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The problem is

Suppose that $\{f_{n}(x)\}$ is a sequence of finite measurable functions on $X$ where $\mu(X)<\infty$. Prove that $\lim\limits_{n\to\infty}f_{n}(x)=0$ a.e. iff $g_{n}\rightarrow0$ in measure, where $g_{n}(x)=\sup\limits_{k\geq n}|f_{k}(x)|$.

I believe that I have done one hand:

Since $\mu(X)<\infty$ and $f_{n}\to0$ a.e. on $X$, we have $f_{n}\to 0$ in measure. That is for every $\sigma>0$,$$\lim_{n\to\infty}\mu(\{x:|f_{n}(x)|\geq\sigma\})=0.$$But$$\{x:|g_{n}(x)|\geq\sigma\}=\bigcup_{k=n}^{\infty}\{x:|f_{n}(x)|\geq\sigma\},$$hence$$\begin{aligned} \lim_{n\to\infty}\mu(\{x:|g_{n}(x)|\geq\sigma\})&=\lim_{n\to\infty}\mu\left( \bigcup_{k=n}^{\infty}\{x:|f_{n}(x)|\geq\sigma\}\right) \\ &=\mu\left( \lim_{n\to\infty}\bigcup_{k=n}^{\infty}\{x:|f_{n}(x)|\geq\sigma\}\right) \\ &=\mu(\lim_{n\to\infty}\{x:|f_{n}(x)|\geq\sigma\})\\ &=\lim_{n\to\infty}\mu(\{x:|f_{n}(x)|\geq\sigma\})=0.\\ \end{aligned}$$This proves that $g_{n}(x)\to0$ in measure.

However, on the other hand, I only know that$$\{x:|f_{n}(x)|\geq\sigma\}\subset\{x:|g_{n}(x)|\geq\sigma\},$$which implies$$\lim_{n\to\infty}\mu(\{x:|f_{n}(x)|\geq\sigma\})\leq\lim_{n\to\infty}\mu(\{x:|g_{n}(x)|\geq\sigma\})=0,$$hence $f_{n}\to0$ in measure. But I'm confused that how to conclude $f_{n}\to0$ a.e.

Any help would be appreciated.

mio
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1 Answers1

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Since $g_n \to 0$ in measure we can find a subsequence $g_{n_i}$ that converges to $0$ a.e (look at this question).

Now, for an arbitrary positive $\epsilon $, consider the set below: $$A=\{ x \in X : \ \ |g_{n_i}|> \epsilon , \ \ for \ \ \ infinitely \ \ many \ \ i \};$$

we must have $\mu (A)=0$ because $g_{n_i} \to 0$ a.e.

On the other hand, let's consider the set below as well:

$$B=\{ x \in X : \ \ |f_n|> \epsilon, \ \ for \ \ \ infinitely \ \ many \ \ n \}.$$

It is very easy to see that $B \subseteq A$ because $g_{n_i}(x)= \sup _{k\geq n_i}|f_k(x)|$. Therefore, $\mu (B)=0$.

Finally, by this link, we conclude that $f_n \to 0$ a.e.

Reza Rajaei
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