Let $f_n$ and $f$ be in $L^1$. We wish to show $f_n \rightarrow f$ in $L^1$. Let $f_n$ and $f$ be in $L^1$. We wish to show $f_n \rightarrow f$ in $L^1$.
Question: Does it follow that $\int |f_n - f| \,d\mu \rightarrow 0$ if it is true for all measurable sets $E$ that $\int_E f_n \rightarrow \int_E f$.
Note that $f_n, f$ are in $L^1$.
The converse of this question is easy.  We are trying to prove this direction.
 A: It isn't true.
Take for example $\Omega = [0,\pi]$ ,$f_n(x) = \sin(n x)$ and $f\equiv0$. Then, for every $E\subset\Omega$ measurable, $$a_n = \int_E f_n = \int_\Omega sin(n x) \mathbb 1_E $$ is the Fourier coeficient (up to a constant) of the $L^2$ function $1_E$ so $a_n\rightarrow 0$.But $$\int |\sin(nx)|dx=2$$

About yor second question:
Define $A_n =\{f_n\geq f\}$ and $B_n = A_n^c$ both measurable. Then
$$ 0\leq\int_{A_n}| f_n-f| = \left | \int_{A_n} |f_n-f |\right |=\left | \int_{A_n} f_n-f \right |\leq \sup_E \left | \int_{E} f_n-f \right |\rightarrow 0$$
in a similar way
$$ 0\leq\int_{B_n}| f_n-f| =\left | \int_{B_n} f_n-f \right |\leq \sup_E \left | \int_{E} f_n-f \right |\rightarrow 0$$
Hence
$$\int_{A_n}| f_n-f|\rightarrow 0$$
$$\int_{B_n}| f_n-f|\rightarrow 0$$
$$\int_{\Omega}| f_n-f|=\int_{A_n}| f_n-f|+\int_{B_n}| f_n-f|\rightarrow 0$$
A: If $f_n \rightarrow f$ a.e. on E, then you can use General Lebesgue Dominated Convergence Theorem to show:
$\int |f_n - f| \,d\mu \rightarrow 0$ if and only if $\int_E |f_n| \rightarrow \int_E |f|$.
Note: even you have $f_n \rightarrow f$ a.e. on E, you can't get $\int |f_n - f| \,d\mu \rightarrow 0$ from $\int_E f_n \rightarrow \int_E f$
Let $E=[0.1]$, define 
$$f_n(x):=\begin{cases}
n, & 0<x<\frac{1}{n}\\
-n, & \frac{1}{n}<x<\frac{2}{n}\\
0 & \text{otherwise}
\end{cases}$$
A: This says that $f_n$ converges $L^1$-weakly to $f$.
