Show that if $f$ is continuous almost everywhere on $A$, then $f$ is measurable on $A$.
We define a set $B$ to be the set of all $x\in A$ such that $f$ is not continuous on these $x$ . By definition of continuous almost everywhere, $f$ is continuous at all $x\in A\backslash B$, $m(B)=0$. 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 $f$ is continuous 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 ?
Edit : Sorry for my wrong statement of problem. I am proving "almost everywhere."
Your proof is not correct. First, $B$ is just the empty set. Second, you state in your proof "since $f$ is continuous on $A\setminus B,$ $f$ is measurable on $A\setminus B.$" It seems here you're using the result you're trying to prove.
A better strategy might be to show that the preimage under $f$ of Borel sets is measurable. You can do this by making use of the fact that $f^{-1}(U)$ is open for open $U.$