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Let $A_n$ be a decreasing sequence of Borel sets with finite but with measure $\geq \epsilon > 0$ for all $n$. Then there exists a compact set $B_n \subset A_n$ for all $n$ such that $\mu(A_n \setminus B_n) < \frac{\epsilon}{2^{n+1}}$ for some $\epsilon >0$. Then define $C_n := \cap_{k=1}^n B_k$. It is clear that $C_n \subset B_n \subset A_n$. Then how come we have $\mu(A_n \setminus C_n) < \frac{\epsilon}{2}$ and $\mu(C_n) > \frac{\epsilon}{2}$, please? I tried the following argument.

$$ \mu(A_n \setminus C_n) = \mu\left( A \setminus \cap_{k=1}^n B_k \right) = \mu \left( A_n \cap \left( \cap_{k=1}^n B_k \right)^C \right) = \mu \left( A_n \cap \left( \cup_{k=1}^n B_k^C \right) \right) = \mu \left( \cup_{k=1}^n \left( A_n \cap B_k^C \right) \right) = \mu \left( \cup_{k=1}^n \left( A_n \setminus B_k \right) \right) \leq \sum_{k=1}^n \mu(A_n \setminus B_k) $$

I cannot see how to proceed from the above to show the two results.

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  • $\begingroup$ This is false unless you assume $\mu(A_n)<\infty$, there is no such compact set for, say $A_n=B(0,n)^c$. $\endgroup$ – Adam Hughes Oct 13 '14 at 11:20
  • $\begingroup$ @AdamHughes Thank you for pointing that out. I forgot to mention it. How to show the results after I assume what you said? $\endgroup$ – LaTeXFan Oct 13 '14 at 11:31
  • $\begingroup$ Also, the condition $\mu(C_n)>{\epsilon\over 2}$ is false. If all the $A_n$ have measure $0$, for example, then sub-additivity says all the $C_n$ have measure $0$. $\endgroup$ – Adam Hughes Oct 13 '14 at 11:36
  • $\begingroup$ @AdamHughes I guess I should add $A_n$ has non-zero measure. And the whole point is to show that $C_n$ is not empty. $\endgroup$ – LaTeXFan Oct 13 '14 at 11:41
  • $\begingroup$ @AdamHughes Thanks. What about the second result? BTW, it is allowed to comment on your answer. $\endgroup$ – LaTeXFan Oct 13 '14 at 11:47
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We use that

$$A_n\setminus C_n=A_n\cap\left(\bigcup_{k=1}^n B_k^c\right)$$ $$=\bigcup_{k=1}^n (A_n\cap B_k^c)\;\subseteq \;\bigcup_{k=1}^n (A_k\cap B_k^c)$$ $$=\bigcup_{k=1}^n (A_k\setminus B_k)$$

then

$$\mu(A_n\setminus C_n)\le\sum_{k=1}^n\mu(A_k\setminus B_k)=\sum_{k=1}^n {\epsilon\over 2^{n+1}}<{\epsilon\over 2}.$$


Edit to reflect edited question

Now note that

$$\mu(B_1\cup (A_1\setminus B_1))=\epsilon\le \mu(A_1)= \mu(B_1)+\mu(A_1\setminus B_1)\le\mu(B_1)+{\epsilon\over 4}$$

so that

$${3\epsilon\over 4}\le\mu(B_1)$$

Now continuing on like this, we see that

$$\epsilon\le \mu(A_n)=\mu(C_n)+\mu(A_n\setminus C_n)< \mu(C_n)+{\epsilon\over 2}$$

i.e.

$${\epsilon\over 2}<\mu(C_n).$$

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