# Half of the real numbers, homogenously

1. Sep 6, 2008

### jostpuur

Does there exist a set $$X\subset\mathbb{R}$$ that has a property

$$m^*(X\cap [0,x]) = \frac{x}{2},\quad\quad\forall x>0,$$

where $$m^*$$ is the Lebesgue outer measure?

My own guess is that this kind of X does not exist, but I don't know why. Anybody knowing proof for the impossibility of this X?

2. Sep 7, 2008

### acarchau

Loosely speaking most sets of positive measure are very close to being unions of intervals and in some sense if $$m(X) > 0$$ then $$m(X\cap[x-\epsilon,x+\epsilon]) \approx 2\epsilon$$ for "most" $$x\in X$$.

There clearly does not exist a X which is lebesgue measurable.
Assume the existence of such a X.
Since we have assumed X to be measurable, we can work with the lebesgue measure $$m()$$ instead of outer measures.

Now,
Choose any $$0 \leq a < b$$. Then $$m (X \cap [0,b]) = m(X \cap [0,a]) +m(X \cap [a,b]) - m(X \cap \{a\})$$
hence,
$$m(X \cap [a,b]) = \frac{b-a}{2}$$

and since points have measure 0, it follows that
$$m(X \cap [a,b]) =m(X \cap (a,b])=m(X \cap (a,b))=m(X \cap [a,b)) =\frac{b-a}{2} (1)$$

The proof hinges on the following: Given any measurable set E and an arbitary $$\epsilon > 0$$, there exists an open set O such that
$$E \subseteq O$$ and $$m(O) \leq m(E) + \epsilon$$ (note: corrected after jostpuur pointed out a typo)

So choose such an O for X.

O being open can be written as $$O=\cup_{i\in A}^{\infty} O_i$$ where $$O_i$$ are disjoint open intervals in $$[0,\infty)$$ and A is a finite or countable index set, clearly $$m(O)= \sum_{i\in A} m(O_i)$$.

From (1) we have for all i, $$m(X \cap O_i) = m(O_i)/2$$

Clearly, $$X = X \cap O = \cup_{i\in A} X\cap O_i$$
and hence,
$$m(X) = \sum_{i\in A} m(X\cap O_i) = \sum_{i\in A} m(O_i)/2 = m(O)/2 \leq (m(X)+\epsilon)/2$$

and we get
$$m(X) \leq \epsilon$$

Since $$\epsilon > 0$$ was arbitary we must have $$m(X) = 0$$

In which case we also must have $$m(X\cap[0,x]) = 0$$ for all x.

A contradiction. So no such X exists.

I won't be surprised if the result holds for arbitrary X.

3. Sep 7, 2008

### jostpuur

I see. Isn't it also true, that for arbitrary set $$X\subset\mathbb{R}$$, there exists a measurable set $$\overline{X}$$ such that $$X\subset\overline{X}$$ and $$m^*(X)=m(\overline{X})$$, so that that deals with the non-measurable case?

edit: Or it could be that this idea starts to lead towards unnecessarily complicated proof. Perhaps one could start dealing with the open covers, used in the definition of the outer measure, directly?

I suppose it would be safest to replace the original $$X$$ with for example $$X\cap [0,1]$$, so that its outer measure becomes finite. It is sufficient to hypothesize the property $$m^*(X\cap[0,t])=t/2$$ for 0<t<1.

Last edited: Sep 7, 2008
4. Sep 7, 2008

### acarchau

I noticed a subtle mistake in my proof which can be corrected, jostpuur perhaps you can figure out how to correct it.

$$m(X) \leq (m(X)+\epsilon)/2$$

implies, $$m(X) \leq \epsilon$$ if $$m(X) < \infty$$

So the proof only holds for measurable sets of finite measure and clearly measure of X cannot be finite.

5. Sep 7, 2008

### jostpuur

Yeah, I just managed to mention about that. It is not a problem, because we can carry out the proof "locally". If the X, that was described in the first post, existed, we could just take its intersection with some finite interval, and it would still have the same strange behavior locally there.

Then there was a typo here:

The intented inequality must have been

$$m(O) \leq m(E) + \epsilon$$

6. Sep 7, 2008

### acarchau

You are correct, thanks!

