Let X be a measurable set with $ \mu (X) < \infty $ and $ 1 \leq p < \infty $
Let $ (f_n)_{n \in \mathbb{N}} \subset X $ and $f \in L^{p}(X)$ with $\lim_{n \to \infty}||f_n - f|| = 0 $
Show that $ \forall$ $\epsilon >0 \ \exists \ \delta >0 $ such that $\forall n \in \mathbb{N} $
$\int_{E}|f_n| \leq \epsilon \ \forall \ E \subset X$ with $\mu(E)<\delta$
Since $ | f_n |^p $ is integrable and positive, we can find $ \delta $ that for every set $ E \subset X $ with $ \mu(E) < \delta $ that $ \int_{E}|f_n |^p <\epsilon $
The problem is that this goes for $ f_n $ specifically for the $ n $ chosen, how can I use the limit of $ ||f_n -f || $ to generalize for all $ n$?
Since the measure space is finite convergence in $L^p$ implies convergence in $L^1$. So for a given $\epsilon$ choose $n_0\in\mathbb{N}$ such that $||f_n-f||_1<\frac{\epsilon}{2}$ for all $n>n_0$. Now, since $f\in L^1$ we know that there is $0<\delta$ such that for all $E\subseteq X$ with $\mu(E)<\delta$ we have $\int_E |f|<\frac{\epsilon}{2}$. Alright, so for all $n>n_0$ and each such set $E$:
$\int_E |f_n|=\int_E |f_n-f+f|\leq\int_E |f_n-f|+\int_E |f|\leq\int_X |f_n-f|+\int_E |f|\leq \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$
So this $\delta$ works for all $n>n_0$. Now, before $n_0$ there are only finitely many functions in the sequence. So for each $n\leq n_0$ choose $\delta_n$ which works for $f_n$ and then take $\min\{\delta_1,...,\delta_{n_0},\delta\}$.