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Consider the unit interval $I=[0,1]$. $B[0,1]$ denotes the sigma-algebra of Borel sets in $[0,1]$ and is generated by the countable subclass $C[0,1]$. $\lambda$ is the usual Lebesgue measure.

Are there a sequence of subsets $E_i$ of $[0,1]$ and a countably additive extension $m$ of $\lambda$ on $\sigma(C[0,1] \cup \{E_i; i=1,2, \cdots\})$ such that

(*) For any $i \geq 0$, $E_{i+1}$ is not $m$-measurable with respect to $B_i:=\sigma(C[0,1] \cup \{E_1, \cdots, E_i\})$, i.e., $$(m\upharpoonright_{B_i})^* (E_{i+1}) = \inf\{m\upharpoonright_{B_i}(E): E \supseteq E_{i+1}\} \neq \sup\{m\upharpoonright_{B_i}(E): E \subseteq E_{i+1}\} = (m\upharpoonright_{B_i})_*(E_{i+1}) $$ ?

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Yes, there are. In the following I give a proof to the affirmative answer. Let $\langle \Omega, \mu, \mathcal{A}\rangle$ be a measurable space. Note that $\mu$ is not necessarily defined on the power set of $\Omega$. We can extend $\mu$ in two standard ways, by defining functions $\mu_\ast$ and $\mu^\ast$, traditionally called inner measure and outer measure induced by $\mu$. For an arbitrary subset $H \subseteq \Omega$, we define

  1. $\mu^\ast(H) := \sup\{\mu(E): E \in \mathcal{A}, E\subseteq H\}$ and
  2. $\mu_\ast(H) :=\inf\{\mu(E): E \in \mathcal{A}, E\subseteq H\}$

We say that a measurable space $\langle \Omega, \mu', \mathcal{A}'\rangle$ is a (countably-additive) extension of $\langle \Omega, \mu, \mathcal{A}\rangle$ if $\mathcal{A}' \supseteq \mathcal{A}$, and $\mu' (A) = \mu(A)$ for all $A\in \mathcal{A}$.

Lemma: Let $H$ be a subset of $\Omega$ which is not in $ \mathcal{A}$.

  1. There is an extension $\langle \Omega, \mu', \sigma(\mathcal{A}\cup\{H\})\rangle$ of $\langle \Omega, \mu, \mathcal{A}\rangle$;
  2. If $\langle \Omega, \mu', \sigma(\mathcal{A}\cup\{H\})\rangle$ is an extension of $\langle \Omega, \mu, \mathcal{A}\rangle$, then $$\mu_\ast(H) \leq \mu'(H)\leq \mu^\ast(H);$$
  3. If $\mu_\ast(H) \neq \mu^\ast(H)$, then there are extensions $\langle \Omega, \mu_1, \sigma(\mathcal{A}\cup\{H\})\rangle$ and $\langle \Omega, \mu_2, \sigma(\mathcal{A}\cup\{H\})\rangle$ such that $\mu_1(H) \neq \mu_2(H)$.

Now we consider the analytical structure on $I:= [0,1]$.

Theorem: (Impossibility Theorem) There exists on $2^{[0, \frac{1}{2}]}$ no probability measure $P$ such that $P(\{x\}) =0 $ for each $x \in [0,1]$ where $\lambda$ is the Lebesgue measure.

Since $\lambda (\{x\}) =0$, this implies that it is impossible to extend $\lambda$ to the power set of $[0, \frac{1}{2}]$ at all. Let $\mathcal{B}[0, \frac{1}{2}]$ denotes the $\sigma$-algebra of Borel subsets of $[0, \frac{1}{2}]$ which is generated by the countably subclass $\mathcal{C} [0, \frac{1}{2}]$. A subset $E$ of $[0, \frac{1}{2}]$ is called $\lambda$-measurable with respect to $\mathcal{B}[0, \frac{1}{2}]$ if $\lambda^* (E)= \lambda_*(E) $ where

  1. $\lambda_*(E) = \sup\{\lambda (E'): E'\in \mathcal{B}[0, \frac{1}{2}], E' \subseteq E\}$;
  2. $\lambda^*(E) = \inf\{\lambda (E'): E'\in \mathcal{B}[0, \frac{1}{2}], E' \supseteq E\}$.

