Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\{f_n\}_{n\in \mathbb{N}}$ be a sequence of measurable functions on a measure space and $f$ measurable.

For $c_n>0$ such that either $\lim_{n\to \infty}c_n=0$, or $c_n\geq c>0$ for all $n$, and measurable sets $E_n$ with $m(E_n)>0$ consider the sequence $f_n(x):=c_n\mathcal{X}_{E_n}(x).$ Then $f_n$ converges in measure to zero, iff $c_n\to 0$, or $m(E_n)\to 0$ as $n\to \infty$.

My approach:

($\Rightarrow$) Suppose that $f_n\to 0$ in measure. For all $\epsilon >0$, let $E:=\{x: |f_n(x)|\geq \epsilon\}.$ I feel that it is obvious, but here is my reasoning anyway: as $n \to \infty$, $\lim_{n\to\infty}f_n(x)=\lim_{n\to\infty}c_n\mathcal{X}_{E_n}(x)$, now if $x\in E_n$ then clearly $m(E_n)\to 0$ as $n\to \infty$ and if $x\notin E_n$ then by assumption we will have $c_n\to 0.$

($\Leftarrow$) I just realized this direction is the obvious one! Isn't it? If $c_n\to 0$ or $m(E_n)\to 0$ as $n\to \infty$ then hmmm... I need help!

share|cite|improve this question
up vote 1 down vote accepted

A quite detailed explanation follows. If you have any questions, do not hesitate to ask me. By definition a function converges in measure to zero iff for every $\epsilon>0$ $$m(x\in X : |f_n(x)|>\epsilon) \rightarrow 0 $$ as $n \rightarrow 0 $. Your function is positive for all $x\in X$. Hence you may "neglect" the absolute value signs.

Let $\epsilon>0$ be given. If $c_n \rightarrow 0$ as $n \rightarrow \infty$, then there is a $N \in \mathbb{N}$ such that $c_n<\epsilon$ for $n\geq N$. Hence if $n\geq N$ then $m(x\in X : f_n(x)>\epsilon) = 0 $. If on the other hand $m(E_n)\rightarrow 0$ as $n \rightarrow \infty$, choosing $N \in \mathbb {N}$ such that $m(E_n)<\epsilon$ for $n\geq N$ we have: $$m(x\in X : f_n(x)>\epsilon)\leq m(E_n)<\epsilon{}$$ for $n\geq N$

For the other direction assume that $f_n(x)$ converges in measure to zero. If $c_n$ does not converge to zero, then by assumption in the exercise $c_n\geq c>0$ all $n$. Choosing $\epsilon<c$, we get $m(x\in X : f_n(x)>\epsilon)=m(E_n) $ all $n$. This is a contradiction unless $m(E_n)\rightarrow 0$ as $n\rightarrow{} \infty$. Lastly assume that $m(E_n)$ does not converge to zero. Then there is a $\delta>0$ such that for all $N$ there is a $n\geq N$ such that $m(E_n)>\delta$. If $c_n\geq c>\epsilon>0$ all $n \in \mathbb{N}$, then $m(x\in X:f_n(x)>\epsilon)=m(E_n)$ all $n\in \mathbb{N}$, and hence for every $N \in \mathbb{N}$ there is a $n\geq N$, $m(x\in X:f_n(x)>\epsilon)>\delta$. Contradicting the convergence in measure of the $f_n$:s.

share|cite|improve this answer

I'll try to give a hint:

concerning $(\Rightarrow)$: So you suppose $f_n \to 0$ in measure. Your task is to prove $c_n \to 0$ or $m(E_n) \to 0$. You define a set $E$ (depending on some $n$, which?, and $\epsilon$) and don't use it anymore. I'd argue as follows: So we have to prove $c_n \to 0$ or $m(E_n) \to 0$. Supposing that $c_n \not\to 0$, we have by assumption $c_n \ge c > 0$. So $f_n \ge c\chi_{E_n}$, but $f_n \to 0$ in measure. We have that for every $\epsilon > 0$ $m(|f_n| > \epsilon) \to 0$. What $\epsilon$ may help (compared to $c$)? Try to picture the set $\{|f_n| > \epsilon\}$ for different $\epsilon$.

concering $(\Leftarrow)$: As you say, this direction is more obvious (IMO). Consider the cases $c_n \to 0$ and $c_n \ge c > 0$ and $m(E_n)\to 0$ separately. If $c_n \to 0$ we have $f_n \to 0$ uniformly, so ... otherwise if $c_n$ is bounded below, you can use your picture of $\{|f_n| > \epsilon\}$ from the other direction. Now use $m(E_n) \to 0$.

Hope that helps.

share|cite|improve this answer
Did you mean to say suppose $c_n$ does not converge to $0$ in your hint for forward direction? – Lyapunov Mar 26 '12 at 9:49
Yes. We have to prove $c_n \to 0$ or $m(E_n) \to 0$. We consider to cases: If $c_n \to 0$ then we are done, otherwise $c_n \not\to 0$ and we can continue as above. – martini Mar 26 '12 at 9:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.