# measure space, sequence

• Dec 6th 2009, 02:02 PM
poincare4223
measure space, sequence
Let $\displaystyle (X, \mathcal{B}, \mu)$ be a measure space. Prove that if $\displaystyle (A_n)_{n \geq 1}$ is a sequence in $\displaystyle \mathcal{B}$ with $\displaystyle \sum_{n=1}^{\infty} \mu(A_n)< \infty$ then there exists a set $\displaystyle E$ with $\displaystyle \mu(E)=0$ such that if $\displaystyle x \not \in E$ then $\displaystyle x$ belongs to at most finitely many of the $\displaystyle A_n$'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.
• Dec 6th 2009, 03:06 PM
Focus
Quote:

Originally Posted by poincare4223
Let $\displaystyle (X, \mathcal{B}, \mu)$ be a measure space. Prove that if $\displaystyle (A_n)_{n \geq 1}$ is a sequence in $\displaystyle \mathcal{B}$ with $\displaystyle \sum_{n=1}^{\infty} \mu(A_n)< \infty$ then there exists a set $\displaystyle E$ with $\displaystyle \mu(E)=0$ such that if $\displaystyle x \not \in E$ then $\displaystyle x$ belongs to at most finitely many of the $\displaystyle A_n$'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 $\displaystyle E=\limsup A_n=\bigcap_{n=1}^\infty \bigcup_{m=n}^\infty A_n$. The proof comes from the observation that $\displaystyle \bigcup_{m=n}^\infty A_n$ is increasing on n, so $\displaystyle \bigcap_{n=1}^N\bigcup_{m=n}^\infty A_n=\bigcup_{m=N}^\infty A_n$.

Use subaditivity on $\displaystyle \mu(\bigcup_{m=N}^\infty A_n)$, 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.