# If two Borel measures coincide on all open sets, are they equal?

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.

• 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. Jul 21, 2014 at 21:21

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.

• Perfect! This works since every (standard) open set has infinitely many points. Jul 21, 2014 at 21:36
• @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?) Sep 13, 2014 at 20:08
• @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. Sep 14, 2014 at 18:19
• @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) Feb 6, 2015 at 4:32
• @RobertCardona I guess maybe Etienne was referring to the $\pi - \lambda$-theorem. Jul 9, 2017 at 7:09

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 $$E$$.

• 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? Jul 21, 2014 at 21:29
• @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 $\infty$.