# Let X be a compact Hausdorff space and &#x03BC;<!-- μ --> be a complex Borel measure on X

Let $X$ be a compact Hausdorff space and $\mu$ be a complex Borel measure on $X$ such that for every $f\in C\left(X\right)$ with $f\ge 0.$ Then show that $\mu$ is non-negative.
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Mateo Carson
Let $E\subseteq X$ Borel. Let $ϵ>0$. By regularity, we can choose a compact subset $K\subseteq X$ and an open subset $U\subseteq X$ with $K\subseteq E\subseteq U$ and $|\mu |\left(U\setminus K\right)<ϵ$. By Urysohn's lemma, we can pick $f\in C\left(X,\left[0,1\right]\right)$ with ${\chi }_{K}\le f\le {\chi }_{U}$. Then
${\int }_{X}|{\chi }_{E}-f|d|\mu |\le |\mu |\left(U\setminus K\right)<ϵ.$
This shows that the function ${\chi }_{E}$ can be approximated by positive continuous functions in the ${L}^{1}\left(|\mu |\right)$-norm.
Next, let $E\subseteq X$ be a Borel subset. Choose a sequence of positive continuous functions $\left\{{f}_{n}{\right\}}_{n=1}^{\mathrm{\infty }}\subseteq C\left(X,\left[0,\mathrm{\infty }\left[\right)$ such that
${\int }_{X}|{f}_{n}-{\chi }_{E}|d|\mu |\stackrel{n\to \mathrm{\infty }}{⟶}0.$
Hence,
$\begin{array}{r}|{\int }_{X}{f}_{n}d\mu -{\int }_{X}{\chi }_{E}d\mu |\le {\int }_{X}|{f}_{n}-{\chi }_{E}|d|\mu |\stackrel{n\to \mathrm{\infty }}{⟶}0\end{array}$
and we deduce that
$\mu \left(E\right)={\int }_{X}{\chi }_{E}d\mu =\underset{n}{lim}\underset{\ge 0}{\underset{⏟}{{\int }_{X}{f}_{n}d\mu }}\ge 0.$
Since $E$ was an arbitrary Borel set, we conclude that $\mu$ is a positive (finite) measure.