Rudin RCA p.32 exercise 10
Let $(X,\mathfrak{M},\mu)$ be a measure space and $\mu(X)<\infty$ and $\{f_n\}$ be a sequence of bounded complex measurable functions on $X$, and $f_n\rightarrow f$ uniformly on $X$. Prove that $\lim_{n\to\infty}\int_X f_n d\mu =\int_X f d\mu$.
Is 'boundedness' essential here? Here's my argument below and please tell me where am i misunderstanding..
Fix $\epsilon>0$. Then, there exists $N\in\mathbb{N} \text{ such that } n≧N \Rightarrow |f_n - f|< \epsilon$.
Let then a sequence $\{|f_n - f|\}_{n≧N}$ is dominated by $\epsilon$ and ,by assumtion, $\int_X \epsilon d\mu <\infty$. Thus, by Lebesgue's dominated theorem, $\lim_{n\to \infty} \int_X |f_n - f| d\mu = \int_X \lim_{n\to\infty} |f_n-f| d\mu = 0$. Q.E.D.
Thank you in advance