Ring of sets

Definition. A ring of sets ${\mathcal R}$ is a collection of sets satisfying the following axioms.

$(\bigcup {\mathcal R}) \in {\mathcal R}$ (contains its universe)
$A, B \in {\mathcal R} \implies A \cap B \in {\mathcal R}$ (closed under intersection)
$A \in {\mathcal R} \implies (\bigcup {\mathcal R}) \setminus A \in {\mathcal R}$ (closed under complement)

Proposition. Rings of sets are closed under finite unions and symmetric differences.

Remark: Rings of sets are typically called algebras in analysis textbooks. But I prefer to say ring due to the fact that they are precisely algebraic rings under $(+, \times) = (\triangle, \cap)$.

Measured ring

Definition. $({\mathcal R}, \mu)$ is a measured ring if ${\mathcal R}$ is a ring of sets and $\mu: {\mathcal R} \to [0, \infty)$ satisfies:

$$\mu(A \cup B) = \mu(A) + \mu(B \setminus A)$$

Proposition. If $({\mathcal R}, \mu)$ is a measured ring, then:

$\mu(\varnothing) = 0$
$A \cap B = \varnothing \implies \mu(A \cup B) = \mu(A) + \mu(B)$
$A \subset B \implies \mu(A) \leq \mu(B)$

Proof. As for $\mu(\varnothing)$, we have

\begin{align*} \mu(\varnothing) &= \mu(\varnothing \cup \varnothing) \\ &= \mu(\varnothing) + \mu(\varnothing \setminus \varnothing) \\ &= 2\mu(\varnothing) \end{align*}

which can only be satisfied by $\mu(\varnothing) = 0$. If $A \cap B = \varnothing$ then $B \setminus A = B$ so $\mu(A \cup B) = \mu(A) + \mu(B)$. Next, if $A \subset B$ then $\mu(B) = \mu(A \cup B) = \mu(A) + \mu(B \setminus A) \geq \mu(A)$. $\square$

Proposition. $d_\mu(A, B) := \mu(A \triangle B)$ is a distance metric.

Proof. Suppose that $x \in (A \triangle C) = (A \setminus C) \cup (C \setminus A)$. Let $x \in (A \setminus C)$. If $x \in B$, then $x \in (B \setminus C)$. And if $x \not\in B$, then $x \in (A \setminus B)$. Therefore, $(A \setminus C) \subset (A \setminus B) \cup (B \setminus C)$, and therefore by symmetry, $(A \triangle C) \subset (A \triangle B) \cup (B \triangle C)$. Therefore by the previous proposition, $\mu(A \triangle C) \leq \mu(A \triangle B) + \mu(B \triangle C)$, which is the triangle inequality for $d_\mu$. $\square$

Proposition. $\mu$ is continuous with respect to $d_\mu$.

Proof. We have

\begin{align*} \mu(A) - \mu(B) &\leq \mu(B \cup (A \setminus B)) - \mu(B) \\ &\leq \mu(B) + \mu(A \setminus B) - \mu(B) \\ &= \mu(A \setminus B) \\ &\leq \mu(A \triangle B) \\ &= d_\mu(A, B) \end{align*}

So by symmetry $|\mu(A) - \mu(B)| \leq d_\mu(A, B)$. $\square$

Corollary. If $(A_n), (B_n) \in \mathrm{Cauchy}({\mathcal R}, d_\mu)$ are two equivalent Cauchy sequences, then $\lim \mu(A_n) = \lim \mu(B_n)$ both exist and are equal.

Definition. The measured ring $({\mathcal R}, \mu)$ is called continuous if

$$A_n \searrow \varnothing \implies \mu(A_n) \to 0$$

Proposition. If $({\mathcal R}, \mu)$ is a continuously measured ring, then:

$A_n \nearrow A \implies \mu(A_n) \to \mu(A)$
$A_n \searrow A \implies \mu(A_n) \to \mu(A)$
$\bigsqcup_{n=1}^\infty A_n = A \implies \sum_{n=1}^\infty \mu(A_n) = \mu(A)$

