 Solution

Consider the inclusion mapIts graph is closed. Indeed, suppose $x_k\to x$ in $X$ and $J{x}_k\to y$ in $L^1$. By hypothesis, some subsequence of $(x_k)$ converges almost everywhere to $x$. From its $L^1$ convergence, that subsequence has a further subsequence converging almost everywhere to $y$. Uniqueness of almost-everywhere limits gives $x=y$ as $L^1$ classes.

Both spaces are Banach, so the closed graph theorem makes $J$ bounded. Therefore some $C>0$ satisfies