Convergence almost everywhere and convergence in measure Let $(\mathbb{R},\mathcal{L},m)$, let $f_{n}(x)=n\chi_{[0,\frac{1}{n}]}$ then the sequence converges to $0$ everywhere except at $x=0$ thus $f_{n}$ converges a.e.
Then in my book (Folland) we have that if $f_{n}\to f$ a.e and $|f_{n}|\le g\in L^{1}$ then $f_{n}\to f$ in $L^{1}$ too. The above conditions are met.
Finally by another proposition we have that if $f_{n}\to f$ in $L^{1}$ then $f_{n}\to f$ in measure. 
I wonder if these relations holds for both finite and infinite measurable spaces? 

It actually work in any measure space as a consequence of the inequality 
$$\mu\{|f_n-f|\gt\varepsilon\}\leqslant \varepsilon\lVert f_n-f\rVert_{L^1}.$$
Since the $f_n$ have a $\sigma$-finite support, we can assume we work on a $\sigma$-finite space, even if it's not needed. 
