# Convergence a.e. and of norms implies that in $L^1$ norm

Suppose $f_n$ is as sequence of functions in $L^1[0,1]$ such that $f_n$ converges pointwise a.e. to $f\in L^1[0,1]$. Suppose also that $\int \vert f_n\vert \rightarrow \int \vert f\vert$. Is it true that $f_n$ converges to $f$ in the $L^1$ norm?

From Javaman's comment below: $|f_n-f|\leqslant |f_n|+|f|$. So DCT applies. Since $f_n\rightarrow f$ a.e. we have $\lim_n\int |f_n-f|=0.$ i.e $\Vert f_n-f\Vert \rightarrow 0.$

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I haven't written it out to check details, but note that $|f_n - f| \leq |f_n| + |f|$ so you can apply Dominated Convergence Theorem. Then, $\lim \int |f_n - f| = \int \lim |f_n - f|$. – JavaMan Nov 17 '11 at 23:27
@JavaMan: Is what I've done right? – steve Nov 18 '11 at 0:00
So DCT applies ... what is the dominating function you want to use? – GEdgar Nov 18 '11 at 1:11
@GEdgar: Would you rather I use Fatou's Lemma instead? – steve Nov 18 '11 at 1:25

Yes you are right. Fatou's lemma can be applied here. As Javaman said, $|f_n-f|\le |f_n|+|f|$ is the key. Then $|f_n|+|f|-|f_n-f|\ge 0$, so Fatou's lemma applies. That is, $$\liminf_{n\to \infty}\int \left(|f_n|+|f|-|f_n-f|\right)\ge \int \liminf_{n\to \infty} \left(|f_n|+|f|-|f_n-f|\right)$$
Hence $$2\int |f|-\limsup_{n\to \infty}\int|f_n-f|\ge 2\int |f|$$
Thus $$\limsup_{n\to \infty}\int|f_n-f|\le 0$$ and we are done.