I'm interested in knowing whether there is a condition for general measure spaces under which we know that we can only achieve the strict inequality of Fatou's lemma. I am working in the situation that $f_n \rightarrow f$, and the limit of the integrals do exist so that Fatou's lemma says $$ \int f \leq \lim_{n \rightarrow \infty} \int f_n \;. $$ Is there a condition on $f_n$ and $f$ which ensures $$ \int f < \lim_{n \rightarrow \infty} \int f_n \;. $$
Strict inequality for Fatou's lemma
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measure-theory
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5A standard example in the real, Lebesgue sense occurs when the pointwise limit of $f_n$ is 0 but not uniformly so, and each $|f_n|$ is positive on a set of measure greater than zero. Consider $f_n(x)$ = {n for 0 $\leq x \leq \frac{1}{n}$; 0 o.w. }. Then $f_n \rightarrow 0$ pointwise, but its integral from 0 to 1 is always 1. This might shed some light on the problem. – 2011-08-26
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0For measure spaces with total mass finite, look up "uniform integrability." – 2011-08-26
1 Answers
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There is a nice discussion of this point in ANALYSIS by Lieb & Loss (section 1.9 of the second edition): It $f_n$ are non-negative and converge a.e. to $f$, then $$ \liminf_n\int f_n = \int f +\liminf_n\int|f-f_n| $$ provided $\sup_n\int f_n<\infty$.