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](https://digitalcommons.wayne.edu/cgi/viewcontent.cgi?article=1050&context=mathfrp)" 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](https://en.wikipedia.org/wiki/Regular_measure#Measures_that_are_neither_inner_nor_outer_regular).


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](https://math.stackexchange.com/questions/4340259/are-there-a-compact-k-and-an-open-o-such-that-k-subseteq-b-subseteq-o-and)) that I asked in math.stackexchange didn't receive any answer.


Thank you for your attention!



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


