Are continuous functions on an LCH space which are equal almost everywhere actually equal everywhere?

Question

This question (which I admit is rather idle) is inspired by this one:

showing almost equal function are actually equal

If $X$ is a locally compact Hausdorff space equipped with a non-trivial (see edit below) Radon measure $\mu$, by which I mean a positive Borel measure which is finite on compact sets, outer regular on all Borel sets, and inner regular on open sets (by the Borel $\sigma$-algebra I mean the $\sigma$-algebra generated by the open subsets of $X$). If $f,g:X\rightarrow\mathbf{C}$ are continuous functions such that the set $\{x\in X:f(x)\neq g(x)\}$ has $\mu$-measure zero, is it necessarily the case that $f(x)=g(x)$ for all $x\in X$?

If $X=G$ is a locally compact (Hausdorff) group and $\mu$ is a Haar measure (non-zero by definition), then using translation invariance one can prove that any non-empty open subset of $G$ has positive $\mu$-measure. Since the set $\{x\in G:f(x)\neq g(x)\}$ is necessarily open by continuity of $f$ and $g$, if it is non-empty, it must have positive measure in this case. But what about an arbitrary LCH space? Can non-empty open sets have measure zero? I guess that if there exists an $X$ with a non-empty open and closed set with measure zero, the zero function together with the characteristic function of such a set would provide an example of the type sought.

Incidentally, the reason I started thinking about this was because I was reviewing some analysis and found the result that for $1\leq p<\infty$, $C_c(X)$ is dense in $L^p(X,\mu)$. So it seemed to be taken for granted that the natural $\mathbf{C}$-linear map from $C_c(X)$ to $L^p(X,\mu)$ was injective, which I believe amounts to a positive answer to my question.

EDIT: I really should have thought more about this before asking, because, as nullUser kindly points out in his\her answer below, the zero measure is a trivial example showing that the answer to my question is a resounding "no." But I would still like to know whether there are non-zero Radon measures for which equality of almost-everywhere equal continuous functions can fail.

Answer 1

Take $X=\mathbb{R}$ with the standard topology, then $X$ is LCH and let $\mu = 0$ be the zero measure which is trivially a Radon measure. All functions are equal almost everywhere [$\mu$], but not all continuous functions are equal, so this contradicts the supposed statement.

Perhaps there is another hypothesis needed that would rule out this trivial counterexample?

EDIT: This also fails if one takes $X=\mathbb{R}$ and $\mu = \delta_x$ the Dirac measure (also trivially a Radon measure). As long as two functions $f,g$ agree at $x$ then they will be equal $\delta_x$-almost everywhere, but again this does not make them equal. Again the problem here (so it seems) is that $\delta_x$ may assign nonempty open sets measure zero.