Some small questions I've worked on.

A Minor Note on the Work of Preiss: July 30th, 2025

D.Preiss [1] constructed in 1981 a measure $\mu$ on an infinite-dimensional Hilbert Space $H$ and a set $C$ so $\mu(C) > 0$ but the density of $\mu$ at any point of $C$ is 0. We can alter this to get a measure $\nu$ of density any fixed value $\beta \in (0,1)$ across $C$. Indeed, Preiss sets up a pair of sequences $a_{k} \in \mathbb{R}_{>0}$ and $N_{k} \in \mathbb{N}$ such that
$ \sum_{k=1}^{\infty} a_{k}N_{1}...N_{k} < \infty \\ \lim_{k \to \infty} a_{k}N_{1}...N_{k+1} = \infty $
He then establishes a set $S$ of all finite sequences $(z_{1},...,z_{k})$ with $z_{i} \leq N_{i}$, and the set $Z$ of all infinite sequences with the same bound. First, define a function $h$ mapping each $(z_{1},...,z_{k})$ to $2^{-k}$ times $e_{z_1,...,z_k}$ (a member of an orthonormal basis indexed by the points of $S$), then a pair of functions
$ g(z) = \sum_{j=1}^{k} h(z_1,...,z_j) \hspace{0.5 cm} z = (z_1,...,z_k) \in S \\ f(z) = \sum_{j=1}^{\infty} h(z_1,...,z_j) \hspace{0.5 cm} z \in Z $
Lastly, consider a measure $\nu$ on $Z$ which is the product of $\nu_{j}$ on $\{1,2,...,N_{j}\}$ where $\nu_j(n) = N_{j}^{-1}$. We may then define
$ u = f\nu + \sum_{S} a_{k}\delta_{g(z_1,...,z_k)} $
where $f\nu$ is the pullback measure and $\delta_{x}$ is the dirac measure at $x$. Setting $C = f(Z)$, Preiss shows that for $2^{-k} \leq r^2 < 2^{-k+1}$,
$ u(B(x,r) \cap C) = (N_{1}...N_{k+1})^{-1} \\ a_{k} \leq u(B(x,r)) $
which together imply the density result.

We now break from Preiss' work and modify the process slightly. Let $N_{1}...N_{n} = b_{n}$. If $k$ is the least number $i$ so $y_i \neq z_i$, then
$ ||g(y_{1},...,y_{m}) - f(z)|| \geq ||f(y)-f(z)|| - ||g(y_1,...,y_m) - f(y)|| \geq 2^{\frac{-k+2}{2}}-2^{\frac{-k}{2}} = 2^{-k/2} $
Then, if $g(y_1,...,y_n) \in B(x,r)$ as above, we must have that $y_i=z_i$ for all $i \leq k$. There is one such $y$ of length $k$, and $N_{k+1}$ of length $k+1$, etc. Hence, we have that
$ u(B(x,r)) \leq (N_{1}...N_{k+1})^{-1} + a_{k} + \sum_{i=k+1}^{\infty} a_{i}b_{i}b_{n}^{-1} $
We alter the assumptions on our requisite sequences. Pick $a_{n}$ and $N_{n}$ inductively so that for $\alpha \in (0,\infty)$,
$ \lim_{n \to \infty} a_{n}b_{n+1} = \alpha \\ \sum_{k=1}^{\infty} a_{k}b_{k} < \infty \\ \sum_{k=n+1}^{\infty} a_{k}b_{k} < \frac{1}{n} $
Using this sequence, for the choice $2^{-k} \leq r^{2} < 2^{-k+1}$
$ \frac{b_{k+1}^{-1}}{b_{k+1}^{-1} + a_{k} + b_{k}^{-1}\sum_{k+1}^{\infty} a_{i}b_{i} } \leq \frac{u(B(x,r) \cap C)}{u(B(x,r))} \leq \frac{b_{k+1}^{-1}}{b_{k+1}^{-1}+a_{k}} $
so that as $k \to \infty$,
$ \lim_{r \to 0} \frac{u(B(x,r) \cap C)}{u(B(x,r))} = \frac{1}{1+\alpha} $
For the choice $\alpha = \frac{1}{\beta} -1$, we then have that the density of any point of $C$ is $\beta$ and $u(C) \geq 1$.

[1] Preiss, D.. "Invalid Vitali theorems." Abstracta. 7th Winter School on Abstract Analysis. Praha: Czechoslovak Academy of Sciences, 1979. 58-60. .

Lower Bounds of Hausdorff Density: July 16th, 2025

It is known from an argument of Besicovitch that for a set $A \subset \mathbb{R}^{n}$ such that $0< H^{s}(A) < \infty$, $$ 2^{-s} \leq \Theta^{*s}(A,x) \leq 1$$ for almost every point in $A$. For example, see Mattila's Geometry of Sets and Measures in Euclidean Space, Chapter 6.
What we seek to show here is an example that the lower bound cannot be increased beyond $2^{-s}$. First, we note a simple fact: in $B(0,1) \subset \mathbb{R}^{n}$, we can place $2n$ balls of radius $0 < \lambda < \frac{\sqrt{2}-1}{2+\sqrt{2}}$. Indeed, placing their centers on the points $\pm (1-\lambda)e_{i}$ keeps them within the circle, and this specific value of $r$ is chosen so that for any pair of these balls $B_1$ and $B_2$ have $d(B_1,B_2) > 1$.
Next, we notice that for this $\lambda$, we may generate a generalized Cantor set as follows.
Fix $s$ and pick some $n$ such that $2n > m$ where $m$ is the constant picked so $m \lambda^{2s} = \lambda^{s}$. Then, set $E_1 = B(0,1)$. Set $E_{2}$ to be some choice of $m$ balls among the choices $\pm (1-\lambda)e_i$. To generate $E_{k}$, on each ball $B_j$ present in $E_{k}$, we pick a sclaed-down arrangement as in the case of the choice for $E_2$, so that the resulting $A_{\lambda} = \bigcap_{i=1}^{\infty} E_{i}$ is self-similar.
It is known that $A_{\lambda}$ has Hausdorff dimension $s$ (see Mattila's Geometry of Sets and Measures in Euclidean Space, Chapter 4) and that $0 < H^{s}(A_{\lambda}) < \infty$. Without loss of generality, we may assume by rescaling that $H^{s}(A_{\lambda}) = 1$.
Furthermore, let us pick some small value $r$ and $x \in A_{\lambda}$. We may take $\lambda^{k+1} \leq r < \lambda^{k}$. Notice that the balls in $E_{k+1}$ have radii $\lambda^{k+1}$, and the distance between the balls is at least $\lambda^{k}$ by the choice of $\lambda$ and the location of the balls. Hence, $B(x,r)$ contains at-most $1$ ball of level $E_{k+1}$ and
$ \frac{H^{s}(A_{\lambda} \cap B(x,r) ) }{(2r)^{s}} \leq \frac{\lambda^{(k+1)s}H^{s}(A)}{(2\lambda^{k+1})^{s}} \leq 2^{-s}$
Finally, $\theta^{*s}(A_{\lambda},x) \leq 2^{-s}$. By Besicovitch's theorem, we have that $\theta^{*s}(A_{\lambda},x) = 2^{-s}$ for almost every $x \in A_{\lambda}$.