Let $A\subset\mathbb{R}$ be bounded with $$\mu^*(A\cap I)<\mu(I)$$ for all open intervals I. Does A have necessarly a measure of $0$?
I actually just did a simple proof on how for every $A$ there should be a cover of disjunct intervals $B_k$ which I then used to get the inequality for arbitrary $I$. (Obviously) it worked well, until I realized that since the inequality must hold for every $I$, one must consider that $I\subset A$. In this case we have equality and I've got a problem.
However I assumed, that the all bounded subsets $A$ of $\mathbb{R}$ and all "not-null-sets" need to be intervals. Are there maybe other "not-null-sets" I should consider to disprove the assertion?
I'd appreciate any help.
No. Let $A$ be a fat Cantor set (this is its official name in math, not kidding). This is like the Cantor set, but instead of removing middle-thirds at each step (as in the usual construction), remove e.g. middle-fourths, or middle-fifths (perhaps the details should be done more carefully, but it works, e.g. at step $n$ remove a family of open intervals the total length of all of them $<\varepsilon/2^n$, so then in the end the fat Cantor set would have measure at least $1-\varepsilon$). Then $\mu(A)>0$ (that is why $A$ is called fat), yet like the usual Cantor set, $A$ is nowhere dense, so for every open interval $I$ there is a smaller open interval $J$ missing $A$, which implies that $\mu(A\cap I)<\mu(I)$.
This example may be given in some real analysis books (e.g. one is listed in the above Wikipedia link). One good book for this topic is by Oxtoby, Measure and Category, very well-known just google it (I think it is easily available online).