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 $[0,1]$".
Best Answer
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 $\mu(A) = 1$.