Let $(X,\mathfrak M,\mu)$ be a measure space and $f_n:X\rightarrow [-\infty,+\infty]$ a measurable function for all $n\in\mathbb N$. Suppose that $$\sum_{n=1}^\infty \left(\int_X\left|f_n\right|\,d\mu\right)<+\infty$$ and consider the series $$\sum_{n=1}^\infty \left|f_n\right|\;,$$ By the monotone convergence theorem we know that: $$\int_X\sum_{n=1}^\infty \left|f_n\right|\,d\mu=\sum_{n=1}^\infty \left(\int_X\left|f_n\right|\,d\mu\right)<+\infty$$ Now can I conclude that $\sum_{n=1}^\infty \left|f_n\right(x)|<+\infty$ for all $x\in X$?
I think that this condition is true only almost everywhere in $X$, because if the series attains infinite value on a set $E$ of measure $0$, then the integral on $E$ is $0$ (we have fixed the convention that $0\cdot\infty=0$).
You can conclude the claim for almost all $x$ because the function $$ \sum_{n=1}^{\infty} \lvert f_n \rvert $$ is integrable. However, the Lebesgue integral doesn't really care what happens on a set of measure $0$, so you can't up that to finiteness everywhere in $X$.