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Let $(X,M,\mu)$ be a measure space and suppose that $\{E_n\}_{1}^{\infty}$ is a sequence in $M$ with the property that $$\lim_{n\to\infty}\mu(X\setminus E_n)= 0$$

Let $G$ be the set of $x$'s that belong to only finitely many of the sets $E_n$. Show that $G\in M$ and $\mu(G) =0$

I am not exactly sure how to proceed with this, any suggestions is greatly appreciated.

2 Answers 2

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If $x \in G$, then $x$ is in only finitely many $E_k$'s. Now if for all $n$, there exists $k\ge n$ such that $x\in E_k$, then for $n=1$, there exists $k_1\ge 1$ such that $x \in E_{k_1}$, and for $n=k_1 + 1$, there exists $k_2 \ge n > k_1$, such that $x\in E_{k_2}$. Continuing in this manner, we can construct a strictly increasing sequence $(k_t)_t$ such that $x\in E_{k_t}$ for all $t \in \Bbb N$, contradicting the fact that $x$ is in only finitely many $E_k$'s. Hence there exists $n \in \Bbb N$ such that for all $k \ge n$, $x \notin E_k$. i.e. $x \in \bigcup_{n \in \Bbb N} \bigcap_{k\ge n} (X \setminus E_k)$. And if $x\in \bigcup_{n \in \Bbb N} \bigcap_{k\ge n} (X \setminus E_k)$, then there exists $n \in \Bbb N$ such that for all $k\ge n$, $x \notin E_k$, hence $x$ is in at most $E_1,\ldots,E_k$, i.e. only finitely many $E_k$'s. Thus:

$$G = \bigcup_{n \in \Bbb N} \bigcap_{k\ge n} (X \setminus E_k)$$

From which it easily follows that $G \in M$. Now we have for each $n$:

$$\mu\left( \bigcap_{k \ge n} (X \setminus E_k) \right) \le \mu(X \setminus E_m), \forall m \ge n$$

hence:

$$\mu\left( \bigcap_{k \ge n} (X \setminus E_k) \right) \le \lim_{m \to \infty} \mu(X \setminus E_m) =0$$

$$\therefore \mu (G) \le \sum_{n\in \Bbb N} \mu\left( \bigcap_{k \ge n} (X \setminus E_k) \right) = 0$$

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    The last equality isn't correct as only the limit of $\mu(X\setminus E_n)$ is zero. But anyway $$\mu\left(\bigcap_{k\ge n}(X\setminus E_k)\right)=0$$ because it's less or equal to $\mu(X\setminus E_k)$ for any $k\ge n$. – 2017-01-07
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    @SergeiGolovan thank you for the correction. – 2017-01-07
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    I am still a bit lost, could you add more details in? – 2017-01-07
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    @Wolfy better now? Where exactly do you want more details? – 2017-01-07
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    Could you do the prove that part? for $G = ...$? – 2017-01-07
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    @Wolfy it's a good exercise. Did you really try proving it? – 2017-01-07
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    I did, and I got stuck. I just want to see how you did it. – 2017-01-07
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    @Wolfy I added a proof. – 2017-01-07
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Hint : suppose that $\mu(G) >0$ and then, because G is the set of x that belong to only finitely many $E_k$, exist $n_0$such that $G\subset X-E_k$for all $k>n_0$

Why this is an absurd ?