If $\int_A f d \mu = \int_A g d\mu$ for continuous positive $f,g$ and a Radon measure $\mu$, is $f=g$ a.e.? Let $X$ be a locally compact Hausdorff space and $\mu$ a Radon measure on $X$. If $f,g: X \to (0, \infty)$ are continuous functions such that
$$\int_A f d \mu = \int_A g d\mu$$ for all Borel subsets $A$,
can we deduce that $f=g$ almost everywhere?
(Note that in general $f(x)\ne g(x)$ for a point $x \in X$ is possible, even if $f,g$ are continuous, because a Radon measure may assign zero measure to an open subset of $X$!).
My attempt:
It is easy enough to see that $f\chi_U = g \chi_U$ $\mu$-almost everywhere for any $U\subseteq X$ compact subset $U$. Indeed, by continuity and compactness $f$ is bounded on $U$ and compact subsets have finite measure. The same can be said about precompact open sets.
However, $X$ may not be $\sigma$-finite so I don't see how to deduce that $f=g$ a.e.

Take $A=\{x: f(x) >g(x)\}$. If $\mu (A) >0$ then there exists a compact set $K \subseteq A$ such that $\mu (K)>0$. Now $\int_{K_N} fd\mu= \int_{K_N} gd\mu$ where $K_N=\{x\in K: f(x) \leq N, g(x) \leq N\}$. If $\mu (K_N)>0$ then  this is clearly a contradiction (since  $\int_{K_N} fd\mu> \int_{K_N} gd\mu$). Hence, $\mu (K_N)=0$ for each $N$ which shows that $\mu(A)=0$. Similarly, $\mu \{x: f(x)<g(x)\}=0$ so $f=g$ a.e.
Remark: Continuity of $f$ and $g$ is not used. Borel measurability is enough.
