# Why does the solution of this Fatou’s Lemma fail

real-analysis

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!

$$\liminf_nf_n=0$$; 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.