This is question question 1.8 in Rudin's Real and Complex Analysis.
Put $f_n= \chi_E$ if n is odd, $f_n= 1-\chi_E$ if n is even. What is the relevance of this example to Fatou's lemma?
My attempt: Suppose $x \in E$. then $f_n$ = 1 if n odd and =0 if n even. Thus $\liminf_nf_n$ =0. Thus, the left handed side of Fatou's lemma =0. Now focus on the RHS of Fatou's lemma. $\int_X f_n d\mu$ = $\mu (E)$ if n odd and =0 if n even. Thus, $\liminf \int_X f_n d\mu$ =0. Well, I prove that equality can occur in Fatou's lemma. (if $x \notin E$ is similar).
But the correct solution shows that strict inequality can occur in Fatou's lemma. So please point out my mistake. Thanks!
Best Answer
$\liminf_nf_n=0$; hence $\int \liminf_n f_n\mu=0$. $$\int f_n\,d\mu=\left\{\begin{array}{lcr} \mu(E) & \text{if} & n\equiv1\mod 2\\ \mu(E^c) &\text{if} &n\equiv0\mod 2 \end{array} \right. $$ $$\liminf_n\int f_n\,d\mu=\min(\mu(E),\mu(E^c))$$
By choosing $E$ with $\mu(E)>0$ and $\mu(E^c)>0$, this exercise shows that strict inequality may occur in Fatou's Lemma.