Show that $f_n\overset{L_p}{\longrightarrow}f\Rightarrow f_n\overset{\mu}{\longrightarrow}f$

Question

$f_n$ is a sequence of functions in $L^p(\mu)$. Show that if $f_n\overset{L_p}{\longrightarrow}f$ for some $0<p<\infty$, then $f_n\overset{\mu}{\longrightarrow}f$.

Suppose $f_n$ converges to $f$ in $L^p(\mu)$ for some $p>0$. Then, $||f_n-f||_p\rightarrow0$ as $\rightarrow\infty$. This implies that $||f_n-f||_p^p\rightarrow0$ as $n\rightarrow\infty$ since $p>0$. For any $\epsilon>0$, by Markov's inequality we have

$$\mu(\{x\in X:|f_n(x)-f(x)|>\epsilon\})\leq \frac{1}{\epsilon^p} ||f_n-f||_p^p\to 0 \text{ as } n\to\infty.$$

Hence,

$$\mu(\{x\in X:|f_n(x)-f(x)|>\epsilon\})\to 0 \text{ as } n\to\infty.$$

Therefore, $f_n$ converges to $f$ in measure.

Is the attempt correct? Am I going wrong somewhere? If yes, how else should I prove this? Another doubt I had was, if instead of going with $||f_n-f||_p$ converging to $0$, I follow the usual step of considering for any $\epsilon$, an $N$ such that for all $n\ge N$, it is less than $\epsilon$, I get stuck at $$\mu(\{x\in X:|f_n(x)-f(x)|>\epsilon\})\leq \frac{1}{\epsilon^p} ||f_n-f||_p^p\le 1 \text{ as } n\to\infty.$$ This is why I was confused.

Answer 1

Your attempt using Markov's Inequality is correct. However, you are confusing the definition of convergence in measure. The $\epsilon$ is fixed and you don't need to (should not) choose $N$ depending on $\epsilon$.

Instead , this is how you should unpack the definition of convergence in Measure.

First fix an $\epsilon>0$.

Then what the definition is telling you is that for each $\delta>0$ (not necessarily depending on $\epsilon$) , you can find $N$ depending on $\delta$ such that $\mu(|f_{n}-f|>\epsilon)<\delta\,,\forall\, n\geq N$ . This is what is meant by $\lim_{n\to\infty}\mu(|f_{n}-f|>\epsilon)\to 0$ and you should be able to do this for all fixed $\epsilon>0$ .

If it is the $\epsilon$ symbol that is confusing you, then you may as well show that for each fixed positive real number $a$ , you have $\mu(|f_{n}-f|>a)\to 0$ .

What the Markov Inequality is telling you is that

$$\mu(|f_{n}-f|>a)=\mu(|f_{n}-f|^{p}>a^{p})\leq \frac{1}{a^{p}}\int_{X}|f_{n}-f|^{p}\,d\mu$$.

Now as $||f_{n}-f||_{p}\to 0$ , you have $\int_{X}|f_{n}-f|^{p}\to 0$ and hence for each $\delta>0$ , there exists $N$ such that $\int_{X}|f_{n}-f|^{p}< a^{p}\delta$

Hence , for all $n\geq N$ , you have $\mu(|f_{n}-f|>a)<\delta$ . This shows that $$\lim_{n\to\infty}\mu(|f_{n}-f|>a)\to 0$$