$\phi:\mathbb{R}^n\rightarrow\mathbb{R}$ is continuous at all points in $\mathbb{R}^n-Y$ with $Y\subset\mathbb{R}^n$ of measure zero. Then $\phi$ is measurable. In the problem, the measure is the Lebesgue measure.
Proof) Let $c\in\mathbb{R}$. We prove $E:=\{x\in\mathbb{R}^n:\phi(x)>c\}=(E-Y)\cup(E\cap Y)$ is measurable. First, $(E\cap Y)\subset Y$, so $m^*(E\cap Y)\leq m^*(Y)=0$ and $E\cap Y$ is measurable as it is measure zero.
Next, for all $x\in E-Y$, $\phi$ is continuous at $x$, so $\exists\delta>0$ such that $B_{\delta}(x)\subset E$. Then $E-Y=(\bigcup\limits_{x\in E-Y}B_{\delta}(x))-Y=(\bigcup\limits_{x_n\in E-Y}B_{\delta}(x_n))-Y$. For the last equality, we used the Lindeloff covering theorem to take countable balls. Then $E-Y$ is measurable as open sets are measurable, the union of countably many open sets is measurable, $Y$ is measurbale and the difference bewteen measurable sets is measurable.
I was wondering if this solutions is correct.