# 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 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 $$p=\infty$$?

• Perhaps I'm missing something, but shouldn't the $L^{\infty}$ case be the simplest? Since $||f_n-f||_{\infty}<\epsilon \implies \mu(|f_n-f|\geq \epsilon) = 0$. Commented May 14, 2015 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.
• A quick question: Tchebychev's inequality can be applied also to measures $\mu$ which are not finite, right? Commented Apr 3, 2021 at 11:28
• @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). Commented Apr 3, 2021 at 17:27