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 $p=\infty$?
Best Answer
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.