Proof. For the limits, if $A_n \nearrow A$, then $A \setminus A_n \to \varnothing$, so $\mu(A_n) = \mu(A) - \mu(A \setminus A_n) \to \mu(A)$. And if $A_n \searrow A$, then $A_n \setminus A \to \varnothing$, so $\mu(A_n) = \mu(A) + \mu(A_n \setminus A) \to \mu(A)$. Lastly, if $\bigsqcup_{n=1}^\infty A_n = A$ is a countable disjoint union, then $B_N := \bigsqcup_{n=1}^N A_n$ satisfies $B_N \nearrow A$, so

\begin{align*} \sum_{n=1}^\infty \mu(A_n) &= \lim_{N \to \infty} \sum_{n=1}^N \mu(A_n) \\ &= \lim_{N \to \infty} \mu\left(\bigsqcup_{n=1}^N A_n\right) \\ &= \lim_{N \to \infty} \mu(B_N) \\ &= \mu(A). \end{align*}

$\square$

Counterexample. Without continuity we could define the measure $\mu([a, b)) := 1\{b = 1\}$ for subsets $[a, b) \subset [0, 1)$. This would satisfy additivity, but not continuity or countable additivity. Or we could in general define $\mu(S) := 1\{S = \Omega\}$, same deal.

Absolutely convergent sequences

We call a sequence $(x_n)$ absolutely convergent if $\sum_{n=1}^\infty d(x_n, x_{n+1}) < \infty$. This will be crucial for the following definition.

Proposition. If $(x_n)$ is absolutely convergent, then it is Cauchy.

Proposition. Every Cauchy sequence has an absolutely convergent subsequence.

Extension to measurable sets

Let $({\mathcal R}, \mu)$ be a measured ring with universe $\Omega := \bigcup {\mathcal R}$ and distance metric $d_\mu(A, B) = \mu(A \triangle B)$.

Definition. An arbitrary set $S \subset \Omega$ is measurable if there exists an absolutely convergent sequence $(A_n) \in \mathrm{Cauchy}({\mathcal R}, d_\mu)$ such that

$$\limsup A_n \subset S \subset \liminf A_n.$$

In this case we say that $S$ is measured by $(A_n)$, and we define $\mu(S) := \lim \mu(A_n)$. We write the collection of all measurable sets as $\mathrm{Measurable}({\mathcal R}, \mu)$.

Theorem. $(\mathrm{Measurable({\mathcal R}, \mu)}, \mu)$ is a well-defined measured ring.

Proof.

Proof: Well-defined

First we show that $\mu(S)$ is well-defined for all measurable sets $S \subset \Omega$. We begin with two lemmas.

Lemma. If $\liminf A_n \subset \limsup B_n$, and both are absolutely convergent sequences, then $\lim \mu(A_n) \leq \lim \mu(B_n)$.

Proof.

Define $C_{N,k} := \bigcap_{n=N}^{N+k} A_n$, and $D_{N,k} := \bigcup_{n=N}^{N+k} B_n$. Both of these are sets in the measured ring. The sequence $(C_{N,k})_{k \geq 1}$ is decreasing and $(D_{N,k})_{k \geq 1}$ is increasing.

For every $N_1,N_2 \geq 1$, $\liminf A_n \subset \limsup B_m$ implies that $\bigcap_{k \geq 1} C_{N_1,k} \subset \bigcup_{k \geq 1} D_{N_2,k}$, because:

\begin{align*} \bigcap_{k \geq 1} C_{N_1,k} &\subset \bigcup_{N \geq 1} \bigcap_{n \geq N} A_n = \liminf A_n \\ & \subset \limsup B_n = \bigcap_{N \geq 1}\bigcup_{n \geq N} B_n\\ &\subset \bigcup_{k \geq 1} D_{N_2,k} \end{align*}

That is: “For any $N_1,N_2$, points always in $A_n$ after $N_1$ must sometimes be in $B_n$ after $N_2$”

Therefore, for any $N_1,N_2$ we have $C_{N_1,k} \setminus D_{N_2,k} \searrow \varnothing$. Hence by the continuity of $\mu$:

\begin{align*} \lim_{k \to \infty} \mu(C_{N_1,k}) &\leq \lim_{k \to \infty}\left(\mu(D_{N_2,k}) + \mu(C_{N_1,k} \setminus D_{N_2,k})\right) \\ &= \lim_{k \to \infty}\mu(D_{N_2,k}) + \lim_{k \to \infty}\mu(C_{N_1,k} \setminus D_{N_2,k}) \\ &= \lim_{k \to \infty}\mu(D_{N_2,k}) \end{align*}

