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 $Supp(\mu)\subset S$?

up vote 2 down vote accepted

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 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 at 14:18

Your Answer


By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.