Folland's Advanced Calculus Chapter 4 Section 2 Exercise 3:
Let $S$ be a bounded set in $\mathbb{R}^2$. Choose a rectangle $Q=[a,b]\times[c,d]$ which includes $S$. Let $P=\{x_0,x_1,\dots,x_{n_1};y_0,y_1,\dots,y_{n_2}\}$ be a partition of $Q$. For $i=1,\dots,n_1$, $j=1,\dots,n_2$, put
$$Q_{ij}=[x_{i-1},x_i]\times[y_{j-1},y_j]\text{,}$$
$$m_{ij}=\inf_{t\in Q_{ij}}\chi_{S}(t)\text{,}$$
$$L(P,\chi_S)=\sum_{i,j}m_{ij}(x_i-x_{i-1})(y_j-y_{j-1})\text{,}$$
where $\chi_S$ is the indicator of $S$. The inner area $\underline{A}(S)$ of $S$ is defined to as
$$\underline{A}(S)=\underline{\int\int_S}dA=\sup L(P,\chi_S)\text{,}$$
the supremum being taken over all partitions $P$.Show that $\underline{A}(S)=\underline{A}(S^\circ)$.
Fix $\epsilon>0$. Suppose we can find a partition $P_0$ of $Q$ such that
$$|L(P_0,\chi_S)-L(P_0,\chi_{S^\circ})|<\epsilon\text{.}$$
Choose partitions $P_1$, $P_2$ so that
$$|\underline{A}(S)-L(P_1,\chi_S)|<\epsilon\text{,}$$
$$|\underline{A}(S^\circ)-L(P_2,\chi_{S^\circ})|<\epsilon\text{.}$$
Take $P^*=P_0\cup P_1\cup P_2$. We then have
$$|\underline{A}(S)-\underline{A}(S^\circ)|<3\epsilon\text{.}$$
It follows that $\underline{A}(S)=\underline{A}(S^\circ)$.
How to find such a $P_0$? It is clear that any rectangle which lies in $S^\circ$ lies in $S$, and that any rectangle which lies in $S$ and consists of interior points of $S$ lies in $S^\circ$. Hence the difference between $L(P,\chi_S)$ and $L(P,\chi_{S^\circ})$ is the sum of areas of those rectangles which lie in $S$ and contain boundary point of $S$. How to make sure the sum is small enough?
Best Answer
Clearly $A(S)\ge A(S^o).$ We show $A(S^o)>A(S)-r$ for any $r>0$. Let $P=\{x_0,x_1,\dots,x_{n_1};y_0,y_1,\dots,y_{n_2}\}$ be a partition of $Q$ such that $L(P,\chi_S)>A(S)-r/2.$
Let $T=\{(i,j):Q_{i,j}\subset S\}.$ There exists $\{a_i,b_i,c_j,d_j: (i,j)\in T$} such that whenever $(i,j)\in T$ we have $x_{i-1}<a_i<b_i<x_i$ and $y_{j-1}<c_j<d_j<y_j$ and also $(b_i-a_i)(d_j-c_j)>(x_i-x_{i-1})(y_i-y_{j-1})-r/(2n_1n_2).$
Let $|T|$ be the number of members of $T$.
Now we can expand $\{a_i,b_i;c_j,d_j:(i,j)\in T\}$ to a partition $P'$ of $Q.$ Since $(i,j)\in T\implies Q_{ij}\subset S\implies [a_i,b_i]\times [c_j,d_j]\subset S^o,$ and since $L(P,\chi_S)=\sum_{(i,j)\in T}(x_i-x_{i-1})(y_i-y_{i-1})$, we have therefore $$A(S^o)\ge L(P',\chi_{S^o})>L(P,\chi_S)-|T|\cdot (r/2n_1n_2)\ge L(P,\chi_S)-r/2>A(S)-r.$$