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

From Rudin, Real and Complex Analysis, Chapter 6, Problem 10, 1st edition

Suppose $\mu$ is a finite positive measure on $X$, $f_{n}$ is a sequence in $L^{1}(\mu)$, $f\in L^{1}(\mu)$, $f_{n}\rightarrow f$ a.e and $$\lim_{n\rightarrow \infty}\int_{E}f_{n}d\mu=\int_{E}fd\mu$$For every measurable set $E\subset X$. Then $f_{n}$ has uniformly absolutely continuous integrals. Prove this (it is suffice to consider the case $f=0$).

My confusions are:

  1. In order to pass from $f\not=0$ to $f=0$ I need to approximate $|\int_{E}|f_{n}-f|d\mu|,\mu(E)\le \delta$ using triangle inequality. So I need to prove that $f$ is uniformly absolutely continuous first. On the other hand the above statement must hold with $f_{n}=f$ instead. Therefore I need to show for any $\epsilon$, there exist some $\delta$ such that for all measurable subset $E$ with $\mu(E)<\delta$, then $$|\int_{E} fd\mu|<\epsilon$$This statement implies the same is true with $f$ replaced by $|f|$. And if I know this holds for $|f|$ I can show the original statement. So it suffice to prove with the assumption $f$ is nonnegative. But then I get stuck. Either $f\in L^{\infty}$ and hence must be bounded, then the proof is trivial. Otherwise I do not know how control the property of $f$. Suppose there exist some $\epsilon$ such that for any $\delta$ there is some $E_{\delta}$ such that the above is not true. I do not know how to deduce a contradicting statement. I can show that $f$'s value will be as close to infinity as I wish on a small enough measurable subset (for example $\tan[x]$). But this does not bring a contradiction to the fact $f$ is absolutely integrable.

  2. Now assuming we have overcome the above difficulty somehow. Then without loss of generality I can assume $f_{n}\rightarrow 0$ pointwise, and $\int_{E}f_{n}\rightarrow 0$ for all measurable subsets of $X$. Further $f_{n}\in L^{1}(\mu)$. I still do not know how to show that $f_{n}$ are uniformly absolutely continuous. It seems I need to prove by contradiction as well. Assuming for any $M$, there is an $\epsilon$ such that for any $\delta$ there is some $\mu(E)<\delta$ and $n>M$ such that $|\int_{E}f_{n}d\mu|\ge \epsilon$. But I do not know how to take a step further from here.

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

$f \in L^1(\mu)$ means $\int_X |f| \ d\mu < \infty$. Let $g_M(x) = |f(x)|$ when $|f(x)| \ge M$, $0$ otherwise. By the Lebesgue dominated convergence theorem, $\lim_{M \to \infty} \int_X g_M(x)\ d\mu = 0$. Take $M$ such that $\int_X g_M(x)\ d\mu < \epsilon/2$. Let $B = \{x: |f(x)| \ge M\}$. Take $\delta > 0$ so that $\delta M < \epsilon /2$. For any measurable set $E$ with $\mu(E) < \delta$, $$ \int_E |f| \ d\mu \le \int_{E \cap B} |f| \ d\mu + M \mu(E \backslash B)< \epsilon$$

share|cite|improve this answer
Can you give a hint how to prove the second part? – Bombyx mori Jan 4 '13 at 2:44
I see. It suffice to use Lusin's theorem to find a subset small enough out of which $f$ is continuous. Then we can apply the hypothesis on the small set together with $f_{n}$ in $L_{1}$. Now on the large set where $f$ is continuous, the positive and negative parts of $f$ are locally fixed; so the above counterexample cannot exist by hypothesis given. This proved the statement. – Bombyx mori Jan 4 '13 at 3:09

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.