Prove that if $A_n$ is a sequence of measurable sets then $\lambda(\liminf A_k) \leq \liminf\lambda(A_k)$

Question

Prove that if $A_n$ is a sequence of measurable sets then $\lambda(\liminf A_k) \leq \liminf\lambda(A_k)$

Note : Can you please also add a example of a sequence of sets such that $A_k \subset [0,1]$, $\lim\lambda(A_k) = 1$, but $\liminf A_k=\emptyset$

Answer 1

I assume $\lambda$ is a measure hence it's continuous from below and so:

$$\lambda(\liminf_{k\to\infty} A_k) = \lambda\left(\bigcup_{k=1}^\infty\bigcap_{m=k} A_m\right) = \lim_{k\to\infty} \lambda\left(\bigcap_{m=k} A_m\right)$$

Because $\lambda\left(\bigcap_{m=k} A_m\right)$ is an increasing sequence it holds $$\lim_{k\to\infty} \lambda\left(\bigcap_{m=k} A_m\right) = \sup_{k\in\Bbb N} \lambda\left(\bigcap_{m=k} A_m\right)$$

On the other hand it holds $$\bigcap_{m=k} A_m \subseteq A_l$$ for all $l\ge k$ and so $$\lambda\left(\bigcap_{m=k} A_m\right) \le \lambda(A_l)$$ for all $l \ge k$ but then also $$\lambda\left(\bigcap_{m=k} A_m\right) \le \inf_{l\ge k} \lambda(A_l)$$

All together we have: $$\lambda(\liminf_{k\to\infty} A_k) = \lambda\left(\bigcup_{k=1}^\infty\bigcap_{m=k} A_m\right) = \sup_{k\in\Bbb N} \lambda\left(\bigcap_{m=k} A_m\right) \le \sup_{k\in\Bbb N} \inf_{l\ge k} \lambda(A_l) = \liminf_{k\to\infty}\lambda(A_k)$$

Try to find an example for the strict inequality yourself by recap the proove…