Note that the class of $\lambda$-measurable with respect to $\mathcal{B}[0, \frac{1}{2}]$ forms a $\sigma$-algebra. It is denoted $\mathcal{M}[0, \frac{1}{2}]$. According to the above Impossibility Theorem, there is a subset $E_1$ of $[0, \frac{1}{2}]$ which is not in $\mathcal{M}[0, \frac{1}{2}]$. Let $I^{(0)}= [0, \frac{1}{2}]$. From the Lemma, we know that there is an extension $\langle I^{(0)}, \lambda_1, \mathcal{B}_1[0, \frac{1}{2}]\rangle$ of $\langle I^{(0)}, \lambda, \mathcal{B}[0, \frac{1}{2}]\rangle$ where $\mathcal{B}_1[0, \frac{1}{2}] = \sigma(\mathcal{C}[0, \frac{1}{2}]\cup \{E_1\})$. Similarly, we can find a sequence of subsets $E_i $ of $I^{(0)}$ and a sequence of $\lambda_i' (i\in \mathbb{N})$ on $\mathcal{B}_{i}[0, \frac{1}{2}]$ where $\mathcal{B}_i [0, \frac{1}{2}]: = \sigma(\mathcal{C}[0, \frac{1}{2}] \cup \{E_1, \cdots, E_i\}) (i \in \mathbb{N})$ such that

  1. $\langle I^{(0)}, \lambda_{i+1}', \mathcal{B}_{i+1}[0, \frac{1}{2}]\rangle$ is an extension of $\langle I^{(0)}, \lambda_{i}', \mathcal{B}_{i}[0, \frac{1}{2}]\rangle$ ;
  2. $(\lambda_i')_\ast(E_{i+1}) \neq (\lambda_i')^\ast (E_{i+1})$.

Let $F_i = E_i \cup (\frac{1}{2}, 1]$ and $\mathcal{B}_i = \sigma(\mathcal{C}[0,1] \cup \{F_1, \cdots, F_i\}) (i\in \mathbb{N})$ where $\mathcal{C}[0,1]$ denotes the countable base that generates the $\sigma$-algebra $\mathcal{B}[0,1]$ of Borel subsets of $[0,1]$. Obviously, $\lambda_*(F_1) \neq \lambda^*(F_1)$. It follows directly from the above results that there is an extension $\langle I, \lambda_1, \sigma(\mathcal{C}[0,1]\cup \{F_1\})\rangle$ of $\langle I, \lambda, \mathcal{C}[0,1]\rangle$. There are a sequence of finite measures $\lambda_i$ on $\mathcal{B}_i$ where $\mathcal{B}_i = \sigma (\mathcal{C}[0,1]\cup \{F_1, \cdots, F_i\})$ such that

  1. $\langle I, \lambda_{i+1}, \mathcal{B}_{i+1}\rangle$ is an extension of $\langle I, \lambda_{i}, \mathcal{B}_{i}\rangle$;
  2. $(\lambda_i)_*(F_{i+1}) \neq (\lambda_i)^* (F_{i+1})$.

Set $\mathcal{B}_{\infty} =\sigma(\mathcal{C}[0,1] \cup \{F_i: i \in \mathbb{N}\})$ and $(\mathcal{B}_{\infty})_0$ to be the \emph{algebra} generated by $\mathcal{C}[0,1] \cup \{F_i: i \in \mathbb{N}\}$. Note that $\mathcal{B}_{\infty}$ is countably-generated. Now we define a set operator $\lambda_{\infty}$ on $(\mathcal{B}_{\infty})_0$: for $E\in (\mathcal{B}_{\infty})_0$, if $E \in \mathcal{B}_i$, then $\lambda_{\infty}(E) = \lambda_i(E)$. In particular, $$\begin{equation*} \lambda_{\infty} (E) = \left\{ \begin{array}{rl} \lambda(E) & \text{if } E\in \mathcal{C}[0,1],\\ \lambda_i(E) & \text{if } E =F_i, \end{array} \right. \end{equation*}$$

Lemma: $\lambda^\ast$ is continuous from above in the sense that $\lim_{i\rightarrow \infty} \lambda^\ast (E_i) =0$ for any decreasing sequence of subsets of $[0,1]$ such that $\bigcap_i E_i = \emptyset$.

 

Lemma: The defined set operator $\lambda_{\infty}$ is a finitely-additive measure on $(\mathcal{B}_{\infty})_0$. And it is continuous from above: for any sequence of $E_i$'s in $(\mathcal{B}_{\infty})_0$, if $E_i \supseteq E_{i+1}$ for all $i \in \mathbb{N}$ and $\bigcap_i E_i = \emptyset$, then $\lim_{i \rightarrow \infty} \lambda_{\infty} (E_i) = 0$.

Proof: The first part comes from the fact that each $\lambda_i$ is actually countably-additive on $\mathcal{B}_i$. And the second one follows from Lemma \ref{OuterMeasure} and the following fact: for any $E\in (\mathcal{B}_{\infty})_0 $, $\lambda_{\infty}(E) \leq \lambda_i^*(E) \leq \lambda^*(E)$ for all $i\in \mathbb{N}$.

Theorem: $\lambda_{\infty}$ on $(\mathcal{B}_{\infty})_0 $ has a unique extension to the generated $\sigma$-algebra $\mathcal{B}_{\infty}$.

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