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Let $(X,M,\mu)$ be a measure space. let $f_n$ be a Cauchy sequence in $L^\infty(X)$. I want to show that there is a mesurable $E\subset X$ such that $\mu(E^c)=0$ and $\forall \epsilon \gt 0$, $|f_n-f_m| \le \epsilon$ on $E$ for some $n,m\gt N$.

Furthermore, I want to use the above to show that $L^\infty(X)$ is complete.

So I know for $n,m\gt N$ $|f_n-f_m|\le \|f_n-f_m\|\le \epsilon $ a.e. Let $$E_k = \{x : |f_k(x) \gt \|f_k\|_\infty \}\qquad E_{n,m} = \{ x : |f_n (x) -f_m(x)|\gt \|f_n-f_m\|_\infty \}$$ Set $$E^c = \bigcup_k E_k \cup \bigcup_{n,m} E_{k,m}$$ then $\mu(E^c)=0$ and so $|f_n-f_m|\leq \epsilon$ on $E$.

For the second part, since $\mathbb R$ is complete, on $E$, $f_n \to f$ uniformly.
Also, $|f(x)|\leq |f_n(x)-f(x)|+|f_n| \lt 1+ |f_n|$, so $f\in L^\infty$. I can also say that $$|f_n-f|=\lim_{n\to\infty} |f_m-f_n|\le \epsilon$$ on $E$.

How do I get that $\|f-f_n\|_\infty\to 0$? Is my approach right?

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From your question, it looks like you are trying to prove that $L^p$-convergence implies uniform convergence a.e. That's not true. – Martin Argerami Apr 21 '12 at 0:24
for $p=\infty$? – Dan Apr 21 '12 at 0:34
Is it the norm $||f||_\infty=\textrm{sup}_X f$ or just $||f||_\infty=\textrm{ess sup}_Xf$? – matgaio Apr 21 '12 at 1:39
essential supremum. – Dan Apr 21 '12 at 1:50
up vote 1 down vote accepted

You know that $f_n$ is cauchy, so choose $\epsilon>0$, and $N$ such that $||f_n-f_m|| < \epsilon$ $\forall n,m \geq N$.

You know that when $x\in E$, $|f_n(x) - f_m(x)| \leq ||f_n-f_m|| < \epsilon$, and $f_n(x) \rightarrow f(x)$. From this you can conclude that $|f_n(x) - f(x)| \leq \epsilon$ (otherwise a quick contradiction). Hence you know $\sup_{x \in E} |f_n(x) - f(x)| \leq \epsilon$.

You can define $f(x) = 0$ on $E^C$, so it is defined everywhere (not that it matters).

Consequently, $\mathbb{ess} \sup_X |f_n(x) - f(x)| \leq \epsilon$, since $\{x |\; | f_n(x) - f(x)| > \epsilon\} \subset E^C$ and $\mu E^C = 0$. In other words, $||f_n-f||_{\infty} \leq \epsilon$.

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which part of the problem did you solve? – Dan Apr 21 '12 at 3:08
You asked how to get $\|f-f_n\|_\infty\to 0$. – copper.hat Apr 21 '12 at 3:19
right. Thanks . Is everything else okay? – Dan Apr 21 '12 at 3:31
Yes. Except your spelling of measurable. – copper.hat Apr 21 '12 at 4:11
haha...that's a good one. Thank you very much. – Dan Apr 21 '12 at 4:57

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