How do I show that an operator on $L^p[0,1]$ taking $a.e.$ convergent sequence to an $a.e.$ convergent sequence is always continuous?

Question

Suppose $T$ is an operator on $L^p[0,1]$ which takes $a.e.$ convergent sequences to $a.e.$ convergent sequences, i.e., if $f_n\rightarrow f$ $a.e.$, then $Tf_n\rightarrow Tf$ $a.e.$. I'm trying to show that $T$ is continuous in $L^p$ norm. I don't think that any one of $a.e.$ or $L^p$ convergence is implied by the other. I tried using the closed graph theorem, but to no avail. How can this be proved?

Answer 1

$L^p$ convergence implies convergence in measure. Convergence in measure implies a.e. convergence of a subsequence. Using this you can apply the Closed Graph Theorem, because if $(f_n,Tf_n)\to(f,g)$, then there is a subsequence $f_{n_k}$ such that $f_{n_k}\to f$ a.e., hence by your hypothesis you have $Tf_{n_k}\to Tf$ a.e.. You also have a subsequence of $Tf_{n_k}$ converging a.e. to $g$, so $Tf=g$.