Say $f_n:[0,1]\to [0,1]^d$ is the $n$-th iteration of a $d$-dimensional Hilbert curve touring its range. Is it true that for any open $S\subset [0,1]^d$, then amount of time $f_n$ spends in $S$ is approximately $\mathrm{vol}(S)$?
That is, for $S\subset [0,1]^d$ open, taking $f^{-1}_n(S)=\{t:f_n(t)\in S\}\subset [0,1],$ is it true that:
\begin{equation} \mu(f^{-1}_n(S))\to_n \mu(S)? \end{equation}
I think so.
Each $f_n$ maps to the boundary of some lattice of cuboids, $(c^{(n)}_\lambda)_{\lambda \in \Lambda^{(n)}}$, where each $c^{(n)}_\lambda$ corresponds to a single cuboidal cell in $[0,1]^d$ with side length $\ell^{(n)}$. Take $\mathcal{S}^{(n)}$ to be the collection of cuboids $c^{(n)}_\lambda$ which come within $\ell^{(n)}$ of $S$ and compute: \begin{align} \mu(S) &\leq \mu(\cup_{\mathcal{S}^{(n)}})\\ &=\left|\mathcal{S}^{(n)}\right|\cdot(\ell^{(n)})^d \\ &\leq \mu(f_n^{-1}(\cup_{\mathcal{S}^{(n)}}))/(\ell^{(n)})^d\cdot (\ell^{(n)})^d \\ &\leq \mu(f^{-1}_n(\cup_{\mathcal{S}^{(n)}}\backslash S))+\mu(f^{-1}_n(S)) \\ &\leq O(1/n)+\mu(f_n^{-1}(S)). \end{align} Get a similar upper bound on $ \mu(f_n^{-1}(S))$ by identically inspecting the collection of cuboids contained in $S$ and at least $\ell^{(n)}$ away from the boundary of $S$.
Is this correct? Is there a better way?