Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

marked as duplicate by Qiaochu Yuan Oct 31 '12 at 7:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
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
1  
@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

1 Answer 1

up vote 11 down vote accepted

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.

In particular, this works without the assumption of finiteness of the measure (we just need a positive measure).

share|improve this answer
1  
+1 for a nice proof of the Scheffé's lemma. –  sos440 Oct 27 '12 at 12:43
    
@sos440 Thanks! I didn't know it was called Scheffé lemma. –  Davide Giraudo Oct 27 '12 at 12:43

Not the answer you're looking for? Browse other questions tagged or ask your own question.