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 $\epsilon$.
Best Answer
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.