# Can $\int|f_n|d\mu \to \int |f|d\mu$ but not $\int|f_n - f|d\mu \to 0$? [duplicate]

Possible Duplicate:
Convergence a.e. and of norms implies that in Lebesgue space

I am trying to show that if $$\int_X |f_n|d\mu \to \int_X|f|d\mu$$ where $f$ and all the $f_n$ have finite integral and $f_n \to f$ pointwise, then $$\int_X |f_n-f|d\mu \to 0.$$

I worked out a proof in the case that $\mu(X) < \infty$, but it relies on Egoroff's theorem which may fail if $\mu(X) = \infty$. I can't find a counterexample in the case $\mu(X) = \infty$ but I suspect that it may not be true. I was thinking of $X=\mathbb{R}$ but maybe there is a good counting measure counterexample on $\mathbb{N}$.

Does anyone know if this is true in the case $\mu(X) = \infty$, and if so, how might I get started in showing it?

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## marked as duplicate by Qiaochu YuanOct 31 '12 at 7:15

What if $f_n=-1$ and $f=1$? Then your conclusion fails on $[0,1]$ using Lebesgue measure unless I am missing something. – Derek Allums Oct 27 '12 at 12:41
@unit3000-21 Missing $f_n\to f$ pointwise. – Did Oct 27 '12 at 12:42
Oh I didn't notice that. Please disregard my comment then. – Derek Allums Oct 27 '12 at 12:44
Exact dupe of this – leo Oct 31 '12 at 3:59

Let $g_n(x):=|f(x)|+|f_n(x)|-|f(x)-f_n(x)|$. It defines an integrable function, and $g_n\to 2|f|$ pointwise. Furthermore, $g_n\geq 0$. By Fatou lemma, $$\int_X\liminf_{n\to+\infty}g_n(x)d\mu(x)\leq\liminf_{n\to+\infty}\int_Xg_n(x)d\mu(x).$$ The LHS is $2\int_X|f(x)|d\mu(x)$, and the RHS is $2\int_X|f(x)|d\mu(x)+\liminf_{n\to +\infty}-\int_X|f-f_n|d\mu$. This gives $$0\leq -\limsup_{n\to +\infty}\int_X|f-f_n|d\mu,$$ which is the wanted result.