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.
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.