I am reading a proof of the existence of Lebesgue measure and am struggling to understand one part. I will first get you up to where I am in the proof.
We define for a set written as a finite disjoint union $A=(a_1,b_1] \cup \cdots \cup (a_n,b_n]$ the set function $\mu(A)=\sum\limits_{k=1}^n (b_k-a_k)$.
Sets of the form of $A$ above form a ring $\mathcal{A}$ and generate the Borel sigma algebra. In order to apply Caratheodory extension theorem, we must show that $\mu$ is countably additive on $\mathcal{A}$. This is equivalent to showing (from an exercise done previously) that if $(A_n)$ is a decreasing sequence of sets in $\mathcal{A}$ with $\bigcap\limits_n A_n=\emptyset$, then $\mu(A_n)\rightarrow 0$ as $n\rightarrow \infty$.
Suppose for contradiction that this fails. Then there exists $\epsilon>0$ such that $\mu(A_n)\geq 2\epsilon$ for all $n=1,2,\ldots$.
THIS IS THE BIT I AM STUCK ON....
But then we can find, for each $n$, a set $B_n\in\mathcal{A}$ such that $\overline{B_n}\subset A_n$ and $\mu(A_n\setminus B_n)\leq \epsilon2^{-n}$.
If somebody could explain the last line or reword it, I will be extremely grateful.
(PS this isn't set work. I am doing it in my holidays as a bit of fun ;)) Thank you in advance! B.
Each $A_n$ is a finite union of some number, say $q$, of half-open intervals of the form $(a,b]$. Shrink each of these just a little from the left end; I'll decide later how much to shrink it but for now I'll just call the shrinkage amount $\delta$. So you're now looking at shrunk intervals $(a+\delta,b]$ in place of $(a,b]$. Do this for each of the $q$ intervals $(a,b]$ that constitute $A_n$, and call the union of the shrunk intervals $B_n$. So $B_n\in\mathcal A$, and the closure of $B_n$ is the union of closed intervals $[a+\delta,b]$. In particular, the closure of $B_n$ is included in $A_n$.
All that we still need, for the statement you asked about, is that $\mu(A_n-B_n)$ is really small, $<\epsilon2^{-n}$. Well, how small is it? That is, how much of $A_n$ was lost in the shrinking process? Each of the $q$ intervals that formed $A_n$ lost $\delta$, so altogether, the loss was $q\delta$. We need that to be $<\epsilon2^{-n}$. We have no control over $\epsilon$, $n$, and $q$; they're given to us. But we can make $\delta$ as small as we need. So take $\delta<\epsilon2^{-n}/q$.