Exercise 1.10: Suppose $\mu(X)<\infty$, $\{f_n\}$ is a sequence of bounded complex measurable functions on $X$, and $f_n \rightarrow f$ uniformly on $X$. Prove that $$ \lim_{n\rightarrow \infty} \int_X f_n d{\mu} = \int_X f d \mu $$ I saw an answer (I think it's right) which is organized as follows:
Proof: Let $\epsilon>0$ as given, and there is an $N$ such that $|f_n (x)-f(x)|<\epsilon/\mu(X)$ for all $n \geq N$ and $x \in X$. Then we have $$ |\int_X f_n - f d \mu| \leq \int_X |f_n - f| d \mu < \epsilon $$ For any $\epsilon$, there is a corresponding $N$, which implies $$ \lim_{n \rightarrow \infty} \int_X (f_n - f) d \mu = 0 $$ It is equivalent to $$ \lim_{n \rightarrow \infty} \int_X f_n d \mu = \int_X f d \mu $$
Q.E.D
My questions are:
- Is the above proof correct?
- If so, which step of the proof uses the condition that $\{f_n\}$ is bounded? Is this condition crucial?
Any help will be greatly appreciated!