# $L^1$ convergence in a finite measure space

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.

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.