# Showing $\lim_{n \to \infty} \int |f_n| - |f - f_n| = \int |f|$

I'm trying to show the following without using the Dominated Convergence Theorem:

Let $$E \subseteq \mathbb{R}^d$$ measurable, and $$\{f_n\}$$ a sequence of integrable functions on $$E$$. Assume that $$\sup \int_E |f_n| < \infty$$ and $$f_n \to f$$ pointwise a.e. Show that

$$\lim_{n \to \infty} \int_E \left(|f_n| - |f - f_n|\right) = \int_E |f|.$$

So far I have that:

Since $$f_n \to f$$ pointwise a.e. we have

\begin{align} \int_E |f| &= \int_E \liminf_{n \to \infty}\left(|f_n| - |f - f_n|\right) \\ &\leq \liminf_{n \to \infty} \int_E (|f_n| - |f - f_n|) \tag{By Fatou's Theorem} \\ &= \int_E |f_n| - \limsup_{n \to \infty} \int_E |f - f_n|. \end{align}

I suppose I now need to show $$\int_E |f| \geq \int_E |f_n| - \limsup_{n \to \infty} \int_E |f - f_n|$$, but I am unsure how to proceed. Also, it's clear that $$f$$ is in $$L^{1}(E)$$ space, but I'm unsure if/how that is helpful either.

• Fatou applies to sequences of non-negative functions, so at best you can do it for $\lvert \lvert f_n\rvert-\lvert f-f_n\rvert\rvert$.
– user239203
Mar 5, 2020 at 0:03
• @Gae. S. Ah, yes you are right. Thus, what I have so far is not valid. Mar 5, 2020 at 0:09

$$||f_n|-|f_n-f|| \leq |f|$$ so DCT can be applied.