In Multivariable Mathematics by Dr. Shifrin, Prop. 1.8 in chapter 7 says
Suppose $f:R\to \mathbb{R}$ is a bounded function and the set $X=\{\mathbf{x}\in R:f \text{ is not continuous at }\mathbf{x}\}$ has volume zero. Then $f$ is integrable on $R$.
Here $R\subset\mathbb{R}^n$ is a closed rectangle. The proof then covers $X$ by finitely many closed rectangles $R_1', \dotsc, R_s'$ due to volume zero and the critical line where I'm stuck is "We can also ensure that no point of $X$ is a frontier point of the union of these rectangles".
My question is why can't a point of discontinuity of $f$ occur on the frontier of the rectangle $R$ and thus necessarily on the frontier of the union of $R_1', \dotsc, R_s' \subset R$?
Good catch! I think you are correct, in fact, but it won't affect the proof that follows. If a frontier point of $R$ is actually in $X$, covered, say, by the rectangle $R_\ell'$, then when we remove $\cup R_j'$ and take the closure, the frontier of $R_\ell'$ does not show up, and $f$ is still continuous on the complement.