$L^1$ convergence in a finite measure space

Question

I encountered this problem while studying a past comprehensive exam.

Let $(X,\mathcal{M}, \mu)$ be a finite measure space. Given measurable $f: X \rightarrow \mathbb{C}$, let $E_f(\lambda) = \mu(\{x: |f(x)| > \lambda\})$ be its distribution function. Suppose that $\{f_n\}$ is a sequence of measurable functions converging to $f$ $\mu$-a.e. such that there exists a lebesgue measurable function $E(\lambda)$ such that $E: (0,\infty) \rightarrow [0,\infty)$ satisfying $E_{f_n}(\lambda) < E(\lambda)$ and $\int_{(0,\infty)}E(\lambda)d\lambda < \infty$ . Prove that $f_n$ converges to $f$ in $L^1$.

I suspect that Egoroff's theorem needs to be used, and that I need to somehow control the integral of $|f-f_n|$ on the exceptional set using $E(\lambda)$ but I can't get much further than that. I'm not even sure for example why $f\in L^1$ even though the conditions on $E_{f_n}$ force $f_n$ to be.

Answer 1

Since $f_n$ converges to $f$ $\mu$-almost everywhere, it suffices to show that $f, f_n\in L^1$ and that $lim_{n \rightarrow \infty}{||f_n||_1} = ||f||_1$. Now $lim_{n \rightarrow \infty}{||f||_1} = lim_{n \rightarrow \infty}{\int_{(0,\infty)}{E_{f_n}(\lambda)~d\lambda}} = \int_{(0,\infty)}{lim_{n\rightarrow \infty}{E_{f_n}(\lambda)}~d\lambda}$ by dominated convergence.

By Egoroff's theorem, $f_n$ converges to $f$ in measure, therefore $E_{f_n}$ converges to $E_{f}$ at every point of continuity. Since $E_{f_n}$ is right continuous, it has at most countably many discontinuities. therefore $lim_{n\rightarrow \infty}{E_{f_n}(\lambda)} = E_f(\lambda)$ almost everywhere.

So $\int_{(0,\infty)}{lim_{n\rightarrow \infty}{E_{f_n}(\lambda)}~d\lambda} = \int_{(0, \infty)}{E_{f}(\lambda)~d\lambda} = ||f||_1$. Since everything is finite, this concludes the proof.