I want to prove the following statement
if $A$ has volume, then $$ \left( \operatorname{vol}(A) = \int_A \chi_A \right) = \inf\left\{ \sum_{i=1}^\infty \operatorname{vol}(S_i) \ | S_1, S_2,\cdots \text{is an open cover of $A$ by boxes} \right\} $$ where $\chi_A$ denotes the characteristic function of $A$ and the volume of a set is the Riemann integral of its characteristic function. i.e. $\int_A \chi_A$. A set has volume simply means the Riemann integral exists.
My updated attempt. Let $S_1,S_2,\cdots$ be a countable cover of $A$. Since $vol(A) \leqslant \sum_{i=1}^N vol(A_i)$ if $A_1,\cdots,A_N$ is a finite cover of $A$, by induction we can show that this is true for any countable cover of $A$. So $\sum_{i=1}^\infty vol(S_i) \geqslant vol(A)$.
Now, let $\epsilon > 0$ be given. Let $B$ be a box containing $A$. Since $vol(A) = \inf\{ U(\chi_A, P) \}$, where the infimum is taken over all partitions of $B$, there exists a partition $P$ of $B$ such that $U(\chi_A, P) < vol(A) + \epsilon$. Let $B_1P$ denote the collection of boxes that intersect $A$, then $B_1P$ covers $A$ and $U(\chi_A, P) = \sum_{S \in B_1P} vol(S) < vol(A) + \epsilon$. Thus, $vol(A) = \inf \left\{ \sum_{i=1}^\infty vol(S_i) | A \subset \bigcup_{i=1}^\infty S_i \right\}$
This time, I tried to prove $vol(A)$ is the infimum of all countable covers of $A$ by applying the property of infimum. Is my proof correct this time? Any comment is appreciated. Thanks!
EDIT: I would like modify the first part of my proof. Let $S_1,S_2,\cdots$ be a countable cover of $A$. Since $A$ has volume and therefore $\partial A$ has measure zero, there exists $T_1,T_2,\cdots$ such that $\partial A \subset \bigcup_{i =1}^\infty T_i, \sum_{i = 1}^\infty v(T_i) < \epsilon$. Then $A \cup \partial A = \overline{A} \subset \left(\bigcup_{i =1}^\infty S_i \right) \cup \left(\bigcup_{i =1}^\infty T_i \right)$. $\overline{A}$ compact implies there exists a finite set $\{ S_1, \cdots, S_n, T_1, \cdots, T_m \} \subset \{S_1, S_2, \cdots, T_1, T_2, \cdots \}$ which covers $\overline{A}$. So $$ vol(A) = vol(\overline{A}) \leqslant \sum_{i = 1}^n v(S_i) + \sum_{i = 1}^m v(T_i) < \sum_{i = 1}^\infty v(S_i) + \epsilon $$ Since $\epsilon$ is arbitary, $\sum_{i=1}^\infty vol(S_i) \geqslant vol(A)$.