Theorem 15.28.94AC87EC7D266

Proof. We construct $f=(1+Q)/M$, where $Q\geq 0$ concentrates on closed nowhere dense sets in such a way that it has essential infimum zero on every nonempty open set.

Step 1: the pieces. Let $(U_ i)_{i\geq 1}$ be a countable basis of nonempty open subsets of $X$. We construct inductively numbers $\alpha _ j>0$, nonzero bounded densities $q_ j$, and closed nowhere dense sets $K_ j\subset X$ such that

Start with $\alpha _1=1/4$. Once $\alpha _ j$ has been chosen, theorem 12.15.4, theorem 12.2.3 and $v_-(h)=0$ give a smooth strictly positive probability density $g_ j$ with $c_ h(g_ j)<\alpha _ j2^{-j}$. Removing from $X$ an open dense set of small volume, we obtain a closed nowhere dense set $K_ j$ for which $q_ j=g_ j\mathbf1_{K_ j}$ is nonzero and, since $g_ j$ is bounded, arbitrarily close to $g_ j$ in $L^2$. By lemma 15.27.2 with $p=2$, we can therefore choose $K_ j$ so that $c_ h(q_ j)\leq \alpha _ j2^{-j}$.

After $K_1,\ldots ,K_ N$ have been chosen, put

each set inside the minimum is open and nonempty, since a finite union of closed nowhere dense sets is closed and nowhere dense. Choose

and continue. In particular,

Step 2: the concentrating part. The function $Q=\sum _ j\alpha _ jq_ j$ is nonnegative and integrable. We claim that its essential infimum on every nonempty open set is zero. Since every nonempty open set contains some $U_ i$, it suffices to consider $U_ i$. Take $N\geq i$ and remove $K_1,\ldots ,K_ N$ from $U_ i$. On the remaining set, whose volume is at least $m_ N$, the first $N$ summands of $Q$ vanish. By Markov's inequality, the set where the tail $\sum _{j>N}\alpha _ jq_ j$ exceeds $2^{-N}$ has volume at most

Hence $Q\leq 2^{-N}$ on a subset of $U_ i$ of positive volume. Letting $N\to \infty $ proves the claim.

Step 3: $b_ h(f)=0$. Set $M=\int _ X(1+Q)\, \mathrm{d}V$ and $f=(1+Q)/M$, a probability density with $f\geq 1/M$. If $\mathrm{MA}^{\mathrm{ext}}_ h(u)\geq af\, \mathrm{d}V$ with $a>0$, then $\mathrm{MA}^{\mathrm{ext}}_ h(u)\geq (a\alpha _ j/M)q_ j\, \mathrm{d}V$, and lemma 15.27.1 gives

This is impossible. Hence $b_ h(f)=0$ by theorem 15.28.4.

Step 4: $C_ h^{\mathrm{sm}}(f)>0$. Let $g$ be a smooth strictly positive probability density and put $R=\| g/f\| _{L^\infty }$. Then $g\leq R(1+Q)/M$ almost everywhere, and this forces $g\leq R/M$ everywhere. Otherwise, by continuity, $g>R(1+\varepsilon )/M$ on a nonempty open set for some $\varepsilon >0$, while $Q<\varepsilon $ on a subset of positive volume of this set by Step 2, a contradiction. The order and homogeneity of the scalar in lemma 15.27.1 now give $c_ h(g)\geq c_ h(R/M)=(M/R)c_ h(1)$, that is,

The constant density $g=1$ attains this value, since $\operatorname *{ess\, inf}_ XQ=0$ gives $\| 1/f\| _{L^\infty }=M$. Thus $C_ h^{\mathrm{sm}}(f)=M c_ h(1)>0$, as required. $\square$