Simple exercise about Fatou's lemma Could anyone help me with this very simple exercise?
Let $(X,\mathcal F,\mu)$ be a measure space, and $A,B\in\mathcal F$ such that $A\cap B=\varnothing$. Consider the sequence
 $$f_n =
\begin{cases}
1_A,  & \text{if $n$ is even} \\
1_B, & \text{if $n$ is odd}
\end{cases}$$
where $1_A$ is the indicator function of the set $A$.
I have to prove that Fatou's Lemma holds, but the inequality should be strict. As usual, I get confused with indicator functions. 
My reasoning is: for all the even terms of my sequence, I get the indicator function on the set A, that is: if n is in A, I get 1, and 0 otherwise. 
The same happens with the odd terms. Since A and B never intersect each other, there is no x in A\capB.
So, the only values of my sequence are 1 and 0, and then the computation of the limit reduces to that of a simple indicator function. Is this right? 
Can I conclude that the liminf is 0? So, the integral of my liminf is simply 0?
But now I get confused. I tried to draw a stupide picture of this, but it gets me nowhere. What does it really mean checking that a point is in the set A? What's wrong with my reasoning?
Thanks. 
 A: This is not true as stated. For instance, take $A=B=\varnothing$. If, on the other hand, you assume that both $A$ and $B$ have positive measure, then it is indeed true, which I'll assume for this answer.
What does Fatou's Lemma say? It says that if $f_n$ is a sequence of non-negative measurable functions on a measure space $(X,\mathcal F,\mu)$, then
$$\int_X \liminf_n f_n\;d\mu\le\liminf_n\int_X f_n\;d\mu.$$
Now, the hypothesis is certainly true in our case because indicator functions are non-negative and measurable, so Fatou's lemma indeed holds. How can we see it is a strict inequality?
Well, what is $\liminf_n f_n$? For any $x\in X$, since $A$ and $B$ are disjoint, there are infinitely many values of $n$ such that $f_n(x)=0$ (why?). This precisely says that $\liminf_n f_n=0$. Therefore the left hand side of the equation is zero.
What about the right side? Well, it's easy to see that
$$\int_X f_n\;d\mu=\begin{cases}\mu(A)&\text{$n$ is even}\\\mu(B)&\text{$n$ is odd}\end{cases}$$
and therefore $\liminf_n\int_X f_n\;d\mu=\min\{\mu(A),\mu(B)\}$. If both $A$ and $B$ have positive measure, then this value is positive, so the inequality above is strict.

Also, I'd like to add that this is the fourth time you've asked a question on this site; if you plan to keep doing so, you should learn to use LaTeX. I edited the first part of your question, and I'd encourage you to try to fix the rest, copying the commands I used in my edit.
