Question:
$(X,\mathcal{M},\mu)$ measure space. Suppose that $\{f_n\}\subset L^1$ and $f_n\rightarrow f$ uniformly. Show that if $\mu(X) <\infty$, then $\int f_n\rightarrow \int f$.
Proof:
I exploit uniform convergence and $\{f_n\}\subset L^1$ to say that $\forall x$ $\exists N$ s.t. $n\geq N \Rightarrow |f_n|\leq M$ where M is finite, then I build the dominating function $g:=M \chi_{[X]}$ and apply the DCT to the sequence $\{f_n\}_{n\geq N}$ and then claim that $\lim_{n\geq N\rightarrow\infty}\int f_n=\lim_{n\rightarrow\infty}\int f_n$.
Makes sense? (I am new to math, please show mercy)
Not quite: The functions $\frac{1}{\sqrt{x}} - \frac{1}{n}$ converge uniformly to $\frac{1}{\sqrt{x}}$ on $(0, 1)$ even though none of these functions is bounded. More generally, there's no reason to expect an $L^1$ function to be bounded.
What you can say, however, is that there exists an $N$ such that whenever $n \ge N$, $|f(x) - f_n(x)| < \epsilon$ for all $x$ in the space. Then estimate
$$\left|\int f_n - \int f\right| \le \int |f_n - f| \le \int \epsilon = \epsilon \mu(X) $$
Alternatively, using uniform continuity, you can show that (eventually) the function $$g(x) = |f(x)| + 1$$ dominates your sequence.