How to prove that outer measure $|A|=\lim_{t\to \infty}|A\cap (-t,t)|$?

Question

Let $A\subset R$. Then, $|A|=\lim_{t\to \infty}|A\cap (-t,t)|\tag 1$

$|A|:=\inf\{\sum_j l(I_j): A\subset \cup I_j\},$ where $I_j$'s are open intervals in $\mathbb R$ and $l(I_j)$ is the length of the interval $I_j$, and infimum is taken over all sequences of open intervals whose union contains $A$.

I am trying to prove $(1)$.

I know that $|A|= |A\cap (-t,t)|+|A\cap (-t,t)^c|, t>0$.

Suppose that $|A|<\infty$. Then, it suffices to prove that $\lim_{t\to\infty}|A\cap(-t,t)^c|=0$.

Define $f(t):=|A\cap (-t,t)^c|,t>0$. Then, $f$ is a monotonically decreasing function. So $\lim_{t\to \infty}f(t)= \inf_{t>0} f(t)=p,$ say.

If it is shown that $p=0$, then the proof is complete. Suppose that $p>0$. I don't understand how to get a contradiction from here.

And if $|A|=\infty$, then I don't know how to proceed.

Please help. Thanks.

Answer 1

In your definition of outer measure you should write $A\subset\cup_j I_j$. You’ve also written $\cup$ a few times where you meant $\cap$.

In the case $A|<\infty$, fix a countable indexing set $J$ of open intervals $(I_j)_{j\in J}$ whose union covers $A$ and whose measure is finite. Let $\lambda$ denote Lebesgue measure.

Define $I=\bigcup_{j\in J}I_j$. I know for any $t>0$, $|A\cap(-t,t)^c|\le\lambda(I\cap(-t,t)^c)$. Because $I$ has finite measure and $I\cap(-t,t)^c\supseteq I\cap(-t’,t’)^c$ when $t<t’$, and because the intersection of all $(-t,t)^c$ is empty and has null measure, it follows from continuity of measure that: $$\lim_{t\to\infty}\lambda(I\cap(-t,t)^c)=0$$Giving the claim.

So what if $|A|=\infty$? Hint: can you prove that: $$|A\cap(-t,t)|=\sum_{n\in\Bbb Z}|A\cap(n,n+1)\cap(-t,t)|$$And then take limits?

---

It was requested that we solve this problem without using continuity of measure and $\lambda$. The infinite case follows from the above summation idea, so consider only $|A|<\infty$.

This method is maybe easier too.

Hint: choose a cover of $A$ by intervals whose length-sum is finite. You can extract large finite subcovers from this which cover "most" of $A$ and you can relate that to $A\cap(-t,t)$ for large $t$. The 'error' term can be made arbitrarily small; think about the length-sum of the intervals not in your chosen finite subcover.

Fix a covering $(I_j)_{j\in\Bbb N}$ of $A$ by intervals where we have $\sum_j\ell(I_j)<|A|+1$. Fix $\epsilon>0$. Because the series is convergent I know for some $j_0\in\Bbb N$, we have $\sum_{j>j_0}\ell(I_j)<\epsilon$. I can find $t_0>0$ so large (why?) that $\bigcup_{j=1}^{j_0}I_j\subseteq(-t,t)$ whenever $t>t_0$. I know that, if $t>t_0$: $$|A\cap(-t,t)^c|\le\sum_{j>j_0}|I_j|<\epsilon$$Which allows me to conclude, given the result $|A|=|A\cap J|+|A\cap J^c|$ for any interval $J$ (this is basically the statement that intervals are Lebesgue measurable).