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I am trying to prove the following:

Let $(A_n)_{n \geq 1}$ be a sequence of measurable sets, then

(i) $|\lim \inf_{n \to \infty} A_n| \leq \lim \inf_{n \to \infty} |A_n|$

(ii) If there is $n \in \mathbb N$, $|\bigcup_{k=n}^{\infty} A_k|< \infty$, then $|\lim_{n \to \infty} \sup A_n| \geq \lim_{n \to \infty} \sup |A_n|$.

I got stuck trying to show both inequalities.

In (i), on one hand we have

$$|\lim \inf_{n \to \infty} A_n|=|\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}A_k|$$$$\leq \sum_{n=1}^{\infty} |\bigcap_{k=n}^{\infty}A_k|$$$$\leq\sum_{n=1}^{\infty}\sup_{k \leq n} |A_k|$$

In (ii), we have $$\lim_{n \to \infty} \sup |A_n|=\inf_{n \geq 0} \sup_{k \geq n}|A_k|$$$$\leq \sup_{k \geq n}|A_k| \space \forall n$$

I would appreciate some help where I got stuck, thanks in advance.

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1 Answer 1

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In your work for part $(i)$, the expression of the second line can be

$$\lim_{n\to \infty} \left|\bigcap_{k = n}^\infty A_k\right|$$

since the sequence $\{\cap_{k \ge n} A_k\}_{n = 1}^\infty$ increases to $\cup_{n\ge 1} \cap_{k\ge n} A_k$. Since, given $n$,

$$\bigcap_{k \ge n} A_k \subseteq A_m \quad \text{for all } \quad m \ge n,$$

we have

$$\left|\bigcap_{k \ge n} A_k\right| \le |A_m| \quad \text{for all}\quad m \ge n.$$

Hence

$$\left|\bigcap_{k\ge n} A_k\right| \le \inf_{m\ge n} |A_m|$$

for all $n\in \Bbb N$. Now

$$|\liminf_{n\to \infty} A_n| = \lim_{n\to \infty} \left|\bigcap_{k\ge n} A_k\right| \le \lim_{n\to \infty} \inf_{k\ge n} |A_k| = \liminf_{n\to \infty} |A_k|.$$

Part $(ii)$ follows from part $(i)$. Take $X = \cup_{n = 1}^\infty A_n$, and note $|X| < \infty$ since for some $N$, $|\cup_{n \ge N} A_n| < \infty$. Consider the sets $B_n = X\setminus A_n$, and use part $(i)$ to deduce $|\limsup A_n| \ge \limsup |A_n|$.

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