1. ## measure space, sequence

Let $(X, \mathcal{B}, \mu)$ be a measure space. Prove that if $(A_n)_{n \geq 1}$ is a sequence in $\mathcal{B}$ with $\sum_{n=1}^{\infty} \mu(A_n)< \infty$ then there exists a set $E$ with $\mu(E)=0$ such that if $x \not \in E$ then $x$ belongs to at most finitely many of the $A_n$'s.

Here, I really don't see how to answer this question. I am really struggling with this one. Any hints on where to start would be appreciated.

2. Originally Posted by poincare4223
Let $(X, \mathcal{B}, \mu)$ be a measure space. Prove that if $(A_n)_{n \geq 1}$ is a sequence in $\mathcal{B}$ with $\sum_{n=1}^{\infty} \mu(A_n)< \infty$ then there exists a set $E$ with $\mu(E)=0$ such that if $x \not \in E$ then $x$ belongs to at most finitely many of the $A_n$'s.

Here, I really don't see how to answer this question. I am really struggling with this one. Any hints on where to start would be appreciated.
I don't know if this will ruin it for you but the result is Borel-Cantelli lemma. The set E is given by $E=\limsup A_n=\bigcap_{n=1}^\infty \bigcup_{m=n}^\infty A_n$. The proof comes from the observation that $\bigcup_{m=n}^\infty A_n$ is increasing on n, so $\bigcap_{n=1}^N\bigcup_{m=n}^\infty A_n=\bigcup_{m=N}^\infty A_n$.

Use subaditivity on $\mu(\bigcup_{m=N}^\infty A_n)$, then use monotone convergence to get an inequality for the limit.

You may want to recall that the tail of a convergent sum converges to zero.