Let $(X, \mathcal{F}, \mu)$ be a measurable space and $(f_n)$ a sequence of nonnegative measurable functions.
Prove that when
$$
\lim_{n \to \infty} \int f_n d \mu \to 0
$$
then $f_n$ converges with measure $\mu$ to 0
My attempt:
Since $\lim_{n \to \infty} \int f_n d \mu \to 0$ is stronger than Lebesgue's Dominated Convergence Theorem we can conclude that $\lim_{n \to \infty}f_n = f \equiv 0$.
We want to show that for every $\epsilon > 0$:
$$
\lim_{n \to \infty} \mu (\{x: |f_n(x)| > \epsilon \}) = 0
$$
Is it then:
$$
\lim_{n \to \infty} \mu (\{x: |f_n(x)| > \epsilon \}) = \mu (\{x: 0 > \epsilon \}) = 0
$$
If not, what is the correct proof?
Best Answer
The first sentence in your attempt has lead you to a wrong direction. Think about it: if we have $f_n\to0$, why the queation asks us to prove the much weaker convergence of $f_n$, which is the convergence in measure?
Note that $$\mu (\{x: |f_n(x)| > \epsilon \})\leq\frac{\int f_n\,d\mu}\epsilon\to0,\ \ n\to\infty.$$
The inequality I used is called Chebyshev's inequality or Markov's inequality. Thanks to @Yanko for pointing out in the comments.