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.
Best Answer
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.