Let $$f_n: X \rightarrow \mathbb{R}$$ be a sequence of integrable functions converging uniformally towards $f$. Prove that for a finite measure, $\mu$
$$\lim_{n\rightarrow \infty}\int f_n d\mu = \int f d\mu $$
My teacher put online a proof of this implication using the dominated convergence theorem with the function
$g(x) = |f_1|+|f_2|+...+|f_n|+2$ where $\forall m \geq n$ we have $|f_m-f|\leq 1$. I understand why such a $g$ exists and why it's integrable, but I have trouble seeing why it would be greater than any $f_m$. (Obviously if $m\leq n$ it's trivial.)
Since you said that you are all right with the case $m\leq n$, I will just focus on the case $m>n$.
Let us set $g'(x):=|f_n(x)|+2$ with the hypothesis that $\forall m\geq n(|f-f_m|<1$ (so $g'<g$). Let $m>n$ be fixed. Let us evaluate $A:=g'(x)-f_m(x)=g'(x)-f(x)+f(x)-f_m(x).$
By hypothesis $f(x)-f_m(x)\geq-1$. Similarly, $f(x)_n-f(x)\geq-1$, which implies $g'(x)-f(x)\geq -1+2=1$. Hence, $A\geq 0$, which means that $g$ dominates $f_m$.