For any point $x \in \mathbb R^n$, denote by $\delta_x$ the Dirac Delta measure centered at $x$.

Let $a_n$ be an sequence of positive numbers with $\lim_{n \to \infty} a_n = 0$, and let $d_i$ be a countable dense subset of $\mathbb R^n$.

Given a rearrangement $a_{n_k}$ of $a_n$, consider the space $\mathbb R^n$ under the Euclidean topology and the measure $\sum_k a_{n_k} \delta_{d_k}$.

Question: For any given $a_n, d_i$ satisfying the above consitions, does there exist a rearrangement such that the above space admits nowhere locally constant continuous functions that are integrable under the given measure?

Note: A function is said to be nowhere locally constant if there exists no open set $S$ for which $f(x) = c$ for all $x$ in $S$.


1 Answer 1


Partition $(a_n)$ into two subsequences $(a_n')$ and $(a_n'')$ with $\sum a_n' < \infty$, and partition $(d_i)$ into two subsequences $(d_i')$ and $(d_i'')$ such that $d_i'' \to \infty$. Pair the $a_i'$s with the $d_i'$s and the $a_i''$s with the $d_i''$s. Then any bounded continuous function that vanishes at every $d_i''$ will be integrable.

  • $\begingroup$ Ah, this was much easier than I expected. Slick solution though! $\endgroup$
    – Nate River
    Sep 28, 2021 at 4:22
  • 2
    $\begingroup$ Thanks! Fun question. $\endgroup$
    – Nik Weaver
    Sep 28, 2021 at 4:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.