Measure zero implies volume zero when volume is defined I need to prove that if a set has measure zero and it has well-defined volume, then its volume is zero. I have tried to bound  the lower sums of the indicator function, but I'm stuck. Given a zero-measure set $A$ which volume is well-defined, if $R$ is a rectangle that contains $A$, we know there exists a partition $P_0$ of $R$ such that:
$$ \int_{-} 1_A \leq L(1_A, P_0) + \frac{\epsilon}2 = \sum_{Q\in P_0| Q\subset A}v(Q)\quad+\frac{\epsilon}2$$
I don't really know how to proceed without assuming things like denumerable subaditivity or monotony of volume function $v$. If you use some of these things to prove this please tell me also how to prove them.
EDIT: when I say "volume of the set $A$" I'm refering to the Riemann integral of the indicator function $1_A$ along $A$: $v(A) = \int_A 1_A$
EDIT2: when I say that a set has measure zero I mean that for all $\epsilon>0$ there exists a denumerable or finite collection of rectangles $R_1,R_2,...$ such that $A\subset\cup R_i$ and $\sum v(A_i)<\epsilon$
 A: The indicator function of $A$ is Riemann integrable iff $A$ is Jordan measurable, so the result linked in the comments applies. A sketch-
A (bounded) set $A$ here is defined to have volume $v(A)$ if for any bounded rectangle $R$ containing $A$, $1_A$ is Riemann integrable, and
$ v(A) = \int_R 1_A$. 
From Riemann integrability and the measure of the boundary , $v(A)$ is well defined iff $\partial A$ has (Lebesgue) measure $m(\partial A)= 0$. (We only need the direction $1_A$ integrable implies $m(\partial A)=0$, whose proof is the first one there.)
From Closure, Interior, and Boundary of Jordan Measurable Sets. , we see that $m(\partial A) = 0$ iff $A$ is Jordan measurable. However this page is a mess so I'll give a direct proof (sketch) of the important direction-
First note that $\partial A$ is a compact set; thus by passing to finite covers, $m(\partial A)=0$ implies its jordan content is $c(\partial A) = 0$. Now it should not be hard to find finite collections of rectangles $L_i,U_i$ such that $L=\bigcup_i L_i \subseteq A \subseteq U = \bigcup_i U_i$ and $c(U\setminus L)$ is arbitrarily small. Thus $A$ is Jordan measurable.
Since for Jordan measurable sets, $c(A) = m(A)$, we now know that $c(A)=0$. Now we can use the result here, Prove Riemann Integral is Zero.
A: Well, I had to solve it for myself.Maybe Calvin Khor's answer was right, but it wasn't was I was looking for. Continuing from where I stopped at the question, we notice that every rectangle $Q\subset A$ , as it is a subset of a zero-measure  set , is also a zero-measure set. As a rectangle, it is compact, and for compact sets zero measure is equivalent to zero volume. Then: $$\int_-1_A =  \sum_{Q\in P_0| Q\subset A} v(Q) = 0 $$ As $A$ has well-defined volume, it is neccessarily $v(A)=0$
