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I am trying to show that if $f_n$ converges to $f$ in $L^p(X,\mu)$ then $f_n\to f$ in measure, where $1\le p \le \infty$.

Here is my attempt for $p\ge 1$: Let $\varepsilon>0$ and define $A_{n,\varepsilon}=\lbrace x: \vert f_n(x)-f(x) \vert \ge \varepsilon\rbrace$. I want to show $\mu (A_{n,\varepsilon})\to 0$. $\Vert f_n-f \Vert_p=\left(\int _X \vert f_n-f\vert ^p\right)^{1/p}\ge \left(\int _{A_{n,\varepsilon}}\vert f_n-f\vert ^p\right)^{1/p}\ge \varepsilon \mu (A_{n,\varepsilon})^{1/p}$ so that $\mu (A_{n,\varepsilon})\le \left(\frac{\Vert f_n-f\Vert_p }{\varepsilon}\right)^{p}$ and the RHS tends to $0$ as $f_n\to f$ in the $L^p$ norm.

How can I deal with the case $p=\infty$?

Fellow InstituteOfMathophile
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cap
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    Perhaps I'm missing something, but shouldn't the $L^{\infty}$ case be the simplest? Since $||f_n-f||_{\infty}<\epsilon \implies \mu(|f_n-f|\geq \epsilon) = 0$. – shalin May 14 '15 at 05:17

1 Answers1

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The case $p = \infty$ is the simplest since $\|f_n-f\|_{\infty} < \epsilon$ means that $|f_n-f|$ is less than $\epsilon$ almost everywhere, and therefore $\mu\big(|f_n-f|\ge\epsilon\big) = 0$.

If $1\le p < \infty$, you can use Tchebychev's inequality: $$ \mu\big(|f_n-f|\ge\epsilon\big) = \mu\big(|f_n-f|^p\ge\epsilon^p\big) \le \frac{1}{\epsilon^p}\int|f_n-f|^p\,d\mu = \frac{1}{\epsilon^p}\|f_n-f\|_p^p, $$ where the last expression can be made arbitrarily small by taking $n$ sufficiently large.

Alex Ortiz
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    A quick question: Tchebychev's inequality can be applied also to measures $\mu$ which are not finite, right? – Jack London Apr 03 '21 at 11:28
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    @JackLondon: Good question, yes it can. Probably the easiest way to see that is to observe that $f/\lambda \ge 1$ on ${f \ge \lambda}$, so Tchebychev's inequality is really only a consequence of monotonicity of integration: $\mu(f\ge \lambda) = \int_{{f \ge \lambda}} 1 \le \int_{{f\ge \lambda}} (f/\lambda) \le (1/\lambda)\int f$ (if $f\ge 0$, say). – Alex Ortiz Apr 03 '21 at 17:27