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.
Please help me to solve this. Thanks in advance.


1 Answer 1


$\int f d\mu=\overline {\int f d\mu}=\int f d\overline \mu$ 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 $\nu=0$.

  • $\begingroup$ 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. $\endgroup$
    – abcdmath
    Commented Mar 5 at 6:07
  • 1
    $\begingroup$ 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 $\endgroup$ Commented Mar 5 at 6:11
  • $\begingroup$ Thank you, this is helpful. $\endgroup$
    – abcdmath
    Commented Mar 5 at 6:15

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