Find a Borel measure such that the closed sets aren't arbitrarily close to the Borel sets with finite measure

Question

I would like example of measures which shows that the following propositions are false:

Proposition 1: Let $\mathfrak{B}$ be the Borel $\sigma$-algebra of a topological space $X$ and $\mu:\mathfrak{B}\to\overline{\mathbb{R}}$ be a measure. Then for all $B\in\mathfrak{B}$ with $\mu (B)<\infty$ and $\varepsilon \in(0,\infty)$ there's a closed set $F\subseteq B$ such that $\mu(B\setminus F)<\varepsilon $.

Proposition 2: Let $\mathfrak{B}$ be the Borel $\sigma$-algebra of a topological space $X$ and $\mu:\mathfrak{B}\to\overline{\mathbb{R}}$ be a measure. Suppose that there's an open cover $\{O_n\}_{n\in\mathbb{N}}$ such that $\mu (O_n)<\infty$ for all $n\in\mathbb{N}$. Then for all $B\in\mathfrak{B}$ and $\varepsilon\in(0,\infty)$ there's an open set $O\supseteq B$ such that $\mu(O\setminus B)<\varepsilon $.

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I tried to find some counterexamples in some books but could not find any. I even looked for one in the book "Measure Theory" written by V.I. Bogachev.

In the Theorem 3.3 (which in the page 74 of the "Measure and Integration" written by John L. Menaldi) the author "proves" the two propositions above. I believe that his proof is flawed since the Theorem 3.3 says that every Borel measure is outer regular which isn't true.

I tried to find the error in that proof but I failed.

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I'm asking this question here since a similar question (see here) that I asked in math.stackexchange didn't receive any answer.

Thank you for your attention!

Answer 1

Let $X$ be the Sierpinski space $\{∅,\{b\},\{a,b\}\}$. Then the point $b$ is Borel but not closed. Let $\mu(a)=\mu(b)=1$. Take $B=\{b\}$, and it's a counterexample to proposition 1. Take $B=\{a\}$, and it's a counterexample to proposition 2.