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

I don't have much experience with measure theory, so I want to make sure that I'm not making any bad mistakes. I also want to be sure that the theorem is true so I can use it.

Theorem: Let $\{u_i\}$ be a Cauchy sequence in $L^p(U)$. Then for all $\epsilon>0$, there exists an $N$ such that $|u_i(y)-u_j(y)|<\epsilon$ almost everywhere in $U$ for all $i,j>N$.

Proof: Suppose not. Then there exists an $\epsilon$ such that for all $N$ there is an $i,j>N$ such that $|u_i(y)-u_j(y)|\ge \epsilon$ for $y\in S \subset U$ with $\mu(S)>0$.

Since $\{u_i\}$ are Cauchy in $L^p$ we have $\int_U |u_i(y)-u_j(y)|^p < |S|\epsilon^p$ for $i,j>M$. But $|S|\epsilon^p = \int_S |u_i(y)-u_j(y)|^p \le \int_U |u_i(y)-u_j(y)|^p$, a contradiction.

Thank you in advance.

share|cite|improve this question
up vote 3 down vote accepted

Your theorem is wrong (you proved convergence even in $L^\infty(U)$). The problem in your proof is, that your set $S$ (and, hence, it's measure) depends on $i,j$. Therefore, you can't choose $M$ as desired.

share|cite|improve this answer
Very good, thank-you! – StuartHa Apr 18 '13 at 21:05

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.