# Difference between volume/measure zero and not having volume

I am taking a real analysis course and am trying to sort out the difference between a set's measure, and a set's volume (sometimes called content).

The motivating example is a practice problem: If $A_1,A_2,...$ have volume and $A=\bigcup_{i=1}^{\infty} A_i$ is bounded, need $A$ have volume? It turns out that this is not true, as a point has volume zero while the countable union of all rational points on $[0,1]$ does not have volume.

Volume is defined as $\int_{A}1_{A}(x) dx$

For measure, I have been using it interchangeably with volume. This this ok? The second question I have is what is the difference between a set having volume zero and a set NOT having volume? Thanks!

• I think you are asking about the difference between a region (of three dimensional space) having zero volume (zero measure) and a region that is not measurable. However your example doesn't illustrate that difference, so I'm not sure what your Question is about. The rational points in all space have zero measure (volume), as does a single point. Mar 29, 2016 at 23:56
• My book makes a distinction, that in $\mathbb{R}$, a single point has volume zero, while the set of all rational points in $[0,1]$ has no volume, which happens to be measure 0. I am trying to sort out the difference between zero volume, measure zero, no volume, un-measurable, Mar 29, 2016 at 23:58
• Your question does not clearly define volume. My best guess if that the integral in your question is Riemann integral (for if it were Lebesgue integral then the rationals would have volume $0$). Mar 30, 2016 at 0:08
• I figured it out. A bounded set $A$ has volume $\iff$ bd($A)$ has measure 0. This is the distinction. Mar 30, 2016 at 0:11
• Yeah, I think my book abuses measure and volume as terms. I figured it out. Thanks though! (And you are completely right, the rationals have measure zero). Mar 30, 2016 at 0:16

$$\text{Non-measurable} \implies \text{Does not have Jordan content}$$ $$\text{Has zero Jordan content} \implies \text{Zero measure}$$