Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.