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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.

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

1 Answer 1

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

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    $\begingroup$ Yes, this is correct. However, I'm looking for a way to prove this without the DCT. $\endgroup$
    – EzioBosso
    Mar 5, 2020 at 0:11

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