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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.

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There seem to be two typos in that, and perhaps a missing step as well. Try this:

$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)$

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  • $\begingroup$ Why $\int_0^1\chi_{[a_i,b_i]}(x)dx = b_i-a_i$ $\endgroup$
    – unit 1991
    May 14, 2022 at 13:11
  • $\begingroup$ Here $\chi_{[0,1]}(x)$ means that if $x\in[0,1],$ then it's either $1$ or $0,$ or I am mistaken? $\endgroup$
    – unit 1991
    May 14, 2022 at 13:14
  • $\begingroup$ 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 $\int_0^1\chi_{[a,b]}(x)dx=b-a$. $\endgroup$
    – TonyK
    May 14, 2022 at 15:52

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