Convergence in $L_1$ implies convergence in measure $f_n \xrightarrow[]{L_1}f \Rightarrow f_n \xrightarrow[]{\mu}f$.

Question

Let $(X, \Sigma, \mu)$ be a measure space. $\mu(X) \le \infty$. Let $f$ and $f_n$ are integrable functions. Prove that $f_n \xrightarrow[]{L_1}f \Rightarrow f_n \xrightarrow[]{\mu}f$.

$f_n \xrightarrow[]{L_1}f := \forall m>0 ,\exists N, \forall n (n > N \Rightarrow \int_X|f_n-f|\mathrm{d}\mu < m)$.

Proof: by definition $$\neg(f_n \xrightarrow[]{\mu}f):=\exists\epsilon>0, \exists \delta>0, \forall N, \exists n(n>N \land \mu( \{x : |f_n - f| \geq \epsilon\}) \geq \delta)$$ Let $m=\epsilon\delta$. Then $$\forall N, \exists n (n>N \land\int_X |f_n - f| \mathrm{d}\mu \geq m)$$ Thus $\neg (f_n \xrightarrow[]{\mu}f) \Rightarrow \neg (f_n \xrightarrow[]{L_1}f)$. And therefore $f_n \xrightarrow[]{L_1}f \Rightarrow f_n \xrightarrow[]{\mu}f$.

Is everything correct? Thanks!

Answer 1

It is perfectly correct.

A bit shorter: assume $f \xrightarrow{L^1} f$ and let $\varepsilon > 0$.

By Chebyshev's inequality we have:

$$0 \le \mu(\left\{x \in X : |f_n - f| \ge \varepsilon\right\}) \le \frac{1}{\varepsilon}\int\limits_{\left\{x \in X : |f_n - f| \ge \varepsilon\right\}}|f-f_n|\,d\mu \le \frac1\varepsilon\int_X |f-f_n|\,d\mu \xrightarrow{n\to\infty} 0$$

Hence, the squeeze theorem implies $\mu(\{x \in X : |f_n - f| \ge \varepsilon\}) \xrightarrow{n\to\infty} 0$.

We conclude $f \xrightarrow{\mu} f$.