# How to prove a $f_{n}$ has uniformly continuous integrals?

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.

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$f \in L^1(\mu)$ 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 $\mu(E) < \delta$, $$\int_E |f| \ d\mu \le \int_{E \cap B} |f| \ d\mu + M \mu(E \backslash B)< \epsilon$$
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