# Determining if $\int f_n\to 0$ implies that $f_n\to 0$ in measure and $f_n(x)\to 0$ a.e.

If $f_n$ is a sequence of measurable functions on $(X,\mu)$ into $[0,1]$ and $\int f_n\to 0$, I am trying to prove (or disprove the following): (i) $f_n$ converges to $0$ in measure. (ii) For almost all $x$, $f_n(x)\to 0$

I believe both claims are true, and here are my attempts but I am not confident in my approach.

(i) If $f_n$ doesn't converge to $0$ in measure, then the set of all $x$ for which $\vert f_n(x)\vert >\varepsilon$ has positive measure for some $\varepsilon>0$. Since $f$ is positive, this means that the integral of $f$ over $X$ would also be positive, contradicting our assumption.

(ii) If $\lim_{n\to \infty}f_n(x)>0$ in a set of positive measure, then once again $\int f_n>0$ which contradicts out assumption.

• Having $\int f_n > 0$, even for all $n$, doesn't contradict the assumption that $\int f_n \to 0$; it could be that $\int f_n = \frac{1}{n}$ or something like that. – Nate Eldredge May 28 '15 at 1:16

Unfortunately there is one problem in your reasoning. Note that $\int f_{n}>0$ is perfectly fine and one might still have $f_{n}\to 0$ almost surely and $\int f_{n}\to 0$. Take for example $f_{n}=\chi_{[0,\frac{1}{n}]}$ for all $n$, i.e. $f_{n}=1$ on $[0,\frac{1}{n}]$ and zero otherwise. Then $\int f_{n} = \frac{1}{n}>0$ for all $n$ but $f_{n}\to 0$ almost surely and $\int f_{n} \to 0$.
• $(i)$ Think about Chebyshev's inequality.
• For $(ii)$ you may consider the following counter example. Take $(X,\mu)=([0,1],m)$ where $m$ is the Lebesgue measure on $[0,1]$, and let $I_{1}=[0,1]$. We then divide $[0,1]$ to two pieces: $I_{2}=[0,\frac{1}{2}]$ $I_{3}=[\frac{1}{2},1]$. Further, we divide $[0,1]$ to three pieces $I_{4}=[0,\frac{1}{3}]$, $I_{5}=[\frac{1}{3},\frac{2}{3}]$ and $I_{6}=[\frac{2}{3},1]$. Continue this process and obtain a sequence of intervals $\{I_{n}\}_{n=1}^{\infty}$. Then consider the sequence of functions $(f_{n})_{n=1}^{\infty}$ where $f_{n}=\chi_{I_{n}}$. Show that $f_{n}$ does not converge to $0$ for any $x\in [0,1]$, but $\int f_{n}\to 0$.