$\int\sum|f_n-f| \, d\mu< \infty \implies \mu( \bigcup\{\omega: |f_n(\omega) - f(\omega)| > \varepsilon\}) < \varepsilon$

Question

Prove that $\sum_{n=1}^\infty\int_\Omega|f_n-f|d\mu<\infty$ implies $$\forall \varepsilon > 0\; \exists N:\; \mu \left( \bigcup_{n=N}^\infty \{\omega: |f_n(\omega) - f(\omega)| > \varepsilon\}\right) < \varepsilon.$$

Could someone help me out? How should I start proving this?

Answer 1

Hint: Use the Markov inequality to show that

$$\begin{align*} \mu \left( \bigcup_{n \geq N} \{|f_n-f| > \epsilon\} \right) \leq \sum_{n \geq N} \mu(|f_n-f|>\epsilon) &\leq \frac{1}{\epsilon} \sum_{n \geq N} \int |f_n-f| \, d\mu . \end{align*}$$

Since the series $\sum_{n \geq 0} \int |f_n-f| \, d\mu$ converges, the right-hand side is less then $\epsilon$ for $N=N(\epsilon)$ sufficiently large.