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I have just encountered the notion of a measure space. I am trying to solve the following problem but I don't really know how to start it:

Let $A_n \in \frak{X}$ where $(X, \frak{X}, \mu)$ is a measure space. Show $\limsup_{n \to \infty} \chi_{A_n}(x) = \chi_A(x)$, where $$A = \bigcap_{m=1}^{\infty} \bigcup_{n \geq m}A_n$$

Further, if $$\sum_{n=1}^{\infty}\mu(A_n) < \infty$$ show $\mu(A) = 0$

Obviously for the first part I need to somehow use the properties of the $\sigma$-algebra, but I don't see where they come into play. Also, I'm not sure whether the two parts of the question are related or not. Using the definition of $\limsup$: $$\limsup \chi_{A_n}(x) = \lim_{m \to \infty} \sup\{\chi_{A_n}(x) | n \geq m \}$$ $$ = \lim_{m \to \infty} \chi_{B_m}(x),$$

where $B_m = \bigcup_{n \geq m}A_n$.

I'm not sure what to do from here. Is it clear that $\lim_{m \to \infty} \chi_{B_m}(x) = \chi_A(x)$?

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2 Answers 2

Showing that $\lim \chi_{B_m}(x) = \chi_A(x)$ shouldn't be too hard. First suppose $x \in A$ and show that $\chi_{B_m}(x) \to 1$, using the definition of limit. Then handle the case $x \notin A$. This doesn't actually have anything to do with $\sigma$-algebras, just elementary set theory.

The second part is sometimes called the Borel-Cantelli lemma. Hints for that:

  1. $\mu(\bigcup C_n) \le \sum \mu(C_n)$, whether or not the $C_n$ are disjoint;

  2. What does it mean for an infinite sum to be finite?

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if $x \in A$, then $x \in B_m\ \forall m \ge 1$, then $$1 = \chi_A(x)= \lim_{m} \chi_{B_m}= \limsup_m \chi_{A_n(x)}.$$ If $x \notin A$, then $x \in B_m$ for some $m$ and $x \notin B_n, \forall n\ge m$. Hence $$0 = \chi_{A}(x) = \lim_{m} \chi_{B_m}= \limsup_m \chi_{A_n(x)}$$ Now, if $\sum_n \mu (A_n) < \infty$, given $\varepsilon > 0$ chose $N$ such that $\sum_{n \ge N} \mu (A_n) < \varepsilon $. Then, $$\mu(A) \le\sum_{n \ge N} \mu (A_n) < \varepsilon.$$

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