almost uniformly convergent implies converge in $L_1$.? I wanted to ask whether the following statements are true : (in all cases there exists $g \in L_1$ such that $|f_n| \le g$ for all n.

(1) almost everywhere convergent implies converge in $L_1$. (2) almost uniformly convergent implies converge in $L_1$.  (3) almost everywhere convergent implies converge in measure.

Only (1) is answered in this site but uses probability space and Scheffé's lemma which i don't know non of them. My question is about general measure space and search of many books or internet did not help me
 A: You'll find that these questions often depend upon whether the space has finite measure or not.
Let $(X, \mathcal{M}, \mu)$ be a measure space and suppose $f_n \to f$ is a.e. convergent.
This means that $\forall \epsilon>0$ and for a.e. $x\in X$ there is an $N(\epsilon, x)\in \mathbb{N}$ so that $|f_n(x)-f(x)|<\epsilon$. Let's call $E=\{x:|f_n(x)-f(x)|<\epsilon\}$ so $\mu(E^c)=0$
From here you can see that if $\mu(X)<\infty$ then we will definitely have $L^1$ convergence since
$$\int_X |f_n(x)-f(x)| d\mu=\int_E |f_n(x)-f(x)| d\mu<\epsilon \mu(X)$$
which means we have $f_n\to f$ in $L^1$.
But what if $\mu(X)=\infty$?
Consider $(\mathbb{R},\mathcal{B}, \lambda)$ and  $f_n=\chi_{[n,n+1]}$ where $\lambda$ is Lebesgue measure and $\mathcal{B}$ the Borel sigma algebra. Then we certainly have pointwise convergence to $0$ which meanse we have a.e. convergence but $\int_\mathbb{R} |\chi_{[n,n+1]}-0| d \lambda =1$ for every $n\in \mathbb{N}$ so we definitely don't have $L^1$ convergence.
However, in your case $|f_n|\leq g\in L^1$ so $\lim_{n\to \infty}\int |f_n-f|d \mu= \int \lim_{n\to \infty}|f_n-f|d \mu=0$ by dominated convergence theorem since $|f_n-f|\leq 2g$. So we actually have $L^1$ convergence in this case!
Suppose $f_n \to f$ is almost uniformly.  Again, if $\mu(X)<\infty$ we have $L^1$ convergence.  But consider $f_n=\frac{1}{n}\chi_{[0,n]}$ then this convergence is actually uniform to $0$ (you should check this) but again $\int_\mathbb{R} |\frac{1}{n}\chi_{[0,n]}-0| d\lambda =1$. so this sequence cannot converge in $L^1$
Suppose again $f_n \to f$ is a.e. so you would think that we have convergence in measure but this is also not true! Again the counterexample being $\chi_{[n,n+1]}$ which converges pointwise but not in measure.  This is because the choice of $N$ depends upon x.  However, uniform and almost uniform convergence implies convergence in measure.
I'll finish this by saying that each mode of convergence is looking at a different property and means something different.  I know they can be confusing but remember, everything is defined for a reason in math.  $L^1$ convergence for example allows us to perform analysis on the space $L^1$.  These are often important in PDE problems as well.
