Borel-Cantelli-like inequality

Question

I’m wondering if the following is true:

If $A_n \subseteq [0,1]$ satisfies $m(A_n) \geq \epsilon > 0$, then $m(\limsup A_n) > 0$.

Here $m$ is the Lebesgue measure.

Answer 1

Yes. Let $B_n = \bigcup_{k\ge n} A_k.$ Then we have $B_1 \supseteq B_{2}\supseteq\ldots$ and $$m(B_n)\ge m(A_n)\ge \epsilon$$ for all $n.$ Thus, using the downward measure continuity for finite measures, $$ m\left(\limsup_j A_j \right)=m\left(\bigcap_{n\ge 1} B_n\right) = \lim_n m(B_n)\ge \epsilon$$