The sharp interval bound for Cantor measure

Here $0<s<1$ and $3^s=2$. For $a,b,c\ge 0$ with $a,c\le b$, one has $a^s+c^s\le (a+b+c{)}^s$. Indeed the function $(x+y{)}^s-x^s$ is nonincreasing in $x>0$ for fixed $y\ge 0$, by its derivative; continuity extends this to $x=0$. Increasing $a$ to $b$ and then $c$ to $b$ can only decrease $(a+b+c{)}^s-a^s-c^s$, whose final value is $b^s(3^s-2)=0$. If $b=0$ all three numbers are zero.

Fix $m$. For each basic interval $J$ at a level $\mathrm{\ell }\le m$ and each interval $I$, let $N_J(I)$ count the level-$m$ basic descendants contained in $I\cap J$. We prove $2^{-m}{N}_J(I)\le \mathrm{diam}(I\cap J{)}^s$ by induction upwards from level $m$. At that level the count is zero or one; a count of one forces diameter at least $3^{-m}$ and hence the desired bound. Empty intersections have count and diameter zero.

For an earlier $J$, its two children have length $b$ and are separated by a gap of length $b$. If $I$ meets neither child the count is zero. If $I$ meets only one child, use its inductive bound and diameter monotonicity. If it meets both, put $a=\mathrm{diam}(I\cap J_{\mathrm{l}\mathrm{e}\mathrm{f}\mathrm{t}})$ and $c=\mathrm{diam}(I\cap J_{\mathrm{r}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}})$. Then $a,c\le b$ and $\mathrm{diam}(I\cap J)\ge a+b+c$. Adding the two inductive bounds and using the first step proves the claim for $J$. Thus for $J=[0,1]$ the total number $N_m(I)$ of level-$m$ intervals contained in $I$ satisfies $2^{-m}{N}_m(I)\le \mathrm{diam}(I{)}^s$.

For a bounded open interval $I$, let $E_m$ be the union of all level-$m$ basic intervals wholly contained in $I$. These sets need not increase as subsets of the line, but $E_m\cap C$ do increase: each point of $C$ lies in a child of its previous interval. Their union is $C\cap I$ because basic diameters tend to zero. They have measure $2^{-m}{N}_m(I)$ by disjointness, concentration and cylinder masses. Continuity from below therefore gives ${\mu }_c(I)\le \mathrm{diam}(I{)}^s$.

Other bounded interval endpoint conventions change at most two points, which have zero mass, so the same bound holds. A singleton has mass zero, and an empty interval has mass zero; an unbounded interval has infinite diameter and the inequality is immediate. Finally a nonempty bounded $U$ lies in $[\inf U,\sup U]$, whose length is exactly its diameter. This Borel superset bounds ${\mu }_c^{\ast }(U)$ as claimed.