# Measure is real if integration of all real valued continuous function is real

Let $$X$$ be a compact Harsdorff space and $$\cal A$$ be a $$\sigma$$-algebra over $$X$$. Let $$\mu$$ be a complex measure on $$(X,\cal A)$$. Let $$C(X)$$ denotes the set of all complex valued continuous function on $$X$$. Let $$C(X)_\mathbb R:=\{f\in C(X)| f=\overline f\}.$$ I want to show that if $$\displaystyle \int fd\mu\in \mathbb R$$ for all $$f\in C(X)_\mathbb R$$, then the masure $$\mu$$ is real.
My approach: Let $$f\in C(X)_\mathbb R$$. Then $$f$$ is real-valued continuous function on $$X$$. Then we can write $$\int f d\mu=\int f d\mu_1+i\int fd\mu_2.$$ Now by the given condition $$\displaystyle \int fd\mu\in \mathbb R$$, so $$\displaystyle \int fd\mu_2=0.$$ Now it is enough to show that $$\displaystyle \int fd\mu_2=0$$ for every $$f\in C(X)_\mathbb R$$ gives $$\mu_2=0$$. If I use indicator function, then for every set $$A$$ the integration $$\displaystyle \int 1_A d\mu_2=0$$ gives $$\mu_2(A)=0$$ and we are done. But indicator functions are not continuous.
$$\int f d\mu=\overline {\int f d\mu}=\int f d\overline \mu$$ if $$f \in C(X)_{\mathbb R}$$ so $$\nu \equiv \mu-\overline \mu$$ is a real measure with $$\int fd\nu=0$$ for every $$f \in C(X)_{\mathbb R}$$. This implies that $$\nu=0$$.
• Thank you for your answer, but my question is exactly the your last line that, how can I show that $\int fd\nu=0$ for every $f\in C(X)_\mathbb R$ implies $\nu=0$? May be this is very trivial, please give me some hint. I think something with indicator function. Commented Mar 5 at 6:07
• That requires regularity of measures. Have you studied Riesz Theorem for $(C(X))^{*}$? Part of the proof of this theprem shows that $\int fd\mu=0$ for any continuous $f$ implies $\mu=0$. Ref: Rudin's RCA. @abcdmath Commented Mar 5 at 6:11