Originally Posted by

**craigmain** Hi,

I am struggling to provide sound mathematical proofs for these. Intuitively it's pretty straight forward.

Prove that the following sets is measurable and has zero area.

1) A set of a finite number of points in a plane.

Let $\displaystyle X:=\{x_1,...,x_n\}\subset\mathbb{R}$ be our finite set, and choose any $\displaystyle \epsilon >0$

Then $\displaystyle \displaystyle{\left\{\left(x_i-\frac{\epsilon}{3n},\,x_i+\frac{\epsilon}{3n}\righ t)\right\}_{i=1,2,...,n}$ is a cover of the

set $\displaystyle X$ with length $\displaystyle <\epsilon\Longrightarrow$ by definition, its measure is zero.

Try now to mimic the above for a finite union of finite sets in the plane, this time taking (open, say) rectangles

instead of open intervals...

Tonio

2) The union of a finite collection of line segments in a plane?

Each finite line segment has width zero, so area zero. A point has length zero and height zero so it's zero.

Given

$\displaystyle

given: h \ge 0, and, k \ge 0

$

$\displaystyle

rectangles\ are\ sets\ such\ that\ : (x,y)\mid \{x \le k, y \le h\}

$

Therefore we can say that for each line k is zero, and h is the length of the line, therefore the area is zero.

For each point h and k are zero, and therefore the area is zero.

Why are the set's measurable though?

How do we know mathematically there is one number c (zero) such that

$\displaystyle

a(S) \le c \le a(T)

$

or is this trivially demonstrated by the fact that h or k are zero above?

I want to be sure I understand this concept properly.

Thanks

Regards

Craig.