Question regarding the sup of an integral I already know the proof to 
let $\epsilon >0 $ there exists $\delta > 0$ such that whenever $m(A) < \delta$ then $\int_A\mid f \mid dm < \epsilon$.
My question is suppose given a subset of $ B \subset L^1$ and I do the same problem but now I consider the supremum, does the proof still go the same? I have been working on it and it seems that it would be the same but now im just working with the supremum, im not sure what extra considerations I need. 
I'm trying to show $\forall \epsilon >0$ there exists $\delta > 0$ such that whenever $m(A) < \delta$ then $\sup_{f \in B}\int_A\mid f\mid dm<\epsilon$. 
Previously I used $\lim_{\lambda \to \infty} \int_{\{\mid f \mid > \lambda\}}\mid f \mid dm = 0$ in the proof of the statement at the top. 
 A: No, you certainly can't say any such thing for an arbitrary subset $B\subseteq L^1$.  For instance, suppose $B$ is all of $L^1$.  Then for any set $A$ of positive measure, $\sup_{f\in B}\int_A |f|dm=\infty$, since $f$ could be arbitrarily large on $A$ while still being integrable.  Explicitly, if $A_0\subseteq A$ is a subset of $A$ which has finite but positive measure, then if $f=c1_{A_0}$ for some $c>0$ then $\int_A|f|dm=cm(A_0)$ which can be made arbitrarily large by making $c$ arbitrarily large.
A: In general it's not true.
Consider $L^1([0;1])$ with Lebesgue measure $m$.
For $n \in \mathbb N$ and $t \in [0;1]$ put $f_n(t) = n$.
Then we have $\widetilde B := \{ f_1, f_2, \ldots \} \subseteq L^1([0;1])$.
Now, for any measurable $A \subseteq [0;1]$ and $n \in \mathbb N$, we have $\int_A\mid f \mid dm = n \cdot m(A)$.
In particular, for $m(A) > 0$, we have $\sup_{f \in \widetilde B}\int_A\mid f\mid dm = \infty$.
Also observe that, given $\delta > 0$, there is $A\subseteq [0;1]$ measurable such that $0 < m(A) < \delta$, for example $A = [0; \frac\delta2]$.
All this together shows that for no $\epsilon > 0$ is there a $\delta > 0$ such that whenever $m(A) < \delta$, then $\sup_{f \in \widetilde B}\int_A\mid f\mid dm < \epsilon$.
This answers your question in the negative.
