# Real Analysis: Uniform Integrability, Convergence in Measure and Convergence Almost Everywhere

I am trying to figure this out for many hours. I am trying to prove that $f_n$ converges to $f$ in $L^1$ norm. I have that $f_n$ converges to $f$ in measure and that $f_n$ are uniformly integrable. I also have that $f_n$ and $f$ are lebesgue measurable functions. I know that I want to use Egorov's theorem but I cannot since I do not have that $f_n$ converge to $f$ almost everywhere.

Indeed, $\int_X|f_n-f|d\mu=\int_{\{|f_n-f|<\epsilon\}}|f_n-f|d\mu+\int_{\{|f_n-f|\geq\epsilon\}}|f_n-f|d\mu\leq\epsilon\mu(X)+\int_{\{|f_n-f|\geq\epsilon\}}|f_n|d\mu+\int_{\{|f_n-f|\geq\epsilon\}}|f|d\mu.$
Now by uniform integrability of $f_n$ and $f$, there exist $\delta$, such that when $\mu(A)<\delta$, $\int_A|f_n|d\mu<\epsilon$ and $\int_A|f|d\mu<\epsilon$.
Because $f_n$ converges in measure to $f$, we can find $n$ sufficiently large so that $\mu(\{|f_n-f|\geq\epsilon\})<\delta$. Therefore, $\int_{\{|f_n-f|\geq\epsilon\}}|f_n|d\mu+\int_{\{|f_n-f|\geq\epsilon\}}|f|d\mu<2\epsilon$.
Putting things together we have $\int_X|f_n-f|d\mu<\epsilon(\mu(X)+2)$ for sufficiently large $n$, which means that $f_n$ converges in $L_1$ to $f$.