Exercise: Suppose that $\mu(X) < +\infty$, $(f_n)_n$ is a sequence of bounded complex measurable functions on $X$ and $f_n \to f$ uniformly on $X$. Prove that: $$ \lim _{n \to +\infty} \int _X f_n d\mu = \int _X f d\mu $$
Now, I noticed that $$\text{bounded} \hspace{2mm} f_n, \hspace{3mm} \mu(X)<+\infty \implies f_n \in L^1(\mu)$$ therefore I tried to prove the theorem by replacing boundedness with $f_n \in L^1(\mu)$.
Proof: By uniform convergence on $X$ and by $\mu(X)<+\infty$, it is easy to see that $f \in L^1(\mu)$: $$\exists N: |f| \leq |f-f_N| + |f_N| \leq 1 + |f_N| \hspace{2mm} \text{in X} \implies \int _X |f| d\mu \leq \mu(X) + \int _X |f_N| d\mu < +\infty$$ Moreover, the uniform convergence implies that: $$\exists M: \forall n \geq M, |f-f_n| \leq 1 \hspace{2mm} \text{in X}$$ so that the sequence $(|f-f_n|)_n$ is a sequence of measurable functions dominated by an $L^1(\mu)$ function (namely the constant $1$ because $\mu(X) < +\infty$); hence we can apply the Dominated Convergence Theorem: $$ |f-f_n| \to 0 \hspace{2mm} \text{in X} \implies \lim _{n \to +\infty} \int _{X} |f-f_n| d\mu = 0$$ Finally, since both $f$ and $f_n$ belong to $L^1(\mu)$, we can apply the following inequality: $$ \left| \int _X f - \int _X f_n \right| d\mu \leq \int _X |f-f_n| d\mu$$ which, together with the previous result, after a proper rearrengement yields: $$ \lim _{n \to +\infty} \int _X f_n d\mu = \int _X f d\mu $$ as desired.
It is clear that $\mu(X)<+\infty$ is essential to the proof and thus I guess also to the statement itself. For this reason I asked myself what would happen without this condition. If I require boundedness as in the original statement, then there is an easy counter-example: $$ f_n = \frac{1}{n}, \hspace{2mm} \text{$\mu$ counting measure on infinite X} $$ however I couldn't find a counter-example in the case in which I require $f_n \in L^1(\mu)$. Does anyone know such a counter-example? In general, what can we say if we have a sequence of Lebesgue Measurable functions?
A counter-example is $f_n=\frac 1 n 1_{(n,2n)}, f=0$ on $\mathbb R$ with Lebesgue mesure.