# convergence in $L^p$ implies convergence in measure

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

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

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

• Perhaps I'm missing something, but shouldn't the $L^{\infty}$ case be the simplest? Since $||f_n-f||_{\infty}<\epsilon \implies \mu(|f_n-f|\geq \epsilon) = 0$. – Shalop May 14 '15 at 5:17

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.