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I came across the following problem in my self-study:

If $f_n, f$ are integrable and $f_n \rightarrow f$ a.e. on $E$, then $\int_E |f_n - f| \rightarrow 0$ iff $\int_E |f_n| \rightarrow \int_E |f|$.

I am trying to prove (1) and the book I am using suggests that it follows from the Generalized Lebesgue Dominated Convergence Theorem:

Let $\{f_n\}_{n=1}^\infty$ be a sequence of measurable functions on $E$ that converge pointwise a.e. on $E$ to $f$. Suppose there is a sequence $\{g_n\}$ of integrable functions on $E$ that converge pointwise a.e. on $E$ to $g$ such that $|f_n| \leq g_n$ for all $n \in \mathbb{N}$. If $\lim\limits_{n \rightarrow \infty}$ $\int_E$ $g_n$ = $\int_E$ $g$, then $\lim\limits_{n \rightarrow \infty}$ $\int_E$ $f_n$ = $\int_E$ $f$.

I suspect that I need the right inequalities to help satisfy the hypothesis of the GLDCT, but I am not certain about what these inequalities should be.

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See this – leo Feb 15 '12 at 1:12
And this – leo Feb 15 '12 at 1:16
up vote 5 down vote accepted

The fact that $$\int_E |f_n - f| \to 0 \Rightarrow \int_E |f_n| \to \int_E |f|$$ is actually a simple consequence of the fact that $f \mapsto \int_E |f|$ is a norm; in other words, it holds independently of the assumptions given. More specifically, in any normed space, the following inequality holds $$ |\| x \| - \| y \| | \leq \| x - y \|.$$ The proof is easy: if $\| x \| \geq \| y \|$, write $\| x \| = \| x - y + y \| \leq \| x - y \| + \| y \|$, so that $0 \leq \|x\| - \| y \| \leq \| x - y \|$. If $\| y \| > \| x \|$, reverse the roles of $x$ and $y$, to get that $\| y \| - \| x \| \leq \| x -y \|$. Taking these inequalities together proves the result.

Hence $\| f_n - f\|_1 \geq |\| f_n\|_1 - \| f \|_1|$, and since the left hand side tends to $0$, so must the right hand side, since it is $\geq 0$. This proves the first implication.

For the converse, assume that $$ \int_E |f_n | \to \int_E |f|$$ and that $f_n \to f$ a.e. on $E$. We can, as Elan B. mentions, take $g_n = |f_n| + |f|$ and conclude using the triangle inequality that $|f_n - f| \leq g_n$. By assumption, we have $|f_n| \to |f|$ a.e. on $E$ (since, again, $|f_n - f | \to 0$ and $|f_n - f| \geq | |f_n| - |f| |$). Furthermore, we have $$ \int_E g_n = \int_E (|f_n| + |f|) \to \int_E 2|f|.$$ From the generalized DCT, we conclude that $$\int_E |f_n - f| \to 0,$$ as claimed. This proves (1).

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Take $g_n = |f_n| + |f|$ and use the triangle inequality to get the bound.

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Suppose that $$\int |f_n|\to |f|.$$ The function $$|f_n|+|f|- |f_n-f|$$ is no negative. The following is just a technique used in other answer on this site. By Fatou's Lemma $$\begin{align*} \liminf \int |f_n|+|f|- |f_n-f| &\leq \int\liminf(|f_n|+|f|- |f_n-f|)\\ \liminf\int |f_n|+\int |f|-\limsup\int |f_n-f| &\leq \liminf \int |f_n|+\int |f|-\int\limsup |f_n-f|\\ 2\int |f|-\int 0 &\leq 2\int |f| - \int\limsup |f_n-f|, \end{align*}$$ the last inequality leads $$\limsup\int |f_n-f|\leq 0,$$ therefore $$\lim_{n\to\infty}\int |f_n-f|= 0.$$ For the other implication see Martin's answer.

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