Support of a regular measure Reg

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 $$Supp(\mu)\subset S$$?

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 $$\text{supp}(\mu)= K \ne S$$
• 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. – Tanmoy Paul Nov 10 '18 at 12:34
• 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 $$1 = \int_{K}fd\mu \ne 0= \int_{S} fd\mu$$ – Taro NGUYEN Nov 10 '18 at 14:18