Frostman's lemma
Frostman’s lemma in metric spaces
Definitions
Let $(X,d)$ be a metric space and $E\subseteq X$. Let $\operatorname{diam}(\cdot)$ be the diameter of a set.
The $s$-dimensional Hausdorff measure $\mathcal H^s(E)$ of $E$, $s \geq 0$ is defined by
\[\begin{align} \mathcal H^s(E) & := \lim_{\epsilon \to 0} \mathcal H^s_\epsilon(E), \\ \mathcal H^s_\epsilon(E) & := \inf\{ \sum_{j=1}^\infty (\operatorname{diam} U_j)^s : U_j \text{ open cover of $E$ with $\operatorname{diam}U_j < \epsilon$} \}, & 0 \leq \epsilon \leq +\infty. \end{align}\]We actually have $\mathcal H^s_\epsilon(E) \nearrow \mathcal H^s(E)$ as $\epsilon\to 0$, and $\mathcal H^s$ is a measure on $X$ for every $s\geq 0$.
Moreover, we have that $\mathcal H^s(E) = +\infty$ for $s < s_0$, $\mathcal H^s(E) = 0$ for $s > s_0$ and the unique $s_0\geq 0$ for which this occurs is the Hausdorff dimension of $E$ , denoted by $\dim_\mathcal H E$. The value of $\mathcal H^s(E)$ at $s = \dim_{\mathcal H} E$ can be any value in $[0,+\infty]$.
Finally, let $P(E)$ denote all the Borel probability measures supported on $E$.
Introduction to Frostman’s lemma
Assume that a set $E$ supports a positive Borel measure $\mu$ such that for every ball $B$ of radius $r$ we have, for some $s>0$,
\[\begin{align} \label{eq:frostman1} \mu(B) & \leq r^s. \end{align}\]It follows then that for any open cover $U_j$ of $E$ we have:
\[\begin{align} 0 < \mu(E) \leq \sum_{j=1}^\infty \mu(U_j) \leq \sum_{j=1}^\infty (\operatorname{diam}U_j)^s, \end{align}\]and by taking the infimum over all such covers, we obtain $\mathcal H^s(E) > 0$.
Frostman’s lemma is the converse: if $\mathcal H^s(E) > 0$, then a positive Borel measure supported on $E$ and satisfying \eqref{eq:frostman1} exists.
Frostman’s lemma on compact metrix spaces. (nonconstructive)
Let $E$ be a compact metric space. If $\mathcal H^s(E) > 0$ then there’s $\mu\in P(E)$ and $C>0$ with $\mu(B) \leq Cr^s$ for all balls $B$ of radius $r$.
Proof
Define $p : C(E) \to \mathbb C$ by
\[\begin{align} p(f) & := \inf\{\sum_{j=1}^\infty c_j (\operatorname{diam}E_j)^s : \text{Numbers } c_j \geq 0 \text{ and sets } E_j \text{ such that } \sum_{j=1}^\infty c_j 1_{E_j} \geq f\}. \end{align}\]The function $p$ is a nonnegative sublinear functional:
- $p(f) \leq |f|_\infty \operatorname{diam}E$.
- $p(f) = 0$ for $f \leq 0$.
- $p(tf) = tp(f)$ for $t\geq 0$.
- $p(f + g) \leq p(f) + p(g)$ since a bound for $f$ and a bound for $g$ adds to a bound for $f + g$.
- \(p(1_{E}) = \mathcal{H}^{s}_{\infty}(E)\), comparing definitions. Thus \(p(1_{E}) > 0\).
Thus by the Hahn-Banach theorem there’s some nonnegative functional $\mu$ (which is a Radon measure by the Riesz-Markov-Kakutani representation theorem) with
\[\begin{align} \mu(E) & = p(1_E), \\ | \int f d\mu| & \leq p(f), & \text{(thus $\mu$ nonnegative),} \\ \mu(B) & \leq \sup_{\substack{f \in C(E) \\ f \leq 1_B}} p(f) \leq (2r)^s. \end{align}\]Normalize to obtain a probability measure.
Frostman’s lemma on complete separable metric spaces.
Frostman’s lemma holds in this case too.
