Dominated convergence theorem and uniformly convergence I try to solve the following task:
Let $(\Omega,\mathfrak{A},\mu)$ be a measurable space and $\mu(\Omega)<\infty$. Let $(f_n)_{n\geq1}$ be a sequence of integrable measurable functions $f_n:\Omega \rightarrow [-\infty,\infty]$ converging uniformly on $\Omega$ to a function $f$. Prove that $$\int f d\mu = \lim\limits_{n\rightarrow \infty} \int f_n d\mu$$.
What I thought: uniformly convergence of $f_n \rightarrow f$ means that $f$ is continuous and therefore measurable. Now I thought that I could use the dominated convergence theorem to show the equality.
uniformly convergence means $\lim\limits_{n\rightarrow \infty} \sup \{|f_n-f(x)|:x\in \Omega\}=0$ so I think I can define a function $s(x):=\sup \{|f_n-f(x)|:x\in \Omega\}=0$ which dominates all the $f_n$ and apply the theorem. But I'm not sure, if this is the correct way.

Uniform convergence and measurability of fn imply that f is measurable. Then |f| is measurable. We claim that f is integrable. Indeed, by uniform convergence, ∃ N s.t ∫|f| ≤ ∫ (|f - fn| + |fn|) ≤ ∫|f - fn| + ∫|fn| < ∞ for all n≥N, so f is integrable. Then by similar argument, ∫f - ∫fn = ∫ (f - fn) ≤ ∫|f - fn| ≤ ε·μ(Ω). Hence, the equality follows. 
