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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.

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1) There always is a maximal open subset of zero measure on a LCH space with a Radon measure (monotone convergence holds for nets of lower semi-continuous functions: prop 1.24 here). Its complement is called the support of $\mu$. 2) The $L^p$-norm is always a semi-norm on $C_c(X)$ and you can define $L^p$ as the completion of the Hausdorff quotient wrt this norm (that you get all of $L^p$ is essentially Lusin's theorem). –  Martin Jan 5 '13 at 4:00
    
@Martin Just an added bonus to your comment. There is a maximal open subset of zero measure in $(X,\tau)$ so long as $\mu$ is a Borel measure such that $\mu(U) = \sup\{ \mu(K): K\subseteq U \text{ is compact} \}$ for all $U$ open. No $X$ being LCH necessary, just a weak form of inner regularity on $\mu$. –  nullUser Jan 5 '13 at 17:12
    
@nullUser: Thanks, you are right LCH is indeed irrelevant. There are endless variations on this theme. A convenient sufficient condition for a measure to have a support is $\tau$-additivity (see Bogachev's book or Fremlin vol 4 for more on this), which (in presence of local finiteness) implies the monotone convergence property of nets of lower-semicontinuous functions I alluded to. $\tau$-additivity in turn follows from the weak form of inner regularity you mention. –  Martin Jan 5 '13 at 23:38

1 Answer 1

up vote 1 down vote accepted

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.

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Wow, thank you @nullUser! I'm ashamed I didn't think of that myself. In the interest of salvaging the question (instead of just deleting it out of vanity) I will edit in the condition of non-triviality. –  Keenan Kidwell Jan 5 '13 at 2:22
    
Dear @nullUser, Thank you again! That's great. –  Keenan Kidwell Jan 5 '13 at 2:34

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