The compactness condition in the above section can be removed. In our construction the sublinear functional $p$ does not take any infinite values, but it could be the case that $\mathcal H^s(X) = +\infty$ for a general metric space without any compact subsets $E\subset X$ with $\mathcal H^s(E) > 0$. However, this does not occur for the standard Hausdorff measure; there are other Hausdorff measures, constructed by functions that are different from the standard power functions $t\mapsto t^s$, which exhibit this peculiar behavior. The counter-example is constructed in Davies, Rogers 1 and the proof that Frostman’s lemma holds for analytic subsets of complete separable metric spaces is Corollary 7 in Howroyd 2, effectively proving (among other things) that for any analytic $A\subseteq X$ there exists some compact $E\subseteq A$ with $0 < \mathcal H^s(E) < +\infty$ if $\mathcal H^s(A) > 0$.
Frostman’s lemma on Euclidean spaces. (constructive, original proof)
Let $E \subset \mathbb R^n$ compact. If $\mathcal H^s(E) > 0$ then there’s $\mu\in P(E)$ and $C>0$ with $\mu(B) \leq Cr^s$ for all balls $B$ of radius $r$.
Definitions
Let $\mathcal Q_k(E)$ denote all the dyadic half-open cubes of side length $2^{-k}$ intersecting $E$.
Proof
Assume $E\subset [0,1)^n$. It suffices to show that $\mu(Q) \leq r^s$ for every dyadic half-open cube $Q\subseteq [0,1)^n$, since any set of diameter at most $2r$ is containted in at most $2^n$ dyadic cubes with side length $r$. We wish to construct measures $\mu_k$ decreasing to $\mu$ such
\[\begin{align} \label{eq:frostman} \mu(Q) \leq r^s \end{align}\]for every dyadic half-open cube of level at most $k$.
Naive construction
In all our constructions, we define our measures to be multiples of the Lebesgue measure. When we define $\mu(Q) = \lambda$, it means that $\mu$ restricted on $Q$ equals $\lambda$ times the normalized Lebesgue measure (i.e. divided by the Euclidean volume of $Q$).
- Idea: Define $\nu_k(Q) := 2^{-ks}$ for $Q \in \mathcal Q_k(E)$.
- Issue: $\nu_{k}([0,1)^n) = 2^{k(n-s)}$. Thus \eqref{eq:frostman} does not hold.
- Why does this happen? Some parts of $E$ may be “too dense” to look $s$-dimensional at the $2^{-k}$-scale.
- Solution: Look at the previous cubes in $Q\in \mathcal Q_{k-1}(E)$ and correct $\nu_k(Q)$ if necessary by making it smaller. One proceeds with a finite induction of $k$ steps correcting all the way up to $\mathcal Q_0(E)$.
Finite induction
We are going to construct $\nu_k, \nu_{k-1}, \dots, \nu_0$ inductively, with $\mu_k := \nu_0$.
For the base case, define $\nu_k(Q) := 2^{-ks}$ for $Q\in \mathcal Q_k(E)$ and $\nu_k := 0$ otherwise.
For the induction, if $\nu_j(Q) \leq 2^{-(j-1)s}$, define
\[\begin{align} \left.\nu_{j-1}\right|_Q := \left.\nu_{j}\right|_Q. \end{align}\]Otherwise, if $\nu_j(Q) > 2^{-(j-1)s}$ for some $Q\in\mathcal Q_{j-1}(E)$, then redefine $\nu_{j-1}(Q) := 2^{-(j-1)s}$. (this impacts all the next-level cubes “under” $Q$; it makes their measures smaller).
Define $\mu_k := \nu_0$ and consider its properties:
- $\mu_k([0,1)^n) \leq 1$, by the correction step.
- $\mu_k(Q) \leq r^s$ for every dyadic half-open cube of radius at least $2^{-k}$, ensured by the initial construction and correction steps.
- $\mu_k(X) \geq \mu_{k+1}(X)$ for all Borel $X\subseteq [0,1)^n$, by comparing the intermediate measures constructed for $\mu_{k}$ and $\mu_{k+1}$.
By the last property, $\mu(X) := \lim \mu_k(X)$ is well-defined and $\mu(Q) \leq r^s$ for all dyadic cubes of side length $r$. Moreover, $\mu(E) \gtrsim \mathcal H^s_\infty(E)$ since the bound holds for each $\mu_k(E)$:
\[\begin{align} \mu_k(E) & = \sum_{Q\text{ initial or corrected}} (\operatorname{side}Q)^s \geq 2^{-s/2}\mathcal H^s_\infty(E). \end{align}\](Note that $\mathcal H^s_\infty(E) > 0$ since $\mathcal H^s_\infty(E) > \min{\mathcal H^s_\epsilon(E), \epsilon^s}$ for all $\epsilon > 0$ and $\mathcal H^s_\epsilon(E) \nearrow \mathcal H^s(E) > 0$ by hypothesis)