I want to finish this proof.
OP has considered the interval differently. I choose here, in $\mathbb{R}^p$, the interval $a_i - (\epsilon/2)^{(1/p)} ≤ x_i ≤ b_i + (\epsilon/2)^{(1/p)}$ (the closed set $F$). I am trying to show that $m(A)≤m(F)+ \epsilon$.
Clearly, $F \subset A$.
We break up the interval into three disjoint parts,
$a_i -\epsilon/2 ≤x_i < a_i $ , $a≤x_i≤b_i$ , $b_i<x_i≤b+ \epsilon/2$ (denoting $I_1, I_2, I_3 $, respectively).
Now, using the additive property of $m$, $m(F)= m(I_1)+m(I_2)+m(I_3) =2( (\epsilon/2)^{1/p})^p+m(A)=\epsilon +m(A)$ which gives us finally, $m(A)≤ m(A)+2 \epsilon \implies m(A) ≤ m(F)+ \epsilon$
The the method valid ? (the other case can be done similarly, I guess)