Problem: If $f_n \to f$ in the $L^1$ norm, and $f_n \in L^1$ for each $n$, show that for every $\epsilon > 0$, there exists $\delta > 0$ such that if $m(A) < \delta$, $\int_A |f_n| < \epsilon$ for all integers $n$.

My attempt: We have already shown that if $f$ is integrable, then for each $\epsilon > 0$ there is a $\delta > 0$ such that whenever $m(A) < \delta$, $\int_A f(x) dx < \epsilon$.

To begin with, I noted that $L^1(R)$ is complete, and therefore $f \in L^1(R)$. Now, it's clear that for any $\epsilon > 0$, I can find a $\delta > 0$ such that for some $N$, we have $\int_A |f_n| < \epsilon$ if $m(A) < \delta$ and $n \ge N$. But I can't figure out how to generalize this to all $n$! A hint in the correct direction would be much appreciated.

Note: I have already tried using that

$$\int_A|f_n| \le \int_A |f_n - f| + \int_A |f|,$$

but this first estimate is where I'm running into problems, since it's only bounded by $\epsilon$ for $n \ge N$, where $N$ is some integer.

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    $\begingroup$ hint: $f_n = f_n - f + f$ $\endgroup$ – user251257 Sep 11 '15 at 1:45
  • $\begingroup$ @user251257 I've tried that, and found that $\int_A |f_n| \le \int_A |f_n-f| + \int_A |f|$. This, along with the fact that for any $\epsilon > 0$, I can find $\delta$ s.t. if $m(A) < \delta$, $\int_A |f| < \epsilon/2$, gives me that $\int_A |f_n| \le \int_A|f_n - f| + \epsilon/2$. I know that I can bound $\int_A|f_n - f|$ as $n\to\infty$, since $f_n \to f$ in the $L^1$ norm, but that only gives me that $\int_A |f_n-f| < \epsilon/2$ for values of $n$ greater than or equal to some integer $N$. What am I missing? $\endgroup$ – poppy3345 Sep 11 '15 at 1:54
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    $\begingroup$ hm. sorry. I missed that you want it for all $n$. Notice the maximum of finitely many $L^1$ functions is $L^1$ again. $\endgroup$ – user251257 Sep 11 '15 at 1:58
  • $\begingroup$ So, set $\epsilon > 0$. First find $N$ such that for all $n \ge N$, $\int_A |f_n| < \epsilon/2$. Then, we write: $\int_A |f_n| \le \int_A |f_n-f| + \epsilon/2$ if $m(A) < \delta$ as before. If $n \ge N$, we are done, we have found that $\int_A |f_n| < \epsilon$. If $n \le N$, we can find $\delta_1$, $\delta_2$,... such that $\int_A|f_n| < \delta_n$ for each $n$. Take $\delta = \min\{\delta, \delta_1,\delta_2,...,\delta_{N-1}\}$? $\endgroup$ – poppy3345 Sep 11 '15 at 2:11
  • $\begingroup$ Oops, guess I should have done that $\delta = \min\{\delta,\delta_1,...,\delta_{N-1}\}$ thing before doing the $n \ge N$ case. But, is that the right idea? $\endgroup$ – poppy3345 Sep 11 '15 at 2:11

First, notice that $L^1(R)$ is complete, and thus $f\in L^1(R)$. Now, $$\int_A|f_n|dm \le \int_A|f_n - f|dm + \int_A|f|dm.$$ Since $f_n \to f$ in the $L^1$ norm, if we fix $\epsilon > 0$, we can find an integer $N$ such that, for all $n \ge N$, we have $$\int|f_n-f|dm < \epsilon/2.$$ Now, for each $f_n$, where $n < N$, $f_n$ is Lebesgue integrable, so we can find $\delta_n > 0$ for each $n$ such that if $m(A) < \delta_n$, $$\int_A|f_n| dm < \epsilon.$$ We also have $\delta$ such that if $m(A) < \delta$, we have $$\int_A |f| dm < \epsilon/2.$$ Let $\tilde{\delta} = \min\{\delta_1,\delta_2,...,\delta_{N-1},\delta\}$. Then, if $n < N$, if $m(A) < \tilde{\delta}$, we have $$\int_A |f_n| dm < \epsilon,$$ and if $n \ge N$, we have $$\int_A |f_n| \le \int_A |f_n - f| + \int_A |f| < \epsilon,$$ as desired.

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    $\begingroup$ Just a little remark. You say "since $L^1$ is complete, $f\in L^1$". I respectfully disagree with this reasoning, because the fact that $f\in L^1$ is already incorporated in the assumption "$f_n \to f$ in $L^1$". No need to invoke completeness. You would need completeness only if the assumption were "$f_n$ is a Cauchy sequence with respect to the $L^1$ norm". $\endgroup$ – Giuseppe Negro Sep 11 '15 at 18:04
  • $\begingroup$ But, except for that minor remark, the proof looks OK to me. $\endgroup$ – Giuseppe Negro Sep 11 '15 at 18:07
  • $\begingroup$ @GiuseppeNegro thank you! I appreciate the helpful insight there, and I agree that $f_n \to f$ in $L^1$ implies $f \in L^1$. However, even though I know it uses too many steps, is the statement still correct, or is there a flaw in the reasoning? Thanks! $\endgroup$ – poppy3345 Sep 11 '15 at 18:14
  • $\begingroup$ It is OK and it does not use too many steps. $\endgroup$ – Giuseppe Negro Sep 11 '15 at 18:19

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