Show the outer measure of $A \subseteq \mathbf{R}$ is equal to the limit as $n\rightarrow \infty$ of $A\cap[-n,n]$

Question

As the title:

Show the outer measure of $A \subseteq \mathbf{R}$ is equal to the limit as $n\rightarrow \infty$ of $A\cap[-n,n]$

I realize since the limit is a monotonic one, and the limit is less than or equal to the outer measure of $A$. I also realize that the limit equals $\sup_{n \in \mathbf{N}}\{$outer measures of$ \ A\cap[-n,n]\}$.

Yet I don't know how to proceed.

Answer 1

An informal observation: A productive way to look at measure theory problems is to look at the basic sets and basic operations on these sets. In this case the basic sets are intervals and the operations are length and partitioning. Things that are true at the basic level tend to percolate upwards to the more general constructs.

If $I$ is any interval, we see that $l(I) = \sum_k l(I \cap [k,k+1))$. This is the basic idea.

I am using $m^*$ to denote the Lebesgue outer measure.

From basic properties of the outer measure you have, for an arbitrary $A$, that $m^* A \le \sum_k m^* (A \cap [k,k+1))$. Furthermore, if $I_n$ is a cover of $A$ by intervals, we see that $I_n \cap [k,k+1)$ is a cover of $A \cap [k,k+1)$ by intervals. Note that $\sum_n l(I_n \cap [k,k+1)) \ge m^* (A \cap [k,k+1))$ and so (see Tonelli to justify the switch) $\sum_n l(I_n) = \sum_n \sum_k l(I_n \cap [k,k+1)) = \sum_k \sum_n l(I_n \cap [k,k+1)) \ge \sum_k m^* (A\cap [k,k+1))$, from which we get $m^*A = \sum_k m^* (A\cap [k,k+1))$ (note the similarity with the comment in the second paragraph).

Note that the latter result is true for any $A$, applying it to $A \cap [-n,n)$ shows that $m^*(A \cap [-n,n) ) = \sum_{k=-n}^{k=n-1} m^* (A\cap [k,k+1))$.

We have $m^*A = \sum_k m^* (A\cap [k,k+1)) = \lim_{n \to \infty} \sum_{k=-n}^{k=n-1} m^* (A\cap [k,k+1)) = \lim_{n \to \infty} m^*(A \cap [-n,n))$.

To finish, note that $m^*A \ge m^* (A \cap [-n,n]) \ge m^* (A \cap [-n,n))$.

Answer 2

As you indicated, because $A\cap [-k,k]\subseteq A$ we can apply monotonicity of the Lebesgue outer measure and take the limit on both sides of the inequality to find lim$_{k\to\infty} m_*(A\cap [-k, k])\leq m_*(A).$

Going the opposite way, first see that lim$_{k\to\infty} m_*(A\cap[-k,k]^c)=0.$ Let $\epsilon>0$ and choose $k\in \mathbb{N}$ such that $m_*(A\cap[-k,k]^c)<\epsilon$ for large $k$. Follow this through by writing $A=(A\cap[-k,k])\bigcup(A\cap[-k,k]^c)$ and apply monotonicity of the Lebesgue outer measure once more.