If $(f_n)_{n \in \Bbb N}$ converges to $f$ uniformly and each $f_n$ integrable would it imply $f$ is integrable and $$\lim_{n \to \infty}\int f_n = \int f$$
In case
each $f_n$ is nonnegative
general integrable functions $f_n$
I am considering Lebesgue integration.
No. Consider the sequence of functions given by $$f_n(x)=\begin{cases} \frac1n&\text{if}\ 0<x<n^2\\ 0&\text{otherwise.} \end{cases}$$ Then for any $x>0$, $|f_n(x)|\leq\frac1n$ so $f_n\to f:=0$ uniformly. However, $$\int_0^\infty f_n(x)\ dx=n\to\infty$$ whereas $$\int_0^\infty f(x)\ dx=0.$$ This counterexample still has $f$ integrable but you get the idea.
In the case where all the $f_n$ are continuous and we are looking at a compact interval $[a,b]$ then your statement holds by a simple application of the dominated convergence theorem.