I Dyadic cubes generate Borel sigma algebra of subset

[psie]

TL;DR Summary

I know the dyadic cubes in $\mathbb R^d$ generate the Borel $\sigma$-algebra of $\mathbb R^d$. Let now $U\subset \mathbb R^d$ be an open subset and the claim is that all dyadic (half-open) cubes whose closure are contained in $U$ generates the Borel $\sigma$-algebra on $U$. How?

Dyadic cubes of order $n\in\mathbb N$ are sets of the form $$C=\prod_{j=1}^d (k_j2^{-n},(k_j+1)2^{-n}],\quad k_j\in\mathbb Z.$$ I know that the dyadic cubes in $\mathbb R^d$ induce a generating collection for the Borel $\sigma$-algebra of $U$ (by intersecting the dyadic cubes in $\mathbb R^d$ with $U$), however, this induced generating collection may be larger than merely the dyadic cubes whose closure is in $U$.

If I can somehow show that every open subset of $U$ is a countable union of dyadic cubes whose closure is in $U$, then I'm done. Note, an open subset of $U$ is of the form $A\cap U$, where $A$ is open in $\mathbb R^d$ (so the open subsets of $U$ are open in $\mathbb R^d$ as well). I feel stuck on how to proceed.

 

[fresh_42]

My 2 ct.

My intuition would be: Consider $U=(0,1)\subseteq \mathbb{R}.$ Then we can exhaust $U$ by dyadic intervals and all can be counted since $\mathbb{Z}\times \mathbb{N}$ is countable.

However, I tend to make mistakes when it comes to topology. Maybe we need Dynkin systems to prove it formally: https://en.wikipedia.org/wiki/Dynkin_system but this could as well be an overreaction and it is as simple as with $(0,1)$.

 

[psie]

Cool. After researching a bit on the wide world web, here are my 2 ct.

If we can show the claim for all closed dyadic cubes that are fully contained in $U$, then it is also true for all half-open dyadic cubes whose closure is contained $U$. This is because the closure of a half-open cube is a closed cube and every closed cube can be written as a countable intersection of half-open cubes. Indeed,\begin{align*}C^n&=\prod_{j=1}^d (k_j2^{-n},(k_j+1)2^{-n}],\\ \text{and}\quad\overline{C^n}&=\prod_{j=1}^d [k_j2^{-n},(k_j+1)2^{-n}].\end{align*} But also if we define \begin{align*}C_i^n&=\prod_{j=1}^d (k_j2^{-n}+1/i,(k_j+1)2^{-n}],\quad i\in\mathbb N\setminus\{1,\ldots,N-1\},\\ \text{then}\quad\overline{C^n}&=\bigcap_{i\geq N}C_i.\end{align*}Of course, we choose $(k_j)_{j=1}^d\subset\mathbb Z$ and $n,N\in\mathbb N$ appropriately so that all sets above are in $U$.

So we proceed by showing all closed dyadic cubes that are fully contained in $U$ generate the Borel $\sigma$-field of $U$. Since the dyadic cubes are Borel sets (they are closed), we only need to show that the open subsets of $U$ are contained in $\sigma(\mathcal C)$, where $\mathcal C$ are the closed dyadic cubes fully contained in $U$. But $U$ is separable as a subset of separable $\mathbb R^d$, hence it has a countable base of open balls. So in turn, the problem reduces to showing that every open ball in $U$ can be written as a union of closed dyadic cubes fully contained in the open ball.

This last claim is just this answer.