Integration against Borel measures on compact Hausdorff spaces

Question

I am studying the properties of integration against Borel measures and Baire measures. And I am not sure whether the following proposition is correct and I tried to give a proof.

Suppose that $X$ is a compact Hausdorff topological space and $\mu$ is a regular Borel measure on $\mathcal B(X)$. Let $\mu_{0}$ be the restriction of $\mu$ to Baire $\sigma$-algebra $\mathcal Ba(X)$. Then we have $\int_{X}fd\mu=\int_{X}fd{\mu}_{0}$ for all $f\in C(X)$.

Define a positive linear functional $I(f)=\int_{X}fd{\mu}_{0}$ for all $f\in C(X)$. Then, by Riesz-Markov Theorem, there is a regular Borel measure $\nu$ (i.e. Radon measure in this case) on $\mathcal B(X)$ such that $\int_{X}fd\nu=\int_{X}fd{\mu}_{0}$ holds for all $f\in C(X)$. Here is a claim saying that every Baire measure can be uniquely extended to a regular Borel measure. So it is sufficient to show that $\nu$ is an extension of ${\mu}_{0}$ and hence $\nu=\mu$.

By the regularity of ${\mu}_{0}$ and $\nu$, we have ${\mu}_{0}(E)=\sup \{ {\mu}_{0}(K): K\subseteq E, K $ is a compact $G_{\delta} $ set $\}$ for all Baire sets $E$ and $\nu(E)=\sup \{ \nu(K): K\subseteq E, K $ is a compact set $\}$ for all Borel sets $E$.

I just showed that for each comapct $G_{\delta}$ set $K$, $\nu(K)={\mu}_{0}(K)$ and I was wondering how to prove $\nu(E)={\mu}_{0}(E)$ for all Baire sets $E$?

Answer 1

Your highlighted proposition is true as a particular case of the following more general proposition.

Let $(S,\Sigma,\mu)$ be a measure space. Let a function $f\colon S\to\mathbb R$ be $\Sigma_0$-measurable and $\mu$-integrable, where $\Sigma_0$ is a sub-sigma-algebra of $\Sigma$. Let $\mu_0$ be the restriction of $\mu$ to $\Sigma_0$. Then $f$ is $\mu_0$-integrable and $\int_S f\, d\mu_0=\int_S f\, d\mu$.

This follows immediately from the definition of the integral of a function as the limit of the integrals of approximating simple functions: if the limit over $\Sigma$-measurable approximating simple functions exists, then the limit over the narrower set of $\Sigma_0$-measurable approximating simple functions exists and equals the "broader" limit.