If $f_k \to f$ a.e. on $B=B(0,1)$ , and $\forall \epsilon >0$ there is $T>0$ such that $\int_{ \{ f_k >T \} } f_k(x) dx < \epsilon$. Show that $f_k \to f$ in $L^1(B)$.
Is my proof correct?
Proof
Let $\epsilon >0$, from the asumption, there $T>0$ such that
$$\int_{ \{|f_k -f| > T\} } |f_k -f| dx \leq \int_{f_k > T} |f_k| dx < \frac{\epsilon}{2}.$$
Let $E = \{|f_k -f| \leq T \}$. By Egoroff's theorem , there is $A \subset E$ such that $|A| \leq \frac{\epsilon}{4T}$ and $f_k \to f$ uniformly on $E \setminus A$. Then we have
$$\int_E |f_k -f| dx \leq \int_{E \setminus A} |f_k-f|dx + \int_A |f_k -f| dx .$$
From the uniform convergence, there is $N >0$ such that $|f_k - f| < \frac{\epsilon}{4 |E \setminus A|}$ for $k > N$. Then
\begin{align} \int_E |f_k -f| dx &\leq \int_{E \setminus A} |f_k -f |dx + \int_A |f_k -f| dx\\ & < \frac{\epsilon}{4} + T \int_A dx < \frac{\epsilon}{2} \end{align}
Now, we have
$$\int_B |f_k - f| dx\leq \int_{|f_k -f| < T} |f_k - f|dx + \int_{|f_k - f| \geq T} |f_k - f|dx< \epsilon,\,\,\,\,\,\, \forall k >N.$$