Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

I am doing some problems off of Terence Tao's blog to study for a measure theory course taught based on Royden and focusing on real valued functions, and am stuck on this problem. Have tried some kinds of countable union type arguments but am not getting very far. Any help is much appreciated.

Using the Lebesgue measure, $\{f_{n}\}$ is a sequence of measurable functions, $f_{n}: E \to \mathbb{R}$. Suppose every subsequence $\{f_{n_{k}}\}$ has a subsequence $\{f_{n_{k_{i}}}\}$ which converges almost uniformly to a measurable function $f$. Then we claim $\{f_{n}\}$ converges in measure to $f$.

share|cite|improve this question
up vote 7 down vote accepted

If $\{f_n\}$ does not converge to $f$ in measure, then there is a $\delta>0$ such that $$ \lim_{n\rightarrow\infty} \mu( \{x\in\Bbb R : |f(x)-f_n(x)|\ge \delta\}\ne 0. $$ Then, one can select an $\alpha>0$ and an increasing sequence of integers $\{n_k\}$ such that $$ \mu( \{x\in\Bbb R : |f(x)-f_{n_k}(x)|\ge\delta \}>\alpha,\quad\text{for all}\ k. $$

No subsequence of $\{f_{n_k}\}$ can converge almost uniformly to $f$.

share|cite|improve this answer

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.