If $m(A_n) > \delta > 0$ for each n, show $m(B) \geq \delta $

Question

From Richard Bass' Real Analysis Text I have the following problem.

Let $m$ be a Lebesgue measure. Suppose for each $n$, $A_n$ is a Lebesgue measurable subset of $[0, 1]$. Let $B$ consist of those points $x$ that is in infinitely many of the $A_n$.

If $m(A_n) > \delta > 0$ for each n, show $m(B) \geq \delta $

I know that $$B = \limsup{A_n} = \bigcap_{m=1}^\infty\bigcup_{n=m}^\infty{A_n}$$ Looking at the fact that $A_n \subseteq \cup_{n=1}^\infty{A_n}$, then $m(A_n) \leq m(\cup_{n=1}^\infty{A_n})$.

So we have $\delta < m(A_n) \leq m(\cup_{n=1}^\infty{A_n})$, but how does the intersections affect the problem?

Answer 1

For each $n$, $m(\bigcup_{k\ge n}A_k)>\delta$. Then $m(B)=\lim_{n\to\infty}m(\bigcup_{k\ge n}A_k)\ge \delta$.