Let $K$ be compact, Hausdorff space but not necessarily metrizable. Let $\mathfrak{M}$ be the Borel $\sigma$ field over $K$ and $\mu$ be a positive Regular Borel measure on $K$. Let $S$ be a subset of $K$ not necessarily in $\mathfrak{M}$. Suppose for all Baire sets $E\subseteq K\setminus S$, $\mu(E)=0$. Can I conclude that $Supp(\mu)\subset S$?
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Let $K=[0,1]$, $\mathfrak{R}$ its usual Borel $\sigma$-field. and $\mu$ Lebesgue measure, $S= K\setminus \mathbb{Q}$. Then, for all Baire sets $E \subset \mathbb{Q}, \mu(E)=0$, while $$\text{supp}(\mu)= K \ne S$$
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$\begingroup$ According to my notations is it possible to conclude that, $\int_Sf(t)d\mu (t)=\int_Kf(t)d\mu(t)$ if $S$ is Borel and $f\in C(K)$. Here remember that $\mu(E)=0$ if $E\subseteq K\setminus S$ is a Baire set. $\endgroup$ Commented Nov 10, 2018 at 12:34
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$\begingroup$ Still false.<br> $ K=[0,1]^{\mathbb{N}}$ with the product topology <br> $ \mathbb{0}= (0,0,...) \in K$<br> $\mu(A) = \mathbb{\delta}_{\mathbb{0}}$<br> $S=K\setminus \{\mathbb{0}\}$<br> $f := 1$<br> Because $\{\mathbb{0}\}$ is not a Baire set (while it is compact), the only Baire set contained in $S^{c}$ is $\emptyset$, and $$1 = \int_{K}fd\mu \ne 0= \int_{S} fd\mu $$ $\endgroup$ Commented Nov 10, 2018 at 14:18