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.?

Question

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.

Answer 1

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.

Answer 2

In fact, a more general thing holds that what I wrote in the question:

If $\mu$ is a Radon measure on a locally compact Hausdorff space and $f,g: X \to [0, \infty[$ are positive continuous functions such that $$\int_K f d \mu = \int_K g d\mu$$ for all compact subsets $K \subseteq X$, then $f=g$ almost everywhere.

Proof: Let $A= \{f > g\}$. If $\mu(A) > 0$, inner regularity implies the existence of a compact subset $K \subseteq A$ such that $\mu(K) > 0$. Note that $f\chi_K$ and $g\chi_K$ are integrable, because these $f,g$ are bounded on $K$ and because $\mu(K) < \infty$ by definition of Radon measure. Then since $\mu(K)>0$, $$\int_K f d \mu > \int_K g d \mu = \int_K f d\mu$$ a contradiction. Thus, $\mu\{f > g\} = \mu(A) = 0$. Similarly, $\mu\{f < g\}=0$ and thus $f = g$ $\mu$-a.e.