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.

Fix an $\varepsilon > 0$, by uniform convergence, we know that there exists $N \in \mathbb{N}$ such that for $n \geq N$,
\begin{equation*}
|f_n| = |f_n - f + f| \leq |f_n - f| + |f| < |f| + \varepsilon
\end{equation*}
Then define the function $g : \Omega \to \mathbb{R}$ by $g(\omega) = |f(\omega)| + \varepsilon$. Then $g$ is integrable, since we work on a finite measure space. If it was an infinite measure space, the $"+ \varepsilon"$ part would give some difficulties. Next, define $h_n = f_{N + n}$, with $\lim\limits_{n \to \infty} h_n = f$, for which it holds that $|h_n| \leq g$. Then, all conditions of the dominated convergence theorem are satisfied, hence we can conclude
\begin{equation*}
\lim_{n \to \infty} \int_\Omega h_n \mathrm{d}\mu= \int_\Omega f \mathrm{d}\mu \tag{$\ast$}
\end{equation*}
Lastly, observe that the difference between $\{f_n\}$ and $\{ h_n \}$ is only an shift in indices, hence it immediately follows from ($\ast$) that
\begin{equation*}
\lim_{n \to \infty} \int_\Omega f_n \mathrm{d}\mu = \lim_{n \to \infty} \int_\Omega h_n \mathrm{d}\mu= \int_\Omega f \mathrm{d}\mu 
\end{equation*}
