Prove $\lim \int \vert f-f_n\vert d\mu=0 \iff \lim \int \vert f_n \vert d\mu=\int \vert f \vert d\mu$

Question

$f_n$ and $f$ are real valued integrable functions and $f_n\xrightarrow{a.e.}f$.

My attempt on $\Rightarrow$: $\vert f_n-f\vert \geq \vert f_n\vert-\vert f \vert$ (and assume that this last one is greater then zero, otherwise we use $\vert f_n-f\vert \geq \vert f \vert - \vert f_n \vert$). Then just integrating and manipulating limits will lead us to what we want.

On $\Leftarrow$ I couldn't do anything =(

I couldn't use any theorem like LDCT or MCT because the hypothesis are just those above. So I got stucked...

Please help!

Answer 1

We need assumption $\int |f| d\mu < \infty$.

We begin with the following claim:

Claim 1: Let $\{g_n\}$ sequence of measurable functions, such that
1. $g_n \geq 0$
2. $g_n \to g$ a.e
3. $\int g d\mu < \infty$ and $\int g_n d\mu \to \int g d\mu$
Then
$\int |g_n -g| d\mu \to 0$

Proof:

Let $h_n = \min\{g_n, g\}$. We have $h_n \to g$ a.e, $0\leq h_n \leq g$.
By the dominated convergence theorem, we have $\int h_n d\mu \to \int g d\mu$.
In other hand, we have $$h_n = \frac{(g_n + g -|g_n -g|)}{2}\text{ or }|g_n -g| = 2h_n -g_n -g$$ Hence the claim is proved.

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By the claim, we have $|f_n| \to |f|$ in $L^1$.

Let $g_n = (|f_n| + f_n)/2$, we have $g_n \geq 0$, $g_n \to (|f| + f)/2$ and $g_n \leq |f_n|$.
By Fatou's lemma, we have $$\liminf \int g_n d\mu \geq \int \frac{|f| +f}2 d\mu.$$ In other hand, $|f_n| -g_n = (|f_n| -f_n) / 2 \geq 0$, hence by Fatou's lemma again, we have $$\liminf \int (|f_n| -g_n) d\mu \geq \int (|f| - \frac{|f| +f}2) d\mu.$$ Since $\int |f_n| d\mu \to \int |f| d\mu$, hence $$\liminf \int (|f_n| -g_n) d\mu = \liminf (\int |f_n| d\mu - \int g_n d\mu) = \liminf \int |f_n| d\mu + \liminf (-\int g_n d\mu) = \int |f| d\mu -\limsup \int g_n d\mu.$$ Hence $$\limsup \int g_n d\mu \leq \int \frac{|f| +f}2 d\mu.$$ Therefore, we have $\int g_n d\mu \to \int \frac{|f| +f}2 d\mu$.
By the claim, we have $g_n \to (|f|+f)/2$ in $L^1$.
Hence $f_n = 2g_n - |f_n| \to 2 \frac{|f| +f}2 -|f| =f$ in $L^1$.

Answer 2

If you know that $L^1$ is a normed space, then by continuity of the norm you obtain that $\Vert f-f_n\Vert_{L^1}\rightarrow 0$ implies that $\Vert f_n\Vert_{L^1}\rightarrow \Vert f\Vert_{L^1}$.

For the other direction, Fatou's lemma is useful with a certain choice of funtions. Specifically define $g_n:=\frac{1}{2}\Big( \vert f\vert +\vert f_n\vert \Big)-\frac{1}{2}\vert f_n-f\vert$. By the triangle inequality these functions are non-negative. Since $f_n\rightarrow f$ almost everywhere you can see that:

$\frac{1}{2}\Big( \vert f\vert +\vert f_n\vert \Big)\overset{a.e}{\rightarrow}\vert f\vert$ and $\vert f_n-f\vert\overset{a.e}{\rightarrow}0$, which implies that $g_n\overset{a.e}{\rightarrow}\vert f\vert$.

Using Fatou's lemma we can conclude:

$ (*) \int \vert f\vert d\mu= \int\underset{n\rightarrow \infty}{\liminf}(g_n)d\mu \leq \underset{n\rightarrow \infty}{\liminf} \int g_n d\mu$

Then:

$\underset{n\rightarrow \infty}{\liminf} \int g_n d\mu \leq \underset{n\rightarrow \infty}{\liminf} \Big( \frac{1}{2}\int \vert f\vert d\mu +\frac{1}{2}\int \vert f_n\vert d\mu \Big)-\frac{1}{2}\underset{n\rightarrow \infty}{\limsup} \int \vert f_n-f\vert d\mu= $

$=\frac{1}{2}\int \vert f\vert d\mu+ \frac{1}{2}\underset{n\rightarrow \infty}{\liminf}\int \vert f_n\vert d\mu-\frac{1}{2}\underset{n\rightarrow \infty}{\limsup} \int \vert f_n-f\vert d\mu=$

$\overset{\int \vert f_n\vert d\mu \rightarrow \int \vert f\vert d\mu}{=} \int \vert f\vert d\mu-\frac{1}{2}\underset{n\rightarrow \infty}{\limsup} \int \vert f_n-f\vert d\mu$

Then from (*) we conclude that:

$\underset{n\rightarrow \infty}{\limsup} \int \vert f_n-f\vert d\mu = 0$, which implies that $\int \vert f_n-f\vert d\mu \rightarrow 0$.