If I have that $f$ is integrable, how show that for all $\varepsilon>0$, there is some $h>0$ for which we have that $$\int_{\{x\in X \colon |f(x)|<h\}} |f(x)|d m<\varepsilon$$ in a general measure space.
And how we can obtain such $h>0$ if a sequence $(f_{n})$ of integrable functions is such that $|f_{n}|<g$ almost everywhere and $g$ is integrable, with the objective to obtain that for all $n$, $$\int_{\{x\in X \colon |f_{n}(x)|<h\}} |f_{n}(x)|dm<\varepsilon?$$
First, find a set $A$ of finite measure such that $\int_A|f|\mathrm dm<\varepsilon/2$. Then $$\int_{\{|f|\lt h\}}|f|\mathrm dm\leqslant \varepsilon/2+m(A)\cdot h.$$
This time, we take the set $A$ such that $\int_A|g|\mathrm dm<\varepsilon/2$.