Sequence of measurable functions converges to 0 if $\lim_{n \to \infty} \int f_n d \mu \to 0$

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?

2 Answers

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.

• Could you elaborate why is the inequality true? – Никита Васильев Jan 12 at 11:51
• @math_beginner you can find it here en.wikipedia.org/wiki/Markov%27s_inequality – Yanko Jan 12 at 11:51
• @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. – 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 $$f_n\not\rightarrow 0$$.