Only one thing is not clear to me.
Proposition. Let $\{f_n\}_{n\in\mathbb {N}}$ a sequence of integrable function no-negative such that $f_{n}\le f_{n+1}$ if \begin{equation} \lim_{n\to\infty}\int_Xf_n\;d\lambda=0, \end{equation} then \begin{equation} \lim_{n\to\infty}f_n<\infty\quad\text{a.e.} \end{equation}
proof. \begin{equation} 0=\lim_{n\to\infty}\int_Xf_n\;d\lambda=\sup_n\int_Xf_n \;d\lambda, \end{equation} then \begin{equation} \int_Xf_n\;d\lambda=0\quad\text{for all $n\in\mathbb{N}$}, \end{equation} therefore $f_n=0$ a.e.for all $n\in\mathbb{N}$. Now \begin{equation} \lambda(\{x:f_n(x)>0\})=0. \end{equation} We observe that \begin{equation} \{x:\lim_nf_n(x)>0\}=\cup_n\{x:f_n(x)>0\}\quad(1) \end{equation} then \begin{equation} \lambda(\{x:\lim_nf_n(x)>0\})=\lambda(\cup_n\{x:f_n(x)>0\})=\lim_n\lambda(\{x:f_n(x)>0\})=0. \end{equation} Is it correct to proceed in this way for 1?
If $x\in\cup_n\{x:f_n(x)>0\}$ $\Rightarrow$ $\exists n_0\in\mathbb{N}$ such that $f_{n_0}(x)>0$ $\Rightarrow$ $\sup_n\{f_n(x)>0\}$ $\Rightarrow$ $\lim_n f_n(x)=\sup_n\{f_n(x)\}>0$. Then \begin{equation} \bigcup_n\{x:f_n(x)>0\}\subseteq \{x:\lim_nf_n(x)>0\}. \end{equation} If $x\in\{x:\lim_nf_n(x)>0\}$ $\Rightarrow$ $\sup_n\{f_n(x)\}>0$ $\Rightarrow$ $\exists n_0$ such that $f_{n_0}(x)>0$ $\Rightarrow$ $x\in\cup_n\{x:f_n(x)>0\}$.
Thanks!
Your argument is correct.
As an aside, you are actually showing that $\lim_{n \to \infty} f_n = 0 \ \mathrm{a.e.}$. So, perhaps one more line could be added, namely that $$ \{ x : \lim_{n \to \infty} f_n(x) = \infty \} \subseteq \{ x : \lim_{n \to \infty} f_n(x) > 0 \}, $$ and from the fact that the measure of any subset of a null set is zero (assuming that $\lambda$ is the Lebesgue measure here), your proof will be (more) complete.