If $\Omega\subseteq\mathbb{R}^n$ is open and connected, then there is a covering $\{B_j\}_{j\geq1}$ with $B_j\cap B_{j+1}\neq\emptyset$, $j\geq1$.

Question

I do not know if the following is true:

Question: If $\Omega\subseteq\mathbb{R}^n$ is open and connected, then there exists a covering by balls $\{B_j\}_{j=1}^{\infty}$ with the property $B_j\subseteq\Omega$ and $B_j\cap B_{j+1}\neq\emptyset$, for all $j\geq1$.

Motivation: I have a function $u$ that is a.e. constant on any ball contained in $\Omega$, and I want to prove that $u$ is a.e. constant on $\Omega$. If my question has a positive answer, then I am done.

Edit: I want an answer to "Question". Although the "Motivation" is important for me and can be proved not using "Question", I am interested on "Question".

Answer 1

There is in fact a useful characterisation of connectedness which is what you want:

A space $X$ is connected iff for every open cover $\mathcal{U}$ of $X$, and for all $x,y\in X$ we have a chain from $\mathcal{O}$ from $x$ to $y$.

I wrote a proof in this answer, for a similar question. I learnt it in my first topology course.

The last thing means there are finitely many $O_1, O_2,\ldots O_n, (n \ge 1)$ from $\mathcal{O}$ such that $x \in O_1, y \in O_n$ and for all $i = 1,\ldots n-1$: $O_i \cap O_{i+1} \neq \emptyset$.

If you know this then any open cover by open balls of your $\Omega$ will do (pick for all $x \in \Omega$ some ball $B(x,r_x) \subset \Omega$, by openness and use $\mathcal{O} = \{B(x,r_x): x \in \Omega\}$). If then your $f$ is constant on these balls, we can for any $x,y$ find a chain and from the non-intersecting property we get that the value on $x$ gets "transported" to $y$, and $f(x) = f(y)$. (I.e. If $a_i \in O_i \cap O_{i+1}$ then $f(x)= f(a_1)$ as both are in $O_1$, $f(a_1) = f(a_2)$ from both being in $O_2$, up to $f(a_{n-1}) = f(y)$ from both being in $O_n$)

Answer 2

Let $B\subset\Omega$ be any ball and let $c\in\mathbb{R}$ be such that $u=c$ a.e. in $B$. Let us define the set $$ A := \{x\in\Omega:\ \exists r>0\ \text{s.t.}\ B_r(x)\subset\Omega,\ u = c\ \text{a.e. in}\ B_r(x)\} $$ The set $A$ is clearly not empty and open. On the other hand, also the set $C := \Omega\setminus A$ is open. Namely, if $x\in C$ and $B_r(x) \subset \Omega$, then there exists a constant $k\neq c$ such that $u = k$ a.e. in $B_r(x)$. As a consequence, $B_r(x) \subset C$.

In conclusion, $A$ is both open and closed in $\Omega$. Since $A\neq \emptyset$ and $\Omega$ is connected, we deduce that $A=\Omega$.

Answer 3

Yes, there is such a covering.

Lemma 1. If $\Omega\subseteq\mathbb{R}^n$ is open and connected, and $B,D$ are two open balls contained in $\Omega$, then there is a finite chain $\{B_j:j=1,...,m\}$ of open balls (for some $m$), each contained in $\Omega$, such that $B_j\cap B_{j+1}\neq\emptyset$ for all $j=1,...,m-1$, and such that $B=B_1$ and $D=B_m$.
Sketch of (standard) proof. Let $A$ be the union of all balls $B_m$ such that for some $m$ there is a chain as in the statement of the lemma. Then $A$ is open and non-empty (as $B\subseteq A$). Its complement $C=\Omega\setminus A$ is also open, for if $x\in C$ then there is some ball $B_x$ with $x\in B_x\subseteq\Omega$, but then $B_x\cap A=\emptyset$, for if $B_x\cap A\not=\emptyset$ then $B_x\cap B_m\not=\emptyset$, for some $B_m$ from a suitable chain, but then we could extend that chain by letting $B_{m+1}=B_x$, a contradiction with the assumption that $x\not\in A$.
Hence $C=\emptyset$ (since $\Omega$ is connected). Pick any $x\in D$, then there is a chain $\{B_j:j=1,...,m\}$ with $x\in B_m$, so we could let $D=B_{m+1}$. End of (sketch of) proof of Lemma 1.

Lemma 2. $\Omega$ is the union of countably many open balls, say $\Omega=\bigcup_{k=0}^\infty C_k$ (each $C_k$ an open ball).
Proof. This is a direct consequence of the existence of a countable basis of open balls for the topology of $\mathbb{R}^n$. (Note. For reasons apparent below, I prefer that $k$ starts at $0$ here.)

Now, Lemma 1 and Lemma 2 answer your question positively, as follows. Fix $\Omega=\bigcup_{k=0}^\infty C_k$ as in Lemma 2. By Lemma 1, there is a finite chain $\{B_j:j=1,...,m_1\}$ of open balls (for some $m_1$), each $B_j$ contained in $\Omega$, such that $B_j\cap B_{j+1}\neq\emptyset$ for all $j=1,...,m_1-1$, and such that $B_1=C_0$ and $C_1=B_{m_1}$. By Lemma 1 again, there is a finite chain $\{B_j:j=m_1,...,m_2\}$ of open balls (for some $m_2\ge m_1$), each contained in $\Omega$, such that $B_j\cap B_{j+1}\neq\emptyset$ for all $j=m_1,...,m_2-1$, and such that $B_{m_1}=C_1$ and $C_2=B_{m_2}$. Then we extend the sequence of the $B_j$ going from $C_2$ to $C_3$, then another finite chain from $C_3$ to $C_4$, etc. That is, in general (if we let $m_0=1$), we have $m_{k-1}\le m_k$, and a finite chain $\{B_j:j=m_{k-1},...,m_k\}$ with $B_{m_{k-1}}=C_{k-1}$ and $C_k=B_{m_k}$.

Answer 4

It is just a short step. If $u(x)=c_j$ for all $x\in B_j$ then we get $c_j=u(y)=c_{j+1}$ for $y\in B_{j+1}\cap B_j\neq\emptyset$ for $j\geq 1$. Then all $c_j$ are the same and $u$ is constant on $\Omega$.