$f_n\in L^1$ and $f_n \to f$ in a.e and $|f_n|\leq g$ where $g\in L^1$ then $f\in L^1$

Question

This is not duplicate I had specific question Please Help me

I want to prove Dominated convergence theorem: I understand the whole proof except for one part.

Why $f_n\in L^1$ and $f_n \to f$ in a.e and $|f_n|\leq g$ where $g\in L^1$ then $f\in L^1$?

I know that $|f|<|f-f_n|+|f_n|<\epsilon+g$ then $$ \int |f|<\int (\epsilon+g). $$ And I know that $$ \int g<\infty $$ However, even if I can choose $\epsilon >0$ as small as needed, I do not control its integral over the space as $\mu(X)=\infty$. Thus how Can I conclude $f\in L^1$?

Any Help will be appreciated

Answer 1

$|f_n| \leq g$ and $f_n \to f$ a.e. imply $|f| \leq g$ a.e. Hence $\int |f| \leq \int g<\infty$.

I have just used the fact that if $|a_n| \leq b$ and $a_n \to a$ then $|a| \leq b$.