Proving m(X)=0: Sequence of Measurable Sets

[e(ho0n3]

Homework Statement

Let $(X_n)$ be a sequence of measurable subsets of $\mathbb R$ such that

$$\sum_{i=1}^\infty m(X_i) < \infty$$

Define

$$X = \bigcap_{i=1}^\infty \left( \bigcup_{j=i}^\infty X_j \right)$$

Prove that m(X) = 0.

Homework Equations

Theorem. Let $(E_n)$ be a sequence of measurable sets such that $E_{n+1} \subseteq E_n$ and $m(E_1) < \infty$. Then

$$m\left(\bigcap_{i=1}^\infty E_i \right) = \lim_{i \to \infty} m(E_i)$$

The Attempt at a Solution

Define $E_i = \bigcup\limits_{j=i}^\infty X_j$. Then by the aforementioned theorem,

$$m(X) = \lim_{i \to \infty} m(E_i)$$

My only problem is showing that the limit is in fact 0. I haven't used that $\sum m(X_i) < \infty$. Any tips?

 

[johnson12]

you can say that m(X) <= m(E_{i}) for each i, and
m(E_{i}) = lim m(X_{j}) = 0 since the sum was finite.

 

[e(ho0n3]

I don't understand why $m(E_i) = \lim m(X_j)$. We have that

$$E_i = \bigcup_{j=i}^\infty X_j$$

so

$$m(E_i) \le \sum_{j=i}^\infty m(X_j)$$

I do agree that $\lim m(X_j) = 0$.

 

[xaos]

are these intervals strictly nested or can there be a smallest interval?

[edit] i need clarifying: what exactly is INT(UNION(X_i)) with two indexes i and j?

 

[johnson12]

sorry that's only true if the E_{i} where increasing, but

$$lim_{i\rightarrow\infty}\left(\sum^{\infty}_{j=i}m(X_{j})\right) =lim_{i\rightarrow\infty}m(X_{i})$$.

recall that if an infinite series converges you can make the remainder sum arbitrarily small.

 

[e(ho0n3]

You're right. That didn't occur to me. Thanks for the tip.