It remains to control $\lim \mu(A_n)$ in terms of $\lim_{k \to \infty} \mu(C_{N_1,k})$ and likewise $\mu(B_n)$ in terms of $\lim_{k \to \infty}\mu(D_{N_2,k})$. Using the absolute convergence of $A_n$:

\begin{align*} \lim_{N \to \infty} \mu(A_N) &\leq \limsup_{N \to \infty} \inf_{k \geq 1} (\mu(C_{N,k}) + \mu(A_N \setminus C_{N,k})) \\ &\leq \limsup_{N \to \infty} \inf_{k \geq 1} \left(\mu(C_{N,k}) + \sum_{n=N}^{N+k} \mu(A_{n+1} \triangle A_n)\right) \\ &\leq \limsup_{N \to \infty} \inf_{k \geq 1} \left(\mu(C_{N,k}) + \sum_{n=N}^\infty \mu(A_{n+1} \triangle A_n)\right) \\ &= \limsup_{N \to \infty} \inf_{k \geq 1} \mu(C_{N,k}) + \lim_{N \to \infty} \sum_{n=N}^\infty \mu(A_{n+1} \triangle A_n) \\ &= \limsup_{N \to \infty} \inf_{k \geq 1} \mu(C_{N,k}) \\ &= \limsup_{N \to \infty} \lim_{k \to \infty} \mu(C_{N,k}) \\ &\leq \sup_{N \geq 1}\lim_{k \to \infty} \mu(C_{N,k}) \end{align*}

And likewise, using the absolute convergence of $B_n$:

\begin{align*} \lim_{N \to \infty} \mu(B_N) &\geq \liminf_{N \to \infty} \sup_{k \geq 1} (\mu(D_{N,k}) - \mu(B_N \setminus D_{N,k})) \\ &\geq \liminf_{N \to \infty} \inf_{k \geq 1} \left(\mu(D_{N,k}) - \sum_{n=N}^{N+k} \mu(B_{n+1} \triangle B_n)\right) \\ &\geq \liminf_{N \to \infty} \inf_{k \geq 1} \left(\mu(D_{N,k}) - \sum_{n=N}^\infty \mu(B_{n+1} \triangle B_n)\right) \\ &= \liminf_{N \to \infty} \inf_{k \geq 1} \mu(D_{N,k}) - \lim_{N \to \infty} \sum_{n=N}^\infty \mu(B_{n+1} \triangle B_n) \\ &= \liminf_{N \to \infty} \inf_{k \geq 1} \mu(D_{N,k}) \\ &= \liminf_{N \to \infty} \lim_{k \to \infty} \mu(D_{N,k}) \\ &\geq \inf_{N \geq 1}\lim_{k \to \infty} \mu(D_{N,k}) \end{align*}

Therefore, putting the previous three results together:

\begin{align*} \lim_{N_1 \to \infty} \mu(A_{N_1}) &\leq \sup_{N_1 \geq 1}\lim_{k \to \infty}\mu(C_{N_1,k}) \\ &\leq \sup_{N_1 \geq 1} \inf_{N_2 \geq 1} \lim_{k \to \infty} \mu(D_{N_2,k}) \\ &= \inf_{N_2 \geq 1} \lim_{k \to \infty} \mu(D_{N_2,k}) \\ &\leq \lim_{N_2 \to \infty} \mu(B_{N_2}) \end{align*}

$\square$

Lemma. If $(A_n), (B_n)$ are absolutely convergent sequences in a measured ring, and $\liminf B_n \subset \liminf A_n \subset \limsup B_n$, then $\lim \mu(A_n) = \lim \mu(B_n)$.

Proof.

The previous lemma immediately gives $\lim \mu(A_n) \leq \lim \mu(B_n)$, so it remains to show that $\lim \mu(B_n) \leq \lim \mu(A_n)$. Observe that

$\Omega \setminus \limsup(\Omega \setminus B_n) = \liminf B_n \subset \liminf A_n \subset \limsup A_n = \Omega \setminus \liminf(\Omega \setminus A_n)$

and therefore, $\liminf (\Omega \setminus A_n) \subset \limsup (\Omega \setminus B_n)$. Both $(\Omega \setminus A_n)$ and $(\Omega \setminus B_n)$ are still absolutely convergent sequences, since $\mu((\Omega \setminus A_n) \triangle (\Omega \setminus A_{n+1})) = \mu(A_n \triangle A_{n+1})$. Therefore, the previous lemma implies that $\lim \mu(\Omega \setminus A_n) \leq \lim \mu(\Omega \setminus B_n)$, so, using the finitude of $\mu(\Omega)$, we have:

\begin{align*} \lim \mu(B_n) &= \lim (\mu(\Omega) - \mu(\Omega \setminus B_n)) \\ &= \mu(\Omega) - \lim \mu(\Omega \setminus B_n) \\ &\leq \mu(\Omega) - \lim \mu(\Omega \setminus A_n) \\ &= \lim(\mu(\Omega) - \mu(\Omega \setminus A_n)) \\ &= \lim \mu(A_n) \end{align*}

