# Jordan measure on the irrationals

1. Sep 21, 2016

### cragar

1. The problem statement, all variables and given/known data
Show that ${[0,1]}^2-{\mathbb{Q}}^2$ has jordan inner measure zero.

2. Relevant equations
Jordan inner measure is defined as
$m_{*}J(E)=sup (m(A))$
where $A \subset E$ E is elementary.
elementary means a rectangular measure in n dimensions.
3. The attempt at a solution
I imagine the proof relies on the fact that there is no sup in E that is elementary that contains all the irrationals in the unit square.
Assume for contradiction that there is a greatest lower bound, call it (x,x) this ordered pair is elementary and is the greatest lower bound for A and A is a subset of the irrationals in the unit sqare. So the sup(A) is an elementary box, so its measure is x*x, now x is some real number between 0 and 1. But it cant be equal to 1 because that is a rational number. So lets assume x is the sup(A).
if x is the sup of A, then $\frac{x+1}{2}=q$ but q is larger than x and less than 1. so x is not the sup(A). because E contains irrationals larger than x , because the irrationals are dense in R.
So E cannot have a positive measure, Since the measure of Q^2 is zero, it means its complement is measureable, so its complement has measure zero,

2. Sep 22, 2016

### andrewkirk

The definition of Jordan Inner Measure with which I am familiar uses the term 'simple set' rather than 'elementary'. But whichever term is used, it is not a rectangular measure or any other sort of measure. It is a set - specifically a set composed of a finite union of disjoint rectangles, where each rectangle is the product of half-open intervals of the form $[a,b)$. Its measure is the sum of the measures of those rectangles.

For the measure to be nonzero, there must be at least one collection of such products of half-open sets that is contained in $S\triangleq [0,1]^2-\mathbb Q^2$. Can there be such a collection? In fact, can there even be a single $[a,b)\times [c,d)$ that is contained in $S$?
Why/why not?

3. Sep 22, 2016

### cragar

oh ok thanks for the response, It can be a union of a finite number of rectangular boxes, because it has holes at the rational points (x,y).
It has to be a finite number of boxes, If it was a box of any width it would not be a subset, because it would contain rational points, which our set does not.
I guess to formalize the proof, Lets assume for contradiction that such a finite set of boxes exist, by defintion these boxes exist over some [a,b)x[c,d).
but this Cartesian product contains rational points, because rationals are dense in R, therefore this is a contradiction that such a set exist, therefore no box exists over this set.

4. Sep 22, 2016

### andrewkirk

Correct. Now remember that our boxes are of the form $[a,b)\times [c,d)$. If they have no width, in either direction, it will be of the form $[a,a)\times [b,b)$. How many points does such a box have in it?

5. Sep 22, 2016

### cragar

each box has one point in it, and the measure of one point is zero. Regular measure theory only allows for countable unions. But there are unaccountably many irrationals. Do we say it breaks down because we can only have finite number of boxes. To me this doesnt prove its measure is zero, it kind of shows the axioms are breaking down.

6. Sep 22, 2016

### andrewkirk

Think about that again. The interval $[a,b)$ is the set of all points $x$ such that $x\geq a$ and $x<b$. How many points will that be if $a=b$?

7. Sep 23, 2016

### cragar

oh ok, so there wont be any points in the box because it is an open interval.
thanks for your response. In my book it defines elementary measure as a possible interval of being open or closed. [a,b), (a,b), so there are 4 possible intervals. So how do you know which interval to use?

Last edited: Sep 23, 2016
8. Sep 23, 2016

### andrewkirk

Good. So the only elementary sets that fit inside $S$ are unions of sets the form $[a,a)\times [b,b)$ which are all empty sets. So the only elementary set that fits inside $S$ is the empty set, whose measure is zero.