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Here is the question:

Prove that for every $\epsilon>0$ and every set $A\in\mathcal{B}(\mathbb{R}^{n})$ there is a compact set $K\subset A$ such that $P(A\setminus K)\leq\epsilon$.

--I have previously shown that there is a closed set $F$ and an open set $G$ such that $F\subset A\subset G$ and $P(G\setminus F)\leq\epsilon$. For the current problem, I think that we can find $K$ as an approximation of the set $F$, such that $K=F_{N}=F\cap[-N,N]^{n}$, which is compact. I think I am just caught up in the set difference algebra, i.e.,$$P(A\setminus K)=P(A\setminus (F\cap[-N,N]^{n}))=P(A\cap(F^{c}\cup[-N,N]^{c})^{c})=\cdots$$

--Any help on how to finish this is appreciated!

Kirk Fogg
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1 Answers1

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You did the hardest part. Now, since you are going to intersect the closed set $F$ with a compact set $M$, you are, indeed, going to get a compact set, but its probability will be less (farther away from $A$).

So, first, assume that $P(G-F)<\epsilon/2$. So, $$P(G)=P((G-F)\cup F)=P(G-F)+P(F)<P(F)+\epsilon/2$$

Then find $N$ large enough so that $P(M)=P([-N,N]^n)>1-\epsilon/2$.

Finally, use $$P(F)=P(F\cap M\cup F\cap M^c)=P(F\cap M)+P(F\cap M^c)\le$$ $$\le P(F\cap M)+P(M^c)< P(F\cap M)+\epsilon/2$$ and, since $F\cap M\subseteq F\subseteq G$, $$P(A-(F\cap M))\le P(G-(F\cap M))=P(G)-P(F\cap M)<\epsilon$$

Vadim
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