Let $\mu$ be a $\sigma$-finite measure and $(f_n)_{n\geq1}$ and $f$ be measurable functions. Show that $f_n \rightarrow f$ in $L^1(\mu)$ iff all of the following three conditions are satisfied
$f_n$ converges to $f$ in measure.
The sequence $(f_n)$ is uniformly integrable, i.e. $\forall \epsilon > 0, \exists \delta > 0$ such that if $\mu(E) < δ$ then $\int_E |f_n(x)| d\mu(x) < \epsilon$ for any $n$.
The sequence $(f_n)$ is uniformly tight, i.e. $\forall\epsilon > 0, \exists E, \mu(E) < \infty$ such that for any $n$ : $\int_{E^c} |f_n(x)| d\mu(x) < \epsilon$.
Remark: If $\mu$ is finite, the last condition is automatically satisfied
Remark: I am able to show the first two parts of implication, I am unable to show that $F_n$ converging in $L^1$ implies uniform tightness; also I am unable to show that given the three conditions are satisfied the sequence of measurable functions converge in $L^1$.