Show that if $\{f_n\}$ a sequence of measurable function on $A$ converges pointwise almost everywhere to $f$, then $f$ is measurable on $A$.
We define a set $B$ to be the set of all $x\in A$ such that $\{f_n\}$ does not converge pointwise to $f$ . By definition, a sequence $\{f_n\}$ of function defined on $A$ is said to converge almost everywhere to a function $f$ if $$\lim_{n\rightarrow\infty}f_n(x)=f(x)$$ for all $x\in A\backslash B$, where $B\subset A$ and $m(B)=0.$
Then $\{x\in A|f(x)>c\} = \{x\in A\backslash B|f(x)>c\}\cup\{x\in B|f(x)>c\}$. Since $\{x\in B|f(x)>c\}\subseteq B$ and $m(B)=0,$ it follows that $\{x\in B|f(x)>c\}$ is measurable and has measure 0. Since $\{f_n\}$ converge pointwise to $f$ on $A\backslash B,$ it follows that $f$ is measurable on $A\backslash B$ and hence $\{x\in A\backslash B|f(x)>c\}$ is measurable. Since union of measurable sets is measurable, it follows that $\{x\in A|f(x)>c\}$ is measurable. Therefore, $f$ is measurable on $A$.
Is my proof correct ? The way I proved this question was similar to the technique I used in Continuous Almost Everywhere on $A$