A dense subset, $E$, of $[0,1]$ with measure $\frac{1}{2}$, and no proper subset of $E$ is an interval?

Question

I'm trying to find a subset of the unit interval that's analogous to the irrationals in some sense; it's dense in $[0,1]$, no subset of it is an interval, but it has a strictly smaller measure than the irrationals, while still having positive measure.

Having measure equal to $\frac{1}{2}$ is not strictly necessary; really any $\alpha \in (0,1)$ will suffice.

I'd be especially interested in seeing ig this set can be constructed, or if we can only know of it's existence via AoC.

edit: After a few replies, I realize that I'm also looking for something that has a certain "uniformity" of measure. in other words, if you give me some interval, $I$, of length $\epsilon$, then $m(E \cap I)$ is the same, regardless of where $I$ is centered (as long as $I$ is fully contained in $[0,1]$, of course). So in the case of $\alpha = \frac{1}{2}$, we might be doing something like "taking the first half of the irrationals, and spreading them out evenly over all of $[0,1]$".

Answer 1

Been dense is quite a weak requirement when we talk about measure (as rational numbers have measure zero while are still dense).

For non-uniform case, one simple example will be $E = \mathbb{Q} \cap [0, \frac{1}{2}] \cup ([\frac{1}{2}, 1] \setminus \mathbb{Q})$ - rationals on left half, irrationals on right half.

By Lebesgue's density theorem, you can't get uniformity you want. Indeed, as our set $A$ has positive measure, it has at least one point $x_0$ with density $1$.

Now, for any $\varepsilon > 0$ for some $n$ we have $\mu(A \cap [x_0 - \frac{1}{2n}; x_0 + \frac{1}{2n}]) > \frac{1 - \varepsilon}{n}$.

By additivity $\mu(A) = \mu(A \cap \bigcup\limits_{k=1}^n[\frac{k - 1}{n}, \frac{k}{n}]) = \sum\limits_{k=1}^n \mu(A \cap [\frac{k-1}{n}, \frac{k}{n}])$, and by uniformity every term of the last sum is equal and greater than $\frac{1 - \varepsilon}{n}$. Thus $\mu(A) > \sum\limits_{k=1}^n \frac{1 - \varepsilon}{n} = 1 - \varepsilon$.

As $\epsilon$ was arbitrary, it means $\mu(A) = 1$.

Answer 2

You cannot get the kind of regularity that you want: see the first result in this question.

Answer 3

I looked at this 15 years ago.

Consider the interval $[0,1]$. Remove from it each point which represents a dyadic fraction (one of the form $\frac a{2^m}$ in lowest terms, i.e. whose binary expression terminates, in this case with $a$ and $m$ positive, $a$ odd and $a<2^mm$) and an interval around it. The remaining set of points (if any remain) is nowhere dense, and if the intervals are chosen suitably then the measure of the remaining points will be between $0$ and $1$. In this case the intervals removed will be of the form

$$\left[\frac a{2^m} - \frac k{2^{2m}}, \frac a{2^m} + \frac k{2^{2m}}\right]$$

with $k$ is a real constant where $0<k<2$.

By changing $k$ you can continuously affect the measure remaining for a set which contains no intervals and is nowhere dense. With $k=1$ it is $0.2677868402178891123766714\ldots$ with the black intervals removed looking like

while with $k=\frac12$ it is double that at $0.5355736804357782247533428\ldots$ with the black intervals removed looking like