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?


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.

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  • $\begingroup$ Could you elaborate why is the inequality true? $\endgroup$ – Никита Васильев Jan 12 at 11:51
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    $\begingroup$ @math_beginner you can find it here en.wikipedia.org/wiki/Markov%27s_inequality $\endgroup$ – Yanko Jan 12 at 11:51
  • $\begingroup$ @Yanko Thank you for the link! In my country I have no access to wiki. Your link will make my answer more readable to the OP. $\endgroup$ – Feng Shao Jan 12 at 11:55

You already have an answer so let me just note that your conclusion that $f_n(x)\rightarrow 0$ for all $x\in X$ is wrong.

To see this, let $f_n:[-1,1]\rightarrow\mathbb{R}$ be the function whose graph is a triangle which begins at $(-\frac{1}{n^2},0)$ increase linearily to $(0,1)$ and then decrease linearily to $(0,\frac{1}{n^2})$ (Draw).

Therefore the integral of $f_n$ is the area of the triangle which is $\frac{1}{n^2}$, hence goes to zero as $n$ goes to infinitey. However $f_n(0)=1$ for all $n$ and so $f_n\not\rightarrow 0$.

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