Lebesgue Measure - positive measure sets not containing intervals

Question

I've encountered two statements regarding the Lebesgue measure that don't exactly contradict each other, but seem to me to be a little bit unintuitive when regarded with respect to one another. The statements are:

1. For each interval $[a,b]$ there exists a set $K\subset[a,b]$ that is compact , totally disconnected and with positive measure (i.e. doesn't contain an interval).
2. Let $A\subset\mathbb{R}$ be some measurable set, and $0\leq\alpha<1$ such that for every interval $I\subset\mathbb{R}$ it holds that $m(A\cap I)\leq\alpha m(I)$ then $m(A)=0$

The second statement seems to imply that any positive measure set must contain an interval up to a set of arbitrary measure $\epsilon>0$ (since a suitable $\alpha$ can't be found for a positive measure set).

On the other hand the first statement says that for any interval I can find a subset of any desired positive "not-full" measure. I realize these statements don't contradict, but they seem a little "contradictory in nature". Intuitively I would expect the first statement to somehow allow me to construct a positive measure set with the property desired in the second statement.

I don't really have a question here but was hoping maybe someone can shed their own perspective on how they see these two properties of the Lebesgue measure co-existing. Maybe you can give me some further intuition.

Thanks a lot!

Answer 1

My reaction is just to trace through some examples. The set $K$ in part $1$ is something like a Cantor set-just fattened up to maintain positive measure, say for concreteness $1/2$. While we're simplifying, may as well take $[a,b]=[0,1]$ as well. $K$ needn't be uniformly distributed-indeed the point of your part 2 is that it can't be-so we may suppose we can remove some interval and increase the proportional measure of $K$. In the most usual construction, we might have $m(K\cap [0,3/8])=1/2$. Then if $K$ is self-symmetrical, as it may certainly be, we immediately get arbitrarily large proportions of $K$ in sufficiently small intervals.

This is all in contrast to the case of the usual Cantor set $C$: one way to prove $m(C)=0$ is to observe that for every interval $I$, $C$ is contained in a subset of $I$ of relative measure no more than $2/3$. For the construction of $C$ is uniform: at every stage we remove intervals of relative measure $2/3$. In contrast again this shows why we don't construct $K$ by uniform removal of even very small intervals.

Answer 2

For 1: There is an open set $U$ containing all rationals such that $m(U)< (b-a).$ Thus $m([a,b]\setminus U) > 0.$ The set $[a,b]\setminus U$ is compact, is a subset of $[a,b],$ and since it contains no rational, has no interior.