Line segment always has content zero

Question

I am new to measure theory, and I got stuck on the first exercise.

Show that every line segment has content zero.

Can someone help me as how to prove that something has content zero?

And one more doubt: Is measure $0$ same as content $0$?

My definition of content zero:

Let $A$ be a bounded subset of a plane. The set $A$ is said to have content zero if $\forall \epsilon>0,\exists $ a finite set of rectangles whose union contains $A$ and whose sum of areas doesnot exceed $\epsilon$.

Answer 1

As $\;A\;$ is a finite, bounded segment of line, we can parametrize it as

$$A=\left\{\;(x,\,f(x))\in\Bbb R^2\;/\;x\in[0,1]\;,\;\;f(x)\;\text{is a linear function}\;\right\}$$

Let us now take any countable sequence of intervals covering $\;\Bbb R\;$ , say $\;\left\{\;(p_n,\,p_n+1)\;\right\}\;$ (for example, you can take the rationals made into a sequence, and let us define now for any $\;\epsilon>0\;$

$$K_n:=\left\{\;\left(x,\,f(x)-\frac\epsilon{2^n}\right)\times\left(x,\,f(x)+\frac\epsilon{2^n}\right)\;/\;x\in(p_n,\,p_n+1)\;\right\}\subset\Bbb R^2$$

It's easy now to see that $\;A\subset\bigcup\limits_{n\in\Bbb N}K_n\;$ and that each $\;K_n\;$ is open. By compacticy of $\;A\;$ (why?) we get that there exists a finite sequence $\;n_1,n_2,...n_k\;$ of indexes s.t. $\;A\subset\bigcup\limits_{i=1}^k K_{n_i}\;$, and

$$|A|\le\left|\bigcup\limits_{i=1}^k K_{n_i}\right|\le\left|\bigcup\limits_{n\in\Bbb N}K_n\right|=\sum_{n=1}^\infty\frac{2\epsilon}{2^n}=\epsilon\,,\,\,\text{and we're done}$$

Note: even if $\;A\;$ is considered an "open" segment of line we can always "close" its endpoint(s) and make it closed.

Answer 2

A rectangle with non-zero finite width and length is a bounded set so a finite union of them is also a bounded set and cannot cover an unbounded set such as a line.

When $S$ is a subset of the plane and $F$ is a family of open rectangles with $\cup F \supset S$ there is a countable (finite or infinite) $G\subset F$ such that $\cup G\supset S.$ For $S$ to have measure $0$ it means that the value of $\sum_{g\in G}A(g)$ can be arbitrarily small (where $A(g)$ is the area of $g.$)

Let $S=\mathbb R\times \{0\}.$ For $r>0$ let $$G(r)=\{(n,n+2)\times (-r2^{-|n|},r2^{-|n|}):n\in \mathbb Z\}.$$ Then $\cup G(r)\supset S$ and $\sum_{g\in G(r)}A(g)=8r$ which can be as close to $0$ as desired. So the measure of $S$ is $0.$ Any line can be obtained by rotating S about some point; the same rotation can be applied to any $G(r).$ So the measure of any line is $0.$

If you want only open rectangles whose sides are parallel to the co-ordinate axis: Let $g\in G(r)$ as above, with $A(g)=r2^{2-|n|}.$ Observe that if $R$ is a rotation then the image $R(g)$ can be covered by no more than $2^{|n|}(1+1/r)$ open rectangles with sides parallel to the axes rectangles, each with width and length $r2^{3-|n|}.$ Their areas add up to at most $$(r2^{3-|n|})^2\cdot (2^{|n|}(1+1/r))=2^{6-|n|}(r^2+r).$$ And note that $\sum_{n\in \mathbb Z}2^{6-|n|}(r^2+r)$ can be as small as desired.