If $f$ is zero in the interior, $f$ has zero integral

Question

Let $f:A\to\mathbb{R}$ be bounded, where $A$ is some box in $\mathbb{R}^d$. Suppose that $f(A^{\circ}) = 0$. I want to show that $\int_{A}fdx = 0$. So my approach is: Let $P$ be a partition of the box $A$ to equi-volume boxes. Let $\mathcal{C}$ be the set of boxes generated by the partition. For $C\in\mathcal{C}$, let $M_C=\sup\limits_{x\in C}f(x)$, $m_C = \inf\limits_{x\in C}f(x)$. Then the upper sum for the partition is $$ U(f,P) = \sum\limits_{C\in\mathcal{C}}M_CVol(C) = \sum\limits_{C\in\mathcal{C},\ C\cap\partial A = \varnothing}M_CVol(C) + \sum\limits_{C\in\mathcal{C},\ C\cap\partial A\neq\varnothing}M_CVol(C) $$ By hypothesis, only the second summand can be non zero, hence $$ U(f,P) = \sum\limits_{C\in\mathcal{C},\ C\cap\partial A\neq\varnothing}M_CVol(C) $$ Since $f$ is globally bounded, then for some $L$, $L\geq M_C$ for all $C\in\mathcal{C}$, hence $$ U(f,P) \leq L\sum\limits_{C\in\mathcal{C}, C\cap\partial A\neq\varnothing}Vol(C) $$ Since the partition is of equi-volume boxes, $Vol(C) = \frac{Vol(A)}{n}$ where $n$ is the number of boxes. We therefore get that $$ U(f,P)\leq LVol(A)\frac{k}{n} $$ where $k$ is the number of boxes in the partition that intersect the boundary. The difference between general dimension and 1D is that in 1D, $k = 2$ always. For higher dimensions, $k = k(n)$, so information on $\frac{k}{n}$ is needed in order to decrease the RHS to be arbitrarily small, and this is where I am stuck

Answer 1

$$\int_A f = \int_{\partial A} f + \int_{A^{\circ}} f = \int_{\partial A} f \leq \textbf{max} (f) \cdot \mu_{\mathbb{R}^d}(\partial A)$$ Since $\textbf{dim}(\partial A) = d-1$, what can you say about $\mu_{\mathbb{R}^d}(\partial A)$?

Answer 2

Because $A$ is a box, you can safely assume that each $C$ is a box and that there are $\sqrt[d]{n}$ boxes along each dimension. Now, in $k$ dimensions, a box has a boundary composed of $2d$ boxes of dimension $d-1$, and each of these boxes can be covered with $\sqrt[d]{n}^{d-1}$ boxes, so $k=2d\cdot\sqrt[d]{n}^{d-1}$ and you can now estimate $\frac{k}{n}$ to see it becomes arbitrarily small.