# if $\int|f_n| \rightarrow \int |f|$, then $\int |f-f_n| \rightarrow 0$

## [Theorem] Dominated Convergence Theorem

Suppose $(f_n)$ measurable, $f_n \rightarrow f$ pointwise and there exists an integrable function $g$ such that $|f_n(x)| \leq g(x)$. Then $\int f = \lim_n \int f_n$.

Infact, $\int |f-f_n| \rightarrow 0$ as $n \rightarrow \infty$

So in class, we porved DCT using Fatou's Lemma.

we showed $\int |f-f_n| \rightarrow 0$ first and then by triangle inequality we got $\int f = \lim_n \int f_n$

Now I have to prove the following

## Suppose $(f_n)$ are measurable, $f$ is integrable and $f_n \rightarrow f$ pointwise. Prove that if $\int|f_n| \rightarrow \int |f|$, then $\int |f-f_n| \rightarrow 0$

How can I prove this? I guess I have to use DCT somehow.

Also, I think if we prove this, we can claim $\int f = \lim_n \int f_n$ as we did in the proof of DCT. Is that right?

• Nope, $f_n \in L^1$ is not given
– Andy
Jun 22 '17 at 14:06
• we must have $f_n \in L^1$ for all but finitely many $n$. Otherwise $\int |f_n| \to \int f$ makes no sense Jun 22 '17 at 14:11

## 1 Answer

Hint:

First prove the following result.

If $f_n, g_n, f, g \in L^1$, $f_n \to f$ and $g_n \to g$ a.e., $|f_n| \leq g_n$ and $\int g_n \to \int g$, then $\int f_n \to f$.

Then apply this to your problem with $g_n = |f| + |f_n|$.

• I found why $f_n |in L^1$ is not required. We can use Fatou's Lemma. Check Xiao's answer here math.stackexchange.com/questions/364059/doubt-in-scheffes-lemma I guess if we have $f_n \in L^1$, then we get Scheffé’s lemma
– Andy
Jun 22 '17 at 16:43
• that is much simpler, thanks Jun 23 '17 at 13:25