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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 $\sum v(A_i)<\epsilon$

Seven
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  • What's your definition of volume, if it is not measure? – Calvin Khor Feb 03 '19 at 11:01
  • sorry I thought it was standard. I'll edit – Seven Feb 03 '19 at 11:03
  • Oh, are you coming at this from the perspective of Riemann integration, rather than Lebesgue integration? – Calvin Khor Feb 03 '19 at 11:05
  • yes, my university does it this way. Maybe other approach would be better, but it's not my choice. – Seven Feb 03 '19 at 11:09
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    I believe this works, https://math.stackexchange.com/questions/2538567/prove-riemann-integral-is-zero – Calvin Khor Feb 03 '19 at 11:22
  • No, but it's my fault. The measure I'm talking about I think it is Lebesgue measure.So a set has measure zero if for all $\epsilon$ there exists a denumerable collection of rectangles which union contains the set and such that the sum of the volumes is less than $\epsilon$ – Seven Feb 03 '19 at 11:36
  • Sorry, but I think a further clarification is in order. On the one hand, if $A$ were also jordan measurable, this would imply that its jordan measure (being equal to the lebesgue measure) is 0, and the above would work. On the other hand, if $A$ were to be not jordan measurable, how do you make sense of $v(A) = \int_R 1_A$? – Calvin Khor Feb 03 '19 at 12:05
  • Well , what I need to prove is that the Jordan measure is zero, because that's equivalent to zero volume. But the question you said has as hypothesis that the set has zero Jordan measure. That has to be my conclusion. – Seven Feb 03 '19 at 12:12

2 Answers2

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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.

Calvin Khor
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  • wow, thank you. However, I hope to find a more direct way. I shouldn't need to use Jordan measure. – Seven Feb 03 '19 at 14:05
  • @Seven The main difficulty in your question is the nebulous way in which "volume is defined". It turns out by the above characterisation to not be nebulous, and also not a generalisation of the linked question. If one wanted a more general result, you can certainly use Lebesgue's measure theory to define $\int 1_A$ /as/ the volume of $A$ for any Lebesgue measurable set, which defines the question away. That said, there's probably a more direct route, I just used the above because my Riemann theory is rusty :) so I relied on quoting results with proofs, which needed to be pieced together. – Calvin Khor Feb 04 '19 at 10:49
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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 $v(A)=0$

Seven
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