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Let $S$ be a Polish space and $\mathcal B$ the Borel $\sigma$-Algebra on $S$. Let $\mu$ be some finite measure on $(S,\mathcal B)$ and $A\in \mathcal B$ such that $\mu (A)>0$. Is the following line of thought right, and if yes, why?

1.: In a Polish space one can find for each set $A\in \mathcal B$ a sequence $(U_n)_{n\in\mathbb N}$ of open sets so that $\cap_{n\in\mathbb N}U_n=A$ (while this feels somehow right, I cannot cite a theorem or something, stating this)

2.: Then for the sequence of sets $(U_n \setminus A)_{n\in\mathbb N}$ we have $\cap_{n\in\mathbb N}U_n\setminus A=\emptyset$

3.: So we have $\mu(\cap_{n\in\mathbb N}U_n\setminus A)\overset{n\rightarrow \infty}{\rightarrow}0$

4.: And hence $\mu(\cap_{n\in\mathbb N}U_n)=\mu(\cap_{n\in\mathbb N}U_n\setminus A)+\mu(A)\overset{n\rightarrow \infty}{\rightarrow}\mu(A)$.

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    $\begingroup$ I don't believe that either 1 or 3 is correct. 1 is equivalent to saying that all Borel sets are $G_\delta$, which is false for the interval. If we consider some sequence of sets $\{U_n\}$ as you defined them, I can see no reason that $U_{2m} =S$ would be invalid for all $m$. If you instead had $\mu(\cup_{n=1}^N U_n\setminus A)$ then yes, you do get the convergence you want. $\endgroup$ – user24142 May 10 '15 at 5:15
  • $\begingroup$ @user24142 I edited 3. and 4. accordingly. Is what you are saying this: "1. is wrong, but one can at least say there is a sequence of open sets $(U_n)_{n\in\mathbb N}$ each containing $A$, so that 3. holds"? I would be glad, if you could point me somewhere to read more on this. $\endgroup$ – Paul May 10 '15 at 18:41
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    $\begingroup$ No, there do not necessarily exist $\{ U_n\}$ tightly containing $A$. If you do have $A$ and $\{ U_n\}$ related as you say, then you can fix 3 as you have. There are set approximation results. Davide Giraudo cites a good example of such. If there is a countable generating algebra, $\mathcal A$ for the $\sigma$- algebra, and a $\sigma$-finite measure, then there is a sequence of sets in $\mathcal A$ such that $\lim \mu( A \Delta A_i) = 0$ for all $A$. I'm not really sure what you're wanting, but trying to tightly contain with open sets is going to be a problem. Consider goo.gl/V3K4mZ. $\endgroup$ – user24142 May 11 '15 at 5:24
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As us user24142 pointed out, it is not true that each Borel set can be written as a countable intersection of open sets. For example, consider the set of rational numbers.

However, the approximation property you want follows from the fact that the collection $$\left\{A\subset S, \forall \varepsilon\gt 0, \exists F\mbox{closed}, O\mbox{ open}\mid F\subset A\subset O,\mu(O\setminus F)\lt \varepsilon \right\} $$ is a $\sigma$-algebra.

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  • $\begingroup$ I don't see the relation between the $\sigma$-Algebra you have given and the Borel-$\sigma$-Algebra. Can I conclude from this, that for each $\epsilon>0$ I can find an open set $U_\epsilon\in\mathcal B$ with $|\mu(U_\epsilon)-\mu(A)|<\epsilon$? Can you point me to a book or other source that would help me to understand the details in this? $\endgroup$ – Paul May 10 '15 at 18:26
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    $\begingroup$ This $\sigma$-algebra contains the open sets, hence the Borel $\sigma$-algebra. The details are left as exercises in the sources I know. You can try and post the details as an answer, or ask an other question if you don't manage to do so. $\endgroup$ – Davide Giraudo May 10 '15 at 18:38

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