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Let $(X,\mathcal{S}, \mu)$ be a measure space with $X$ a locally compact Hausdorff space, $\mathcal{S}$ the Borel subsets of $X$ and $\mu$ a complex measure. Suppose that $$ \int_X f \ d\mu \in \mathbb{R} $$ for all $f\in C( X,\mathbb{R} )$ i.e. the continous functions from $X$ to $\mathbb{R}$. I want to know if this implies that $\mu$ is a real measure? Of course with $f\equiv1$ we have that $\mu(X) \in \mathbb{R}$, but does this necessarily implies that $\mu(E)\in \mathbb{R}$ for all $E \in \mathcal{S}?$

I am incline to think that $\mu$ must be a regular measure for this to happen.

Leo Sera
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  • What is $X$? Locally compact Hausdorff space? I did not know that the concept of abstract integration with respect to "complex measure" actually exists, but whatever the definition of that integration is, is continuous function always integrable under that definition? If $\mu= a+ib$ ($a,b$ are real and imaginary part respectively), is $\int f d\mu$ defined as $\int f da^+ - \int f da^- + i( \int f db^+ - \int f db^-)$? – Rubertos Jun 28 '16 at 17:07
  • @PhoemueX I missed to point out that $X$ is a topological space, and yes we can say that $X$ is a locally compact Hausdorff space. I editing. – Leo Sera Jun 28 '16 at 17:12
  • @Rubertos look here http://math.stackexchange.com/questions/119448/integration-in-respect-to-a-complex-measure . And of course every continuos function is measurable so the integration makes sense. – Leo Sera Jun 28 '16 at 17:19

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Seems to me that you need to add a couple more hypothesis. As you say the fact that $\mu$ is regular is crucial, but you also need that $\int f d\mu \in \mathbb{R}$ for all $f \in C_c(X)$ (of course if $X$ is compact it self, then $C_c(X)=C(X)$)

With all this, then indeed $\mu$ is a real measure. To prove it, take any $E \in \mathcal{S}$, now since $C_c(X)$ is dense in $L^1(X, \mu)$, we can find a sequence $(f_n)_n \subset C_c(X)$ such that $\|f_n-\chi_{E}\|_1 \to 0$ as $n \to \infty$. Moreover since $\mu$ is a complex regular measure, then $|\mu|(X)<\infty$, so $$ \left| \int (f_n-\chi_{E}) d\mu \right| \leq \|f_n-\chi_{E}\|_1 |\mu|(X) \to 0\text{ as $n \to \infty$} $$ and therefore $$ \mu(E) = \int \chi_E d\mu = \lim_{n \to \infty} \int f_n d\mu \in \mathbb{R} $$ since by the hypothesis each $\int f_n d\mu \in \mathbb{R}$.

Alonso Delfín
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