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

Solution 1:

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.

Solution 2:

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