$\lim\limits_{n\rightarrow \infty} f_n = f$ ,$f_n\in L^1(X,\mathcal{F},\mu)$ and $\int|f_n|\,\mathrm{d}\mu<M$ for all $f_n$ then $f\in L^1$

Question

Given $f_n: X \rightarrow \mathbb{C}$ and $\lim\limits_{n\rightarrow \infty} f_n = f$ (In the question it says point wise but I assume the intention was $\mu .a.s$ cause we work with $L^1$ functions). Such that $f_n\in L^1(X,\mathcal{F},\mu)$ and $\int_X|f_n|\,\mathrm{d}\mu<M$ for all $f_n$ then the limit function $f\in L^1$.

I was trying to search for uniformly convergence, and also using Fatu's lemma:

$\int \liminf |f_n| \le \lim\inf\int|f_n| < M$, But I don't see whether it may lead to any conclusions about $f$. Also the questions doesn't say anyting about $\mu$ finality, so Egoroff's theorem can't be used...

Answer 1

So $\displaystyle\int|f|=\int\liminf_{n}|f_{n}|\leq\liminf_{n}\int|f_{n}|\leq M<\infty$, so $f\in L^{1}$.

Answer 2

Note that $$\lim_{n\to\infty} f_n(x) = f(x)~~ \mu-a.e\implies \liminf_{n\to\infty} f_n(x) =\lim_{n\to\infty} f_n(x) = f(x) ~~\mu-a.e $$

Hence, $$\int|f|d\mu=\int\liminf_{n}|f_{n}|d\mu\leq\liminf_{n}\int|f_{n}|d\mu\leq M<\infty$$ that is $f\in L^1(\mu)$