# Strict inequality for Fatou's lemma

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 $$\int f < \lim_{n \rightarrow \infty} \int f_n \;.$$

• A 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. – Rachel Aug 26 '11 at 12:35
• For measure spaces with total mass finite, look up "uniform integrability." – ShawnD Aug 26 '11 at 14:32

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 $\sup_n\int f_n<\infty$.