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Let $X$ be a compact Hausdorff topological space. Suppose $X$ is not a singleton set and $C(X)$ denotes the space of continuous functions on $X$. Do we have that for all $L \subset C(X)$ a nondense subspace, there exist two probability measures which agree in integration against all elements of $L$ but not on $C(X)$? Here $L$ is assumed to contain the constants. (This last sentence was added after my comment, but before the answer.)

Jeff
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  • Apparently the answer is no, because to any $X$ you can adjoin 1 more point disjointly, and the same issue as with the singleton set arises. So the singleton set is a legitimate counterexample, not a "cheap" one. So now I'm wondering if the answer is "yes" if $X$ is connected, say. – Jeff Dec 21 '13 at 04:01

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I don't see any counterexamples. If $L$ is a subspace that is not dense in $C(X)$, then (by Hahn-Banach) there is a nonzero linear function whose kernel contains $L$. By Riesz representation, this functional is given by integration against a signed measure $\nu$. By Hahn-Jordan decomposition theorem, $\nu$ is $\nu^+-\nu^-$ where both measures are positive (and finite, since $\nu$ is). Since $L$ contains constants, $\nu^+(X)=\nu^-(X)$. Normalize the measures, and you are done.

In the case of $X$ being a singleton, the statement remains true, vacuously: there is no $L$ that satisfies the assumptions.

Post No Bulls
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  • Ah I see yes, just as soon as I added that $L$ contains all constants, the singletons are no longer a problem. So it's not singleton sets or connectedness issues that pose a problem. The pathologies are all within the case where $L$ may not contain the constants. – Jeff Dec 21 '13 at 07:29
  • @Jeff The pathologies are not so pathological either; since you insist on two measures being normalized, they automatically agree on constants. Thus, you are actually assuming that the measures agree on the span of $L\cup{1}$. The rest goes through as above, unless of course this span is dense. – Post No Bulls Dec 21 '13 at 08:05