# Example of a set that has zero measure but nonzero volume. [duplicate]

A set $$E\subset R^n$$ is said to have volume $$0$$ if; for every $$\epsilon>0,$$ there exists a finite family of rectangles $$U_1,..,U_m$$ such that; $$E\subset \bigcup_{i=1}^mU_i$$ $$\text{and }\sum_{i=1}^m V(U_i)< \epsilon.$$

My book constructs an example of a set that has zero measure but a non-zero volume. It is given below;

Let $$E$$ be the set of rational numbers in the closed interval $$[0,1]$$. Then $$E$$ is countable, and so it has measure $$0$$. If the sets $$U_i=[a_i,b_i]$$ for $$i=1,2,\dots,m$$ cover $$E$$ then they will also cover $$[0,1]$$. Therefore, $$\chi_{[0,1]}(x)\leq\sum_{i=1}^m\chi_{[a_i,b_i]}(x).$$

Here $$\chi_{[0,1]}(x)$$ means that if $$x\in[0,1],$$ then it's either $$1$$ or $$0,$$ or I am mistaken?

Then;

$$1=\int_0^1\chi_{[0,1]}(x)dx\leq\int_0\sum_i\chi_{[a_i,b_i]}(x)dx$$ (why upper value of integral is not written?) $$=\sum_{i=1}^m(b_i-a_i)$$(how author get this?) $$=\sum_{i=1}^m V(U_i)$$

Can you please explain above parts.

• Presumably measure = Lebesgue measure here. What is your definition of volume? May 14, 2022 at 12:39
• edited my question May 14, 2022 at 12:43
• Does this answer your question? Example of a countably infinite set of measure zero that has a positive volume May 14, 2022 at 12:45
• @DietrichBurde No because I am not familar with Jordan measure yet. May 14, 2022 at 12:46
• See this post. May 14, 2022 at 12:47

$$1=\int_0^1\chi_{[0,1]}(x)dx\leq\int_0^1\sum_i\chi_{[a_i,b_i]}(x)dx=\sum_{i=1}^m\int_0^1\chi_{[a_i,b_i]}(x)dx=\sum_{i=1}^m(b_i-a_i)=\sum_{i=1}^m V(U_i)$$
• Why $\int_0^1\chi_{[a_i,b_i]}(x)dx = b_i-a_i$ May 14, 2022 at 13:11
• Here $\chi_{[0,1]}(x)$ means that if $x\in[0,1],$ then it's either $1$ or $0,$ or I am mistaken? May 14, 2022 at 13:14
• I can't make sense of that! $\chi_{[0,1]}(x)$ means $1$ if $x\in[0,1]$, and $0$ if $x\notin[0,1]$. So if $[a,b]\subseteq[0,1]$, then clearly $\int_0^1\chi_{[a,b]}(x)dx=b-a$. May 14, 2022 at 15:52