# Prove that $|A| = \lim_{t\rightarrow \infty}| A \cap (-t,t)|$ for all $A \subset \mathbb{R}$

Problem taken from the books sheldon Axler Measure , integration Real analysis

Prove that $$|A| = \lim_{t\rightarrow \infty}| A \cap (-t,t)|$$ for all $$A \subset \mathbb{R}$$

My attempt : $$\lim_{t\rightarrow \infty}| A \cap (-t,t)|=|A \cap (-\infty,\infty)|=\min| A|$$

Im newly learning measure theory

• How did you get that equality in your attempt? Seems like you assumed what you want to prove. Commented Jul 27, 2020 at 16:15
• @DonThousand i directly write the infinity in place of $t$ Commented Jul 27, 2020 at 16:17
• can't do that without justification. Commented Jul 27, 2020 at 16:17

Take a sequence $$t_n\nearrow \infty$$ (i.e., $$t_n>t_{n-1}$$) and let $$A_n:=A\cap(-t_n,t_n)$$. By monotonicity, $$\lim_{n\to\infty}|A_n|\le |A|$$ ($$\because A_n\subseteq A$$ for each $$n\ge 1$$ and $$|A_n|$$ is nondecreasing). Thus, it remains to show that $$\lim_{n\to\infty}|A_n|\ge |A|.\tag{1}\label{1}$$ Let $$B_n:=A_n\setminus A_{n-1}$$, where $$A_0\equiv \emptyset$$. Using Problem 2A.8 in Axler's book, we have \begin{align} |A_n|&=\left|\bigcup_{i=1}^{n} B_{i}\right|=\left|\bigcup_{i=1}^{n} B_{i}\cap(-t_{n-1},t_{n-1})\right|+\left|\bigcup_{i=1}^{n} B_{i}\cap (\mathbb{R}\setminus(-t_{n-1},t_{n-1}))\right| \\ &=\left|\bigcup_{i=1}^{n-1} B_{i}\right|+\left|B_n\right|=\cdots= \sum_{i=1}^n |B_i|. \end{align}

Consequently, $$\sum_{n\ge 1}|B_i|=\lim_{n\to\infty}|A_n|$$. If the sum on the RHS is infinite, the inequality $$\eqref{1}$$ is trivially satisfied. Otherwise, since $$|\cdot|$$ is subadditive, for each $$m\ge 1$$, $$|A|\le |A_m|+\sum_{n> m}|B_i|,$$ and the second term on the RHS converges to $$0$$ as $$m\to\infty$$.

• How are you using continuity of measure without $A$ being measurable? Commented Jul 27, 2020 at 16:45
• @Sorfosh You're right. I incorrectly assumed that $A$ is measurable. I'll update my answer.
– user140541
Commented Jul 27, 2020 at 17:45
• If $$A$$ is measurable then monotone convergence gives the desire result, just as in d.k.o.'s proof.

• If $$A$$ is not measurable and $$|A|^*=\infty$$, then $$\lim_{t\rightarrow\infty}|A\cap(-t,t)|^*\geq\lim_{t\rightarrow\infty}|(-t,t)|=\infty$$ where $$|\cdot|^*$$ is the outer measure induced by $$|\cdot|$$ and the collection of countable unions of intervals of the form $$(a,b]$$, $$-\infty (Caratheodory's construction)

• If $$|A|^*<\infty$$, then there is a measurable set $$B$$ such that $$A\subset B$$ and $$|B\cap E|=|A\cap E|^*$$ for every measurable set $$E$$. (This is Caratheodory's construction of measurability). Then $$\lim_{t\rightarrow\infty}|A\cap(-t,t)|^*=\lim_{t\rightarrow\infty}|B\cap(-t,t)|=|B|=|A|^*$$

Comment:

The limit $$\lim_{t\rightarrow\infty}$$ can be understood as taking a limit over any increasing sequence $$t_n\nearrow\infty$$ as $$n\rightarrow\infty$$. That is to make use of monotone convergence (or $$\sigma$$-continuity) of the measure $$|\cdot|$$.

I very slightly changed your proof and I wrote a proof for me.

Let $$f(t):=|A\cap (-t,t)|$$ for any $$t\in (0,\infty)$$.
Then, $$f(t)\leq |(-t,t)|=2t$$.
So, $$f(t)\in\mathbb{R}$$ for any $$t\in (0,\infty)$$.
If $$s, then $$f(s)\leq f(t)$$.
If $$\{f(t):t\in (0,\infty)\}$$ is not bounded above, then $$\lim_{t\to\infty} f(t)=\infty$$.
If $$\{f(t):t\in (0,\infty)\}$$ is bounded above, then $$\lim_{t\to\infty} f(t)=\sup\{f(t):t\in (0,\infty)\}\leq |A|$$.

Let $$A_0:=\emptyset$$.
Let $$A_n:=A\cap (-n,n)$$ for any $$n\in\{1,2,\dots\}$$.
Let $$B_n:=A_n\setminus A_{n-1}$$ for any $$n\in\{1,2,\dots\}$$.
Then, $$A_n=\bigcup_{k=1}^{n} B_k$$.
By Exercise 8 in Exercises 2A in the book, $$|A_n|=|A_n\cap (-(n-1),n-1)|+|A_n\cap (\mathbb{R}\setminus (-(n-1),n-1))|\\=|A_{n-1}|+|B_n|\\=|A_1|+|B_2|+\dots+|B_n|\\=|B_1|+|B_2|+\dots+|B_n|.$$

If $$\lim_{n\to\infty} f(n) = \lim_{n\to\infty} |A_n|=\infty$$, then $$|A|=\infty$$ since $$|A_n|\leq |A|$$.
In this case, $$\{f(t):t\in (0,\infty)\}$$ is not bounded above.
So, $$\lim_{t\to\infty} f(t)=\infty=|A|$$.

Suppose $$\lim_{n\to\infty} f(n) = \lim_{n\to\infty} |A_n|\in\mathbb{R}$$.
Obviously, $$A=\bigcup_{k=1}^\infty B_k$$.
By countable subadditivity of outer measure, $$|A|\leq\sum_{k=1}^\infty |B_k|=\lim_{n\to\infty} |A_n|.$$
Therefore $$\lim_{n\to\infty} f(n) = \lim_{n\to\infty} |A_n|=|A|$$.

Let $$\epsilon$$ be an arbitrary positive real number.
There exists $$N\in\{1,2,\dots,\}$$ such that $$|A|-\epsilon.
If $$N, then $$|A|-\epsilon.
Therefore $$\lim_{x\to\infty} f(x)=|A|$$.

Hint:

Prove that the limit is equal to $$\lim_{n\to \infty}|A\cap(-n,n)|$$. Now use the fact that $$\big(A\cap (-n,n)\big)$$ is an increasing sequence.