Thus we have both $\lim \mu(A_n) \leq \lim \mu(B_n)$ and $\lim \mu(B_n) \leq \lim \mu(A_n)$, so that $\lim \mu(A_n) = \lim \mu(B_n)$. $\square$

Now we are ready to prove that $\mu$ is well-defined. Let $(A_n)$ and $(B_n)$ be two sequences in ${\mathcal R}$ which measure $S$. Without loss of generality let $\liminf B_n \subset \liminf A_n$. By definition we also have $\liminf A_n \subset S \subset \limsup B_n$. Therefore $\liminf B_n \subset \liminf A_n \subset \limsup B_n$, so the preceding lemma yields $\lim \mu(A_n) = \lim \mu(B_n)$.

Proof: Ring of sets

It remains to prove that $(\mathrm{Measurable({\mathcal R}, \mu)}, \mu)$ is a valid measured ring.

Clearly the first property is satisfied, since $\Omega \in \mathrm{Measurable({\mathcal R}, \mu)}$ via the absolutely sequence $(A_n)$ where every $A_n = \Omega$.

For the second property, let $E, F \in \mathrm{Measurable({\mathcal R}, \mu)}$ be two subsets measured by $(A_n)$ and $(B_n)$. Define $C_n := A_n \cap B_n$; we need to show that $(C_n)$ measures $E \cap F$. It’s clear that $\liminf C_n \subset E \cap F \subset \limsup C_n$. It remains to show that $C_n$ is an absolutely convergent sequence. We have:

\begin{align*} \mu(C_n \setminus C_{n+1}) &= \mu((A_n \cap B_n) \setminus (A_{n+1} \cap B_{n+1})) \\ &= \mu(((A_n \cap B_n)\setminus A_{n+1}) \cup ((A_n \cap B_n) \setminus B_{n+1})) \\ &\leq \mu((A_n \setminus A_{n+1}) \cup (B_n \setminus B_{n+1})) \\ &\leq \mu(A_n \setminus A_{n+1}) + \mu(B_n \setminus B_{n+1}) \\ \end{align*}

so that by symmetry $\mu(C_n \triangle C_{n+1}) \leq \mu(A_n \triangle A_{n+1}) + \mu(B_n \triangle B_{n+1})$, and therefore

$$\sum_{n=1}^\infty \mu(C_n \triangle C_{n+1}) \leq \sum_{n=1}^\infty \mu(A_n \triangle A_{n+1}) + \sum_{n=1}^\infty \mu(B_n \triangle B_{n+1}) < \infty$$

Finally, for the last property, suppose that a subset $E$ is measured by $(A_n)$; we want to show that $\Omega \setminus E$ is measured by $(\Omega \setminus A_n)$. It’s clear that $\liminf (\Omega \setminus A_n) \subset E \subset \limsup (\Omega \ A_n)$. And we already showed that $(\Omega \setminus A_n)$ remains absolutely convergent in the proof of the lemmas, so we are done.

Proof: Finite additivity

We have now proven that $\mathrm{Measurable({\mathcal R}, \mu)}$ is a ring of sets, and the extention of the measure $\mu$ to this ring is well-defined. It remains to prove that it is a valid measure.

