If $m(A) \to 0,$ then $\int_A f\,dx \to 0$ 
Let $f$ be an integrable non-negative function defined on a measurable set $E.$ 
  Then, for $A \subset E,$ if $m(A) \to 0,$ then $$\int_A f\,dx \to 0$$ 

Very intuitively (cause I'm thinkng of a bounded function defined on a closed domain) I beleive i understand what I'm being asked to prove: If the area of the domain is smaller and smaller, then the integral wil be smaller and smaller, until the interval is just a point, and so the function would also be a point, so the area underneath would be zero. 
But formally, I don't know how to use and reason with the hipothesis $m(A) \to 0.$ In the other excercises that I'm supposed to solve, I should use some convergence thoerems available. I have looked at the proofs about some of them, like the monotone convergence theorem, Fatou's Lemma, Lebesgue Convergence theorem, and none of them use explicitly something like the domain is tending to zero. They are using properties about pointwise convergence, or seem to be dealing mainly with images of the functions. 
Convergence in measure uses something like: $m\{x : |f(x) -f_n(x)| \ge \epsilon\} < \epsilon .$
Definition: A sequence $<f_n>$ of measurable functions is said to converge to $f$ in measure if, given $\epsilon > 0,$ there is an $N$ such that for all $n \ge N$ we have $m\{x : |f(x) -f_n(x)| \ge \epsilon\} < \epsilon .$
Then, Fatou's Lemma or Monotone Convergence theorem are valid for convergence in measure (it is a proposition, but I have not proved it), and so maybe that is a way to involve a the measure in the domain converging to zero and and integral converging to a funcion $f = 0.$
More overly Im talking to much because it is not clear to me how to relate $m(A) \to 0$ with an integral converging to zero. 
I would appreciate some help... 
 A: Hint: This is clear if $f$ is also bounded. But given $\epsilon>0,$ there exists a bounded $g\in L^1(E)$ such that $\int_E |f-g| < \epsilon.$
A: Say $(A_n)_n \subseteq E$ satisfy $m(A_n) \downarrow 0$. Suppose for some $\epsilon > 0$, $\int_{A_n} f > \epsilon$ for all large $n$. Take a subsequence $(n_k)_k$ so that $\sum_k m(A_{n_k}) < \infty$. Then if $A = \limsup_k A_{n_k}$, $\int_A f > \epsilon$ but $m(A) = 0$ by Borel Cantelli. This is a contradiction.
The motivation for the proof above comes from the fact that $v(A) := \int_A f$ is a measure which is absolutely continuous with respect to $m$. It is a general fact that for measures $v_1,v_2$, if $v_1 \ll v_2$, then for all $\epsilon > 0$ there is a $\delta > 0$ so that if $v_2(A) < \delta$, 
then $v_1(A) < \epsilon$, which is your problem statement.
A: By Lebesgue Dominated Convergence Theorem we have 
\begin{align*}
\int|f|1_{|f|\geq n}dx\rightarrow\int|f|1_{|f|=\infty}dx=0
\end{align*}
since $m(\{|f|=\infty\})=0$ and the convention that $\infty\cdot 0=0$. Or we can consider the sequence $\left(|f|1_{|f|\leq n}\right)$ and apply Monotone Convergence Theorem to get the convergence to $\displaystyle\int|f|dx$, and then we do the subtraction. Given $\epsilon>0$, we have some $n_{0}\in{\bf{N}}$ such that 
\begin{align*}
\int|f|1_{|f|\geq n_{0}}dx<\epsilon.
\end{align*}
For any measurable set $A$ with $m(A)<\epsilon n_{0}^{-1}$ then we have 
\begin{align*}
\int|f|1_{A}dx&=\int|f|1_{A\cap\{|f|\geq n_{0}\}}dx+\int|f|1_{A\cap\{|f|<n_{0}\}}dx\\
&<\epsilon+\int|f|1_{A\cap\{|f|<n_{0}\}}dx\\
&\leq\epsilon+n_{0}\int 1_{A}dx\\
&=\epsilon+n_{0}m(A)\\
&<2\epsilon.
\end{align*}
