Construct a measure space $(X,\mathcal{T},\mu)$ such that there is a $\{A_i\mid i \in I\}$ and a $r>0$ so that $\mu(A_i)<r$ but $\mu(\bigcup A_i)>r$.

Question

I found the following problem:

Construct a measure space $(X,\mathcal{T},\mu)$ such that: there is a totally ordered $I$ along with a family $\{A_i\mid i\in I\}\subset \mathcal{T}$ of increasing sets, that is, for each $i\leq j$ one has $A_i\subseteq A_j$ and $r>0$ such that for any $i\in I$, $\mu(A_i)<r$ but $\mu\left(\bigcup_{i\in I}A_i\right)>r$.

I have no idea where to start: all my attempts lead to $\mu(\bigcup_i A_i)=r$. I tried using $\mu$ as the counting measure letting $A_i$ be finite and $I=\Bbb{N}$ which again leads to the same result. I tried $\mu=\arctan\circ \nu$ (and $\mu(A)=\infty$ when $|A|=\infty$) where $\nu$ is the counting measure but it's not a measure. I have no idea how to procede. Any hints or solutions are welcomed.

Answer 1

Consider the space $(\Omega,\mathscr{F},\mu)$ where $\Omega$ is the first non-countable ordinal, $$\mathscr{F}=\{A\subseteq \Omega\mid \aleph(A)\leq\aleph_0\lor \aleph(\mathscr{C}A)\leq \aleph_0\}$$ and $$\begin{align}\mu\colon \mathscr{F}&\to\overline{\mathbb{R}}_+\\ A&\mapsto \begin{cases}0, & \aleph(A)\leq \aleph_0,\\ 1, & \aleph(A)>\aleph_0.\end{cases}\end{align}$$ We will use $I$ as the well ordered set $\Omega$.

Let $\{A_\omega\mid \omega\in\Omega\}$ be defined by $A_\omega=\bigcup_{\omicron\leq\omega}\{\omicron\}$ and note that $\aleph(A_\omega)\leq \aleph_0$ and thus $\{A_\omega\mid \omega\in\Omega\}\subseteq \mathscr{F}$ and $\forall \omega\in\Omega,\; \mu(A_\omega)=0<1/2$. It is clear that $\omicron\leq \omega\implies A_\omicron\subseteq A_\omega$. We now consider $\bigcup_{\omega\in\Omega}A_\omega$: clearly $\bigcup_{\omega\in\Omega}A_\omega\subseteq \Omega$; let $\omega\in \Omega$, then $\omega\in A_\omega\subseteq\bigcup_{\omega\in\Omega}A_\omega$ and $\bigcup_{\omega\in\Omega}A_\omega=\Omega$. From this, $\bigcup_{\omega\in\Omega}A_\omega\in\mathscr{F}$ and $\mu(\bigcup_{\omega\in\Omega}A_\omega)=1>1/2$.

That is, $$(\omicron\leq \omega\implies A_\omicron\subseteq A_\omega)\land(\forall \omega\in\Omega,\; \mu(A_\omega)<1/2)\land \left(\mu\left(\bigcup_{\omega\in\Omega}A_\omega\right)>1/2\right)\;\square.$$

Answer 2

In the real line with Lebesgue measure consider the family of all finite subsets odered by itself: $A_i=i$) ordered by inclusion: $F_1 \leq F_2$ if $F_1 \subseteq F_2$. Take $r=1$. Note that $\bigcup_i A_i=\mathbb R$.