Suppose $A $ is a measurable set, and $\{h _n \} $ is a sequence of nonnegative simple functions such that $h _n \uparrow \chi _A $, where $\chi _A $ is the charachteristic function.
I wonder why in proving that $\int h _n d \mu \uparrow \mu (A) $, I need to show that $\lim \inf \int h _n d \mu \ge \mu(A )$, and not just $\lim \int h _n d \mu \ge \mu(A )$. What is the difference?
Trivially I have that $\lim \int h_n d \mu \le \mu (A) $ , since the integral is monotone. And since $h _n $ is an increasing sequence $\int h _n d \mu$, also is increasing.
For the opposite inequality,
I can define sets $A _n= \{s \in A : h _n (s ) > 1 - \epsilon \} $
And I would believe that $\lim \int _{A _n } h _n d \mu \ge (1-\epsilon )\lim \mu(A _n) \to (1-\epsilon )\mu (A) $
Am I missing something? Do I have to thake the inferior limit in this last expression?
Thanks in advance!