For finite additivity, let $E, F \in \mathrm{Measurable({\mathcal R}, \mu)}$ be measured by $(A_n)$ and $(B_n)$. Define $C_n := B_n \setminus A_n$. Then $C_n$ is absolutely convergent via a similar argument to the previous section. And we have

\begin{align*} \liminf C_n &= \bigcup_{N \geq 1} \bigcap_{n \geq N} (B_n \setminus A_n) \\ &\subset \bigcup_{N \geq 1} \left(\bigcap_{n \geq N} B_n \right)\setminus \left(\bigcup_{n \geq N} A_n\right) \\ &\subset \bigcup_{N \geq 1} \left(\bigcap_{ \geq N} B_n \right)\setminus \limsup A_n \\ &= \liminf B_n \setminus \limsup A_n \\ &\subset \liminf B_n \setminus E \\ &\subset F \setminus E \end{align*}

and symmetrically

\begin{align*} \limsup C_n &= \bigcap_{N \geq 1} \bigcup_{n \geq N} (B_n \setminus A_n) \\ &\supset \bigcap_{N \geq 1} \left(\bigcup_{n \geq N} B_n\right) \setminus \left(\bigcap_{n \geq N} A_n\right) \\ &\supset \limsup B_n \setminus \liminf A_n \\ &\supset F \setminus E \end{align*}

therefore $\liminf C_n \subset F \setminus E \subset \limsup C_n$, so $(C_n)$ measures $F \setminus E$. By a similar argument to this, $(A_n \cup B_n)$ measures $E \cup F$. And we trivially have $\lim \mu(A_n \cup B_n) = \lim \mu(A_n) + \lim \mu(B_n \setminus A_n)$, so we have proved that $\mu(E \cup F) = \mu(E) + \mu(F \setminus E)$.

Real half-intervals measured by length

Let ${\mathcal H} := \{[a, b) \mid 0 \leq a < b < 1\}$ be the set of half-open intervals in $[0, 1]$. The closure of ${\mathcal H}$ under finite disjoint unions is a ring ${\mathcal R}_\mathrm{half}$.

Define $\mu: {\mathcal D} \to [0, \infty)$ via $\mu([a, b)) := (b-a)$ to be the length measure. And extend it to ${\mathcal R}_\mathrm{half}$ via $\mu(\bigsqcup_{i=1}^m I_i) := \sum_{i=1}^m \mu(I_i)$ where $\bigsqcup_{i=1}^m I_i$ is a finite disjoint union of half-intervals.

Theorem. $({\mathcal R}_\mathrm{half}, \mu)$ is a measured ring.

Proof. We omit the proof that $\mu$ is well-defined and additive. To show that $\mu$ is continuous, let $B_n \searrow \varnothing$. Then proving $\lim \mu(B_n) = 0$ is equivalent to proving $\lim \mu(A_n) = 1$ where $A_n := [0, 1) \setminus B_n$.

We always have $\lim \mu(A_n) \leq \mu([0, 1))$, so it remains to show the other direction. Define $A_0 = \varnothing$ for notational convenience. Then each difference $A_{n+1} \setminus A_n = \bigsqcup_{k=1}^{K_n} [a_{n,k}, b_{n,k})$ is a finite disjoint union of intervals, which can be canonically ordered by their lower bound. Therefore the overall limit is

\begin{align*} \lim_{N \to \infty} \mu(A_N) &= \lim_{N \to \infty} \mu\left(\bigsqcup_{n=1}^N A_n \setminus A_{n-1}\right) \\ &= \lim_{N \to \infty} \mu\left(\bigsqcup_{n=1}^N \bigsqcup_{k=1}^{K_n} [a_{n,k}, b_{n,k})\right) \\ &= \lim_{N \to \infty} \sum_{n=1}^N \sum_{k=1}^{K_n} \mu([a_{n,k}, b_{n,k})) \\ &= \sum_{n=1}^\infty \sum_{k=1}^{K_n} (b_{n,k} - a_{n,k}) \end{align*}

Therefore, the proof reduces to showing that if a countable disjoint union of half-intervals is equal to $[0, 1)$, then the sum of its measure is at least $1$.

Let $[a_n, b_n)$ be disjoint half-intervals such that $\bigsqcup_{n=1}^\infty [a_n, b_n) = [0, 1)$. We want to show that $\sum_{n=1}^\infty (b_n - a_n) \geq 1$.

