Let $E$ be a measurable set with $m(E)<\infty$, $(f_n)$ a sequence of real-valued measurable functions on $E$ and $f$ be a real_valued measurable function on $E$. Suppose $(f_n)$ converges to $f$ pointwise a.e. in $E$. Now it is required to prove that $(f_n)$ converges to $f$ in measure. Here is my attempt.

Assume $(f_n)$ does not converge to $f$ in measure. Then there exists $\eta_0>0$ such that $\lim_{n\to\infty}m(\{x\in E:|f_n(x)-f(x)|\geq\eta_0\})\neq0$. Therefore there exists $\epsilon_0>0$ such that for all $n\in\mathbb{N},$ there exists $N>n$ such that $m(\{x\in E:|f_N(x)-f(x)|\geq\eta_0\})\geq\epsilon_0$. Then there is a subsequence $(f_{n_k})$ of $(f_n)$ such that $m(\{x\in E:|f_{n_k}(x)-f(x)|\geq\eta_0\})\geq\epsilon_0$ for each $k\in\mathbb{N}$. Now $(f_{n_k})$ is a subsequence of $(f_n)$ that does not converge to $f$ on a set of positive measure; contradiction.

Is this proof alright? The stipulation $m(E)<\infty$ worries me. Is there something wrong? Thanks.

2nd attempt-added later:

Suppose $(f_n)$ converges to $f$ pointwise a.e. in $E$. Let $\epsilon>0$. Then by Egoroff's theorem, there is a measurable set $F\subseteq E$ such that $m(F)<\epsilon$ and $(f_n)$ converges to $f$ uniformly on $E\setminus F$. Now $(|f_n-f|)$ converges to $0$ uniformly on $E\setminus F$. Thus there exists $N\in\mathbb{N}$ such that for each $n>N$ and $x\in E\setminus F$, $|f_n(x)-f(x)|<\epsilon$. Let $n>N$. Then $m(\{x\in E:|f_n(x)-f(x)|\geq \epsilon\})=m(\{x\in E\setminus F:|f_n(x)-f(x)|\geq \epsilon\})+m(\{x\in F:|f_n(x)-f(x)|\geq \epsilon\})<0+\epsilon=\epsilon.$

@Ian Is this argument alright?

  • $\begingroup$ Yes, this is correct now. $\endgroup$
    – Ian
    Jun 6, 2017 at 20:13

2 Answers 2


You need that stipulation because a sequence that converges a.e. by "moving mass to infinity" does not converge in measure. A concrete example of that is $f_n(x)=\chi_{[n,n+1]}(x)$ on $\mathbb{R}$ with the Lebesgue measure.

In your particular proof, first there is one very minor error. Namely, the negation of "the measures go to zero" is "the measures don't go to zero", not "the measures go to something nonzero". But this error is not a big deal; you can get your "bad" subsequence $f_{n_k}$ regardless. The bigger problem is the last sentence where you claim that there is a set of positive measure where $f_{n_k}$ does not converge. This is not true: all that you have is a sequence of "bad" sets $A_k$ whose measures are bounded below away from zero, but they may not have points in common (in my example, they don't).

Doing this correctly will require a totally different approach, one that exploits the finite measure assumption at its center. There's a certain "named" theorem that will help you a great deal...

  • $\begingroup$ Is that the Egoroff's theorem? $\endgroup$
    – Janitha357
    Jun 6, 2017 at 19:46
  • $\begingroup$ @Janitha357 Yes. $\endgroup$
    – Ian
    Jun 6, 2017 at 19:57
  • $\begingroup$ I edited my post. Could you please check? $\endgroup$
    – Janitha357
    Jun 6, 2017 at 20:11

I suggest not using contradiction when possible.

Fix $\epsilon>0$. For almost every $x$, there exists $n$ such that $|f_m (x) -f(x)|\le\epsilon$ for all $m\ge n$, where $m$ may depend on $x$. Let $A_n = \{x: |f_m(x)-f(x)| \le\epsilon,~\mbox{for all }m \ge n\}$. Then $A_n\subset A_{n+1} \subset \dots$ and the above shows that $A_n \nearrow E$ a.e. If $|f_n(x)-f(x)|\ge \epsilon$, then clearly $x \in E-A_n$. This gives
$$m (\{x:|f_n (x)-f(x)|>\epsilon\})\le m (E-A_n) \underset{n\to\infty}{\to} 0,$$

where the limit holds because $A_n \nearrow E$ and $m(E)<\infty$.

As for your proof, there are several problems as already pointed out. The first and most minor is that you assume that the limit $\lim_{n\to\infty}m(\{x\in E:|f_n(x)-f(x)|\geq\eta_0\})\neq0$ exists. The main problem is that you did not show that there is ONE set of positive measure on which the pointwise convergence fails. What you do show is that there is a sequence of subsets $E_{n_k}$ such that on $E_{n_k}$ $|f_{n_k}(x) -f(x) |>\epsilon_0$ (this is fundamentally different than what you claim: $f_{n_k}$ does not converge on some fixed set. This you did not show).

Finally, here's a quick proof if you're familiar with integration. WLOG $f=0$. Since $\min\{|f_n|(x),\epsilon\}/\epsilon$ is $1$ when $f_n(x) \ge \epsilon$ and between $0$ and $1$ otherwise, its integral is larger than the measure of the set $\{x:|f_n|(x)>\epsilon\}$. Therefore

$$ \mu(\{x:|f_n(x)|>\epsilon\}) \le \frac{1}{\epsilon}\int \min\{|f_n|(x),\epsilon\}d m (x).$$

Now apply bounded convergence ($m(E)<\infty$) to show that the righthand side tends to zero.


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