the collection of all set where the 2 measures agree? Does that collection form a monotone class? Does it contains all open set?

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    $\begingroup$ Link for those (like me) that didn't know what a monotone class was: en.wikipedia.org/wiki/Monotone_class_theorem. $\endgroup$
    – Mike F
    Jul 21, 2014 at 21:24
  • $\begingroup$ Looking at the article above, it seems I need to show that $\mu$ and $\eta$ agree on the algebra of sets generated by the opens to apply the theorem. Are you suggesting this holds? $\endgroup$
    – Mike F
    Jul 21, 2014 at 21:29
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    $\begingroup$ @MikeF: sorry, it appeared that I made the classic mistake of calculating $\infty-\infty$. Among the requirement that it is an algebra, only complement is problematic. Since $X$ is open, they must agree on it, so I thought if they agree on $A$ they must agree on $X\backslash A$ too. But does not work if both are $\infty$. $\endgroup$
    – Gina
    Jul 21, 2014 at 21:36
  • $\begingroup$ That's fine, feel free to leave the answer in spite of the error. It is still interesting information. $\endgroup$
    – Mike F
    Jul 21, 2014 at 21:37

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