Let $\varepsilon > 0$ be arbitrary. Define $U_n := (a_n-\varepsilon/2^n, b_n+\varepsilon/2^n)$, so that $[a_n, b_n) \subset U_n$ and $U_n$ is an open set in ${\mathbb R}$. This implies that $\bigcup_{n=1}^\infty U_n$ covers $[0, 1)$. In particular, it covers $[0, 1-\varepsilon]$, which is compact. Therefore, we can extract a finite subcover $U_{k_1}, \dots, U_{k_N}$ such that

$$[0, 1-\varepsilon) \subset [0, 1-\varepsilon] \subset \textstyle\bigcup_{i=1}^N U_{k_i} \subset \textstyle\bigcup_{i=1}^N [a_{k_i}-\varepsilon/2^{k_i}, b_{k_i}+\varepsilon/2^{k_i}).$$

and therefore,

\begin{align*} (1-\varepsilon) &= \mu([0, 1-\varepsilon)) \\ &\leq \mu\left(\bigcup_{i=1}^N [a_{k_i}-\varepsilon/2^{k_i}, b_{k_i}+\varepsilon/2^{k_i})\right) \\ &= \sum_{i=1}^N (b_{k_i}+\varepsilon/2^{k_i}) - (a_{k_i}-\varepsilon/2^{k_i}) \\ &= \sum_{i=1}^N (b_{k_i} - a_{k_i}) + 2\varepsilon/2^{k_i} \\ &\leq 2\varepsilon + \sum_{i=1}^N (b_{k_i} - a_{k_i}) \\ &\leq 2\varepsilon + \sum_{n=1}^\infty (b_n - a_n) \end{align*}

Hence $\sum_{n=1}^\infty (b_n - a_n) \geq 1 - 3\varepsilon$. The choice of $\varepsilon > 0$ was arbitrary, so we have proved that $\sum_{n=1}^\infty (b_n - a_n) \geq 1$ as desired.

$\square$

Corollary. $(\mathrm{Measurable}({\mathcal R}_\mathrm{half}, \mu), \mu)$ is a measured ring.

Proposition. If $E \subset [0, 1)$ is countable, then $E \in \mathrm{Measurable}({\mathcal R}_\mathrm{half}, \mu)$ and $\mu(E) = 0$.

Properties of ${\small (\mathrm{Measurable}({\mathcal R}, \mu), \mu)}$

Continuity?

Metatheorem. In ZF it is not possible to prove that $(\mathrm{Measurable({\mathcal R}, \mu)}, \mu)$ is continuously measured.

Proof.

The following statement is provably false in ZF.

$$(\exists f: {\mathbb N} \to ({\mathbb N} \to {\mathbb R})) : {\mathbb R} = \bigcup_{n \geq 1} f(n)({\mathbb N})$$

That is, ${\mathbb R}$ is not countable union of counted sets.

However, I have been informed that the following related statement is consistent with ZF, in particular it is “true in the Feferman-Levy model”.

$$(\exists f: {\mathbb N} \to {\mathcal P}({\mathbb R}))(\forall n \geq 1) (\exists g_n: {\mathbb N} \to f(n))(g_n({\mathbb N}) = f(n)) : {\mathbb R} = \bigcup_{n \geq 1} f(n)$$

That is, ${\mathbb R}$ may be a countable union of countable sets.

Define $E_N := \bigcup_{n=1}^N f(n)$. Then certainly each $E_N$ is countable and therefore measurable. And it has zero measure. It satisfies $E_N \nearrow \Omega$, but $\lim \mu(E_N) = 0$, so $\mu$ may not be continuous on $\mathrm{Measurable({\mathcal R}, \mu)}$. $\square$

To proceed further we need to develop an abstraction like “codable Borel sets,” cf. Chapter 56 of Measure Theory by D.H. Fremlin. <– What an amazing work of scholarship this treatise is.

So here our alignment with standard real analysis must end.

In ZF there are many scary possibilities. In ZFC we get rid of some of those scary possibilities, at the cost of turning others into scary actualities.

Codable continuity

Definition. We say that a sequence of sets $(E_m)$ is measurable if there exists an enumeration of sequences $((A_{m,n})_{n\geq1})_{m\geq1}$ such that each $E_m$ is measured by $(A_{m,n})_{n\geq1}$. And in this case we say that $((A_{m,n}))$ measures $(E_m)$.

Lemma. Let $(E_n)$ be any sequence in a measured ring which converges absolutely. Then there canonically exists another sequence $(F_n)$ in the ring with the same $\liminf$ and $\limsup$ which converges at an exponential rate.