7. Sep 7, 2008

### jostpuur

Hehe. Thanks to you for the proof

8. Sep 7, 2008

### jostpuur

I think I managed modifying the proof so that it can be carried out with outer measures. This equation

would require measure, but outer measure has the property

$$m^*(\bigcup_{n=1}^{\infty} A_n) \leq \sum_{n=1}^{\infty} m^*(A_n)$$

which is enough for the proof.

So if there is $$X\subset [0,1]$$ with the property $$m^*(X\cap [0,t])=t/2$$ for all $$0<t<1$$, then let $$\epsilon >0$$ be arbitrary. By definition of the outer measure, and by the fact that if two open intervals are not disjoint, then they together are one interval, there exists a sequence of open intervals $$]a_k,b_k[$$, $$k=1,2,3,\ldots$$, so that

$$X\subset \bigcup_{k=1}^{\infty}\; ]b_k,a_k[$$

and

$$\sum_{k=1}^{\infty} (b_k-a_k) \leq m^*(X) + \epsilon.$$

Then

$$X = \bigcup_{k=1}^{\infty} \big(X\cap \;]b_k,a_k[\big)$$

so that by the property mentioned earlier we get

$$m^*(X) \leq \sum_{k=1}^{\infty} m^*(X\cap\; ]b_k,a_k[) = \sum_{k=1}^{\infty} \frac{b_k-a_k}{2} \leq \frac{1}{2}(m^*(X) \;+\; \epsilon).$$

Isn't this right now too?

hmhmh.... I see some little difficulties. The equation

$$m^*(X\cap \;]b_k,a_k[) = \frac{1}{2}(b_k-a_k)$$

requires some explanation, and it could happen that some of these intervals go outside the $$[0,1]$$.... but these don't look fatal problems.

Last edited: Sep 7, 2008
9. Sep 7, 2008

### acarchau

the approach appears sound

10. Sep 7, 2008

### jostpuur

I'm still puzzled by this all. If I define following sequence of sets:

$$X_1 = [0,\frac{1}{2}[$$

$$X_2 = [0,\frac{1}{4}[\;\cup\;[\frac{1}{2},\frac{3}{4}[$$

$$X_3 = [0,\frac{1}{8}[\;\cup\;[\frac{1}{4},\frac{3}{8}[\;\cup\;[\frac{1}{2},\frac{5}{8}[\;\cup\;[\frac{3}{4},\frac{7}{8}[$$

...

Then for example $$m^*(X_{10^{10^{100000000}}}\cap [0,t])\approx \frac{t}{2}$$ would be true to great accuracy. One might think, that if you can device a set that has that property approximately, but with arbitrary preciseness, then we could also device, as some kind of limit, a set that has this property precisely

What's happening there?

11. Sep 7, 2008

### acarchau

Is $$X_n$$ the set of all x in [0,1] whose nth bit, after the "decimal" point, is 0 in the binary representation?

12. Sep 7, 2008

### Hurkyl

Staff Emeritus
Whatever notion of limit you come up with, at least one of the following will be true:

1. The limit isn't a set, but instead some generalized object
2. Lesbegue outer measure won't be continuous
3. The limit won't exist

13. Sep 7, 2008

### acarchau

My take is this:

1. $$X_n$$ is constructed by dividing [0,1] into $$2^n$$ equal intervals and dropping alternate ones, the idea being that only half of any "good" set lies in $$X_n$$, upto a resolution. This makes $$X_n$$ to be the set of points having their nth bit 0, it is a disjoint union of $$2^{n-1}$$ intervals having dyadic rationals as their endpoints.

2. Limit of $$X_n$$ does not exist, for instance 0.1010101010... lies in $$X_2, X_4 \ldots$$ but not in $$X_1, X_3 \ldots$$

3. $$\sup_{t \in [0,1]} | m(X_n \cap [0,t]) -t/2 | \leq \frac{1}{2^n}$$, hence $$\lim_{n \to \infty} \sup_{t \in [0,1]} | m(X_n \cap [0,t]) -t/2 | = 0$$

14. Sep 7, 2008

### jostpuur

The remark with disjointness was useless, btw. It stayed there from the previous proof that did not use outer measures.

Very clear thinking.