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.



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> 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 $f\in C(X)$.
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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](http://www.allresearchjournal.com/archives/2015/vol1issue7/PartG/1-7-47.pdf) 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 $E$?



Integration against Borel measures on compact Hausdorff spaces
Your highlighted proposition is true as a particular case of the following more general proposition. 



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> 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
>  $\int\_S f\, d\mu\_0=\int\_S f\, d\mu$.
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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.


