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If $X$ is a locally compact space. And $\mu, \nu$ are two Radon measures on $X$. If for any continuous compactly supported function $f \in C_c ( X \to \mathbb R )$ we have $ \int f d\mu = \int f d\nu $, does this implies $ \mu = \nu $?

This question just come up in my mind when I reading my analysis book. I know the result holds if $X$ is also Hausdorff and $\sigma$-compact by Riesz representation theorem. But does the result still hold without assuming this? Any help is appreciated.

user112564
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  • It is sufficient for $X$ to be locally compact and Hausdorff. Note that in this case, the Riesz representation theorem still holds (see e.g. Folland's Real analysis).

    The main ingredients of the proof that $\int f,\mathrm{d}\mu=\int f,\mathrm{d}\nu\Rightarrow\mu=\nu$ are Urysohn's lemma (on a LCH space) and regularity properties of the measures $\mu$.

    If $X$ is not Hausdorff, I therefore doubt this property still holds, although I do not have a counter-example in mind.

     – Ian May 01 '14 at 07:24

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