Borel measure of point is 1

Question

Let $\Sigma$ be the Borel $\sigma$-algebra on $\mathbb{R^n}$ and $\mu: \Sigma \to [0,1]$ a measure. Suppose the existence of a Borel set $E\subset\mathbb{R^n}$ s.t. its measure is $1$ and the measure of every subset of $E$ is either $0$ or $1$. Then there exists a point $x\in\mathbb{R^n}$ with measure $1$.

At first I wanted to subdivide $E$ into always smaller and smaller subset to reach single point sets, but since $E$ contains possibly uncountably many points I can't really do that. Then I realized I could maybe show that having such properties forces $E$ to be itself a single point set, but I don't see how I could do that.

If anyone can tell me if I'm going in the right direction or if there is a better and/or simpler solution I'd be very grateful.

Answer 1

Assume there exists none. Let $\mathcal{C}=\{B_{\delta}(x): x\in\mathcal{D},\delta\in{\bf{Q}}^{+}\}$, $\mathcal{D}$ is countable dense in ${\bf{R}}^{n}$. For each $x\in E$, let $B_{\delta_{n,x}}(x)$ such that $\delta_{n,x}$ strictly decreasing to zero. Since $\mu(B_{\delta_{n,x}}(x)\cap E)\rightarrow\mu\{x\}=0$ and $\mu(B_{\delta_{n,x}}(x)\cap E)$ is either $0$ or $1$, then $\mu(B_{\delta_{n,x}}(x)\cap E)$ is eventually zero, pick some $n_{x}$ such that $\mu(B_{\delta_{n,x}}(x)\cap E)=0$. Now, pick some $c_{x}\in\mathcal{D}$ and $\eta_{n,x}\in{\bf{Q}}^{+}$ such that $x\in B_{\eta_{n,x}}(c_{x})\subseteq B_{\delta_{n,x}}(x)$. Now $I=\{(\eta_{n,x},c_{x})\}\subseteq{\bf{Q}}^{+}\times\mathcal{D}$ is a countable set and \begin{align*} E&=\bigcup_{x\in E}\{x\}\\ &=\bigcup_{(\eta_{n,x},c_{x})\in I}B_{\eta_{n,x}}(c_{x})\cap E, \end{align*} so $\mu(E)\leq\displaystyle\sum_{(\eta_{n,x},c_{x})\in I}\mu(B_{\eta_{n,x}}(c_{x})\cap E)=0$, a contradiction.

Answer 2

A correct version of a deleted answer: First define a new measure $\nu$ by $\nu(S)=\mu(S\cap E)$. Now every Borel $S\subset\mathbb R^n$ has $\nu(S)$ either $0$ or $1$.

Suppose $\nu(\{x\})=0$ for every $x$. For every $x$ let $B_x=B(x,r(x))$, where $r(x)>0$ is so small that $\nu(B_x)=0$.

If $K$ is compact then $K$ is covered by finitely many $B_x$, so $\nu(K)=0$. Hence $\nu(\mathbb R^n)=0$ (since $\mathbb R^n$ is a countable union of compact sets), contradiction.