$|f_n|\le M$ and $f_n\to f$ in measure implies $|f|\le M$ a.e.?

Question

Is it true that $|f_n|\le M$ and $f_n\to f$ in measure implies $|f|\le M$ a.e.? It appears so to me, as justified below. I'd appreciate it if someone can confirm or refute it for me.

Let $A\triangleq \{x:|f(x)|>M\}$. We want to show $\mu(A)=0$. Let $A_k\triangleq \{x:|f(x)|>M+\frac{1}{k}\}, k\in\mathbb N.$ Then $A=\bigcup_{k=1}^\infty A_k$, and $A_{k}\subset A_{k+1}$. Therefore, $\mu(A)=\lim_{k\to \infty}\mu(A_k).$

Now consider $E_n\triangleq\{x:|f_n(x)-f(x)|>\frac{1}{k}\}$. Clearly $A_k \subset E_n$, so $\mu(A_k)\le \mu(E_n)$ for any $n$. But $\mu(E_n)\to 0$, since $f_n\to f$ in measure. So $\mu(A_k)=0$ for any $k$, and hence $\mu(A)=0.$

Is this proof correct? Are there flaws or simpler proof? Thanks a lot!

A sketchy idea: convergence in measure implies convergence a.e. of a subsequence. Restricting yourself to this subsequence $(f_{\varphi(n)})n$, you have for almost every $x$ that $$M\geq \lvert f{\varphi(n)}(x)\rvert \xrightarrow[n\to\infty]{} \lvert f(x)\rvert$$ giving the conclusion. (I may have missed something.) — Clement C., Aug 24 '17 at 16:13
When I saw the title of your question, I wanted to post the same proof, but that's not necessary, yours is fine. — , Aug 24 '17 at 16:42

score 6 · Accepted Answer · answered Aug 24 '17 at 16:49

6

A simpler argument:

Recall that convergence in measure implies convergence a.e. of a subsequence. Restricting yourself to this subsequence $(f_{\varphi(n)})_n$, you have for almost every $x$ that $$ M\geq \lvert f_{\varphi(n)}(x)\rvert \xrightarrow[n\to\infty]{} \lvert f(x)\rvert $$ giving the conclusion.

answered Aug 24 '17 at 16:49

Clement C.

67,323

NB: I may have missed something. Feel free to point out any gap in the above argument, if any. – Clement C. Aug 24 '17 at 16:49
Thank you for the suggestion. Do we need a finite measure space for the a.e. convergence of a subsequence to hold? c.f. https://math.stackexchange.com/q/1006091/235690 – syeh_106 Aug 25 '17 at 08:10
1

@syeh_106 Not necessarily -- see e.g. Theorem 0.3 of these lecture notes. – Clement C. Aug 25 '17 at 14:05

score 3 · Answer 2 · answered Aug 24 '17 at 16:47

Let $\epsilon>0$ and consider

$A_\epsilon = \{ x | |f(x)| > M + \epsilon\}$.

Let $\delta>0$ and choose $N$ such that for $n \ge N$ we have $\mu \{ x | |f(x)-f_n(x)| > \epsilon \} < \delta$.

Note that if $|f(x)| > M+\epsilon$ then we have $|f(x)-f_n(x)| > \epsilon $ and so $\mu A_\epsilon < \delta$. Since $\delta$ was arbitrary, we have $\mu A_\epsilon = 0$.

It follows that $|f(x)| \le M$ for ae. $x$.

$|f_n|\le M$ and $f_n\to f$ in measure implies $|f|\le M$ a.e.?

2 Answers2