Given the proposition (*), I ask you whether the outline of the reasoning is correct.
Proposition (*)
If a sequence of measurable functions $\{f_n\}_{n\ge1}:X\mapsto\bar R\text{ or }\bar C$ converges to the measurable function $f:X\mapsto R\text{ or } C$ almost uniformly (a.u.) on $E\in{\cal S}(X)$ and each $f_n$ is almost everywhere (a.e.) bounded on $E$, then $f$ and, for every $n$, $f_n$ are almost uniformly (a.u) bounded on $E$.
Proof
Because of $\{f_n\}_{n\ge1}$ converges to $f$ a.u. on $E$, given $\delta>0$ there exist $E_\delta\in{\cal S}(X)$, $E_\delta\subseteq E$ with $m(E\setminus E_\delta)<\delta$, and $\bar n$ such that for all $n\ge \bar n$ we have
\begin{equation}(**)\sup_{x\in E_\delta}|f_n(x)-f(x)|<1.\end{equation}
Combining the fact that each $f_n$ is a.e. bounded on $E$ with the equation (**), we can say that there exists a null set $N_\delta\subseteq (E\setminus E_\delta)$ such that for all $n$ $|f_n|<M(n)$ on $(E\setminus N_\delta)\supseteq E_\delta$. Then
\begin{align} &\sup_{x\in E_{\delta}}\big(|f(x)|-|f_{\bar n}(x)|\big)\le\sup_{x\in E_{\delta}}|f_{\bar n}(x)-f(x)|<1\nonumber\\ \Rightarrow&\sup_{x\in E_{\delta}}|f(x)|<\sup_{x\in E_{\delta}}|f_{\bar n}(x)|+1< M(\bar n)+1.\nonumber \end{align}
Let $|f_1(x)|< M(1),\ldots,|f_{\bar n-1}(x)|< M(\bar n-1)$ for all $x\in E_\delta$, so
\begin{align} &\sup_{x\in E_\delta}\big(|f_n(x)|-|f(x)|\big)\le\sup_{x\in E_\delta}|f_n(x)-f(x)|<1\nonumber\\ \Rightarrow &\sup_{x\in E_\delta}|f_n(x)|<\sup_{x\in E_\delta}|f(x)|+1<M(\bar n)+2.\nonumber \end{align}
Put $M=\max\big(M(1),\ldots,M(\bar n-1),M(\bar n)+2\big)$. Then
$$\sup_{x\in E_\delta}|f(x)|<M,\quad\text{ e }\quad\sup_{x\in E_\delta}|f_n(x)|<M, \forall n$$
that is $f$ and the sequence $\{f_n\}_{n\ge1}$ are a.u. bounded on $E$.
Note that the set $(E\setminus N_\delta)$ contains points belonging to $(E\setminus E_\delta)$, where the sequence $\{f_n\}_{n\ge1}$ does not converge uniformly to $f$, and so
$$\sup_{x\in (E\setminus N_\delta)\setminus E_\delta}|f_n(x)-f(x)|\ge1.$$
Moreover $\sup_{x\in (E\setminus N_\delta)\setminus E_\delta}|f_n|< M'(n)$ for all $n$. By these inequalities we can't say if $f$ is bounded on $(E\setminus N_\delta)\setminus E_\delta$. Anyhow we could have said this if the sequence $\{f_n\}_{n\ge1}$ were converging to $f$ uniformly a.e. on $E$.
Thank you so much for the time deserved.
My reasoning is incorrect, as the formula
$$(**)\sup_{x\in E_\delta}|f_n(x)-f(x)|<1,$$
dealing with the supremum, only implies that the difference between $f_n$ and $f$ is finite on $E_\delta$ and we cannot state that $f$ and each $f_n$ are bounded on that set.
Putting together the hypothesis that $f_n$ is a.e. bounded on $E$ with equation (**) we can only state that there exists a null set of $N_\delta\subseteq E$ such that for all $n\ge\bar n$ we have $|f_n|< M(n)$ on $(E\setminus N_\delta)\cap E_\delta$ and $\sup_{x\in (E_\delta\setminus N_\delta)}|f_n(x)-f(x)|<1$. Therefore
\begin{align} &\sup_{x\in(E_\delta\setminus N_\delta)}\big(|f(x)|-|f_{\bar n}(x)|\big)\le\sup_{x\in (E_\delta\setminus N_\delta)}|f_{\bar n}(x)-f(x)|<1\nonumber\\ \Rightarrow&\sup_{x\in(E_\delta\setminus N_\delta)}|f(x)|<\sup_{x\in(E_\delta\setminus N_\delta)}|f_{\bar n}(x)|+1< M(\bar n)+1.\nonumber \end{align}
Let now $|f_1(x)|< M(1),\ldots,|f_{\bar n-1}(x)|< M(\bar n-1)$ forall $x\in (E_\delta\setminus N_\delta)$, then
\begin{align} &\sup_{x\in (E_\delta\setminus N_\delta)}\big(|f_n(x)|-|f(x)|\big)\le\sup_{x\in (E_\delta\setminus N_\delta)}|f_n(x)-f(x)|<1\nonumber\\ \Rightarrow &\sup_{x\in (E_\delta\setminus N_\delta)}|f_n(x)|<\sup_{x\in (E_\delta\setminus N_\delta)}|f(x)|+1<M(\bar n)+2.\nonumber \end{align}
Taking $M=\max\big(M(1),\ldots,M(\bar n-1),M(\bar n)+2\big)$, then
$$\sup_{x\in (E_\delta\setminus N_\delta)}|f(x)|<M, \quad\text{ e }\quad\sup_{x\in (E_\delta\setminus N_\delta)}|f_n(x)|<M, \forall n$$
so $f$ and the sequence $\{f_n\}_{n\ge1}$ are a.e. uniformly bounded on $E_\delta$ and uniformly bounded on $(E_\delta\setminus N_\delta)$.
Observe that the set $(E\setminus N_\delta)$ contains points on $(E\setminus E_\delta)$, where the sequence $\{f_n\}_{n\ge1}$ does not converge uniformly to $f$, where therefore
$$\sup_{x\in (E\setminus N_\delta)\setminus E_\delta}|f_n(x)-f(x)|\ge1$$
and $\sup_{x\in (E\setminus N_\delta)\setminus E_\delta}|f_n|< M'(n)$ for all $n$. Thus from these inequalities we cannot claim that $f$ is bounded on $(E\setminus N_\delta)\setminus E_\delta$.
Eventually, the following claim is not true:
``Anyhow we could have said this if the sequence $\{f_n\}_{n\ge1}$ were converging to $f$ uniformly a.e. on $E$.''.
In the light of the former reasoning sketch, the starting proposition (*) will be as follows:
Let a sequence of measurable functions $\{f_n\}_{n\ge1}:X\mapsto \bar R\text{ or }\bar C$ be convergent to the measurable function $f:X\mapsto R\text{ or }C$ almost uniformly (a.u.) on $E\in{\cal S}(X)$ and each $f_n$ be almost everywhere (a.e.) bounded on $E$. Given $\delta>0$, there exist a null set $N_\delta$ and a set $E_\delta\in{\cal S}(X)$, $E_\delta\subseteq E$, $m(E\setminus E_\delta)$, such that $f$ and, for every $n$, $f_n$ are uniformly bounded on $E_\delta\setminus N_\delta$ and a.e. uniformly bounded on $E_\delta$.