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
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.