Proof. Recursively define the subsequence of indices $N_k \in {\mathbb N}$ via

$$N_1 := 1$$

$$N_{k+1} := \min(N_k + 1, \min\{N \geq 1 \mid \textstyle\sum_{n \geq N} \mu(E_{N_k} \triangle E_n) < 2^{-(k+1)}\})$$

This definition implies that $N_{k+1} \geq N_k+1$, in particular, $N_k \nearrow \infty$ as $k \to \infty$. Therefore, if we define $B_n$ via

$$F_{2k} := \bigcup_{n=N_k}^{N_{k+1}-1} E_n$$

$$F_{2k+1} := \bigcap_{n=N_k}^{N_{k+1}-1} E_n$$

then $(F_n)$ preserves the $\limsup$ and $\liminf$ of $(E_n)$, while now converging at a uniform rate, because, firstly:

\begin{align*} \limsup_{k \to \infty} F_j &= \bigcap_{N \geq 1} \bigcup_{j \geq N} F_j \\ &= \bigcap_{N \geq 1} \bigcup_{j \geq 2N} F_j \\ &= \bigcap_{N \geq 1} \bigcup_{k \geq N} (F_{2k} \cup F_{2k+1}) \\ &= \bigcap_{N \geq 1} \bigcup_{k \geq N} \bigcup_{n=N_k}^{N_{k+1}-1} E_n \\ &= \bigcap_{N \geq 1} \bigcup_{n \geq N} E_n \\ &= \limsup_{n \to \infty} E_n \end{align*}

Likewise, $\liminf_{k \to \infty} F_k = \liminf_{n \to \infty} E_n$. To calculate the rate of absolute convergence, first observe that

$$d_\mu(E_{N_k}, F_{2k}) = d_\mu\left(E_{N_k},\, \textstyle\bigcup_{n=N_k}^{N_{k+1}-1} E_n\right) \leq 2^{-k}$$

$$d_\mu(E_{N_k}, F_{2k+1}) = d_\mu\left(E_{N_k},\, \textstyle\bigcap_{n=N_k}^{N_{k+1}-1} E_n\right) \leq 2^{-k}$$

Therefore the rate of absolute convergence is yielded by the calculations

\begin{align*} \sum_{k \geq K} d_\mu(F_{2k}, F_{2k+1}) &= \sum_{k \geq K} d_\mu(F_{2k}, E_{N_k}) + d_\mu(E_{N_k},F_{2k+1}) \\ &\leq \sum_{k \geq K} 2^{-k} + 2^{-k} \\ &= 2 \cdot 2^{-(K-1)} \end{align*}

and

\begin{align*} &\sum_{k \geq K} d_\mu(F_{2k+1}, F_{2k+2}) \\ &\quad= \sum_{k \geq K} d(F_{2k+1}, E_{N_k}) + d(E_{N_k}, E_{N_{k+1}}) + d(E_{N_{k+1}}, F_{2k+2}) \\ &\quad\leq \sum_{k \geq K} 2^{-k} + 2^{-k} + 2^{-k} \\ &\quad= 3 \cdot 2^{-(K-1)} \end{align*}

hence

\begin{align*} \sum_{n \geq N} d_\mu(F_n, F_{n+1}) &\leq \sum_{k \geq \lfloor N/2\rfloor} d_\mu(F_{2k}, F_{2k+1}) + \sum_{k \geq \lfloor N/2\rfloor} d_\mu(F_{2k+1}, F_{2k+2}) \\ &\leq 5 \cdot 2^{-(\lfloor N/2 \rfloor-1)} \\ &\leq 2^{-N/2+1} \end{align*}

Therefore, the sequence $(F_n)$ has the same $\liminf$ and $\limsup$ as $(E_n)$ while converging at the rate $\sum_{n \geq N} d_\mu(F_n, F_{n+1}) \leq 2^{-N/2+1}$. This rate can be made arbitrarily faster by further accelerating the sequence in a similar fashion.

$\square$

Lemma. If $(E_m)$ is a measurable sequence in $\mathrm{Measurable}({\mathcal R}, \mu)$, then it can be measured by an enumeration of sequences with uniform rate of absolute convergence.

Proof. Given that $(E_m)$ is measured by $((A_{m,n}))$, use the previous lemma to canonically replace each $(A_{m,n})_{n\geq1}$ with a new sequence $(B_{m,n})_{n\geq1}$ such that every $(B_{m,n})$ converges at an exponential rate. $\square$

Theorem. Let $(E_m)$ be a measurable sequence such that $E_m \searrow \varnothing$. Then $\mu(E_m) \to 0$.

Proof. Let $((A_{m,n}))$ measure $(E_m)$. By the previous lemma, we can assume without loss of generality that it has uniform rate of convergence $\sum_{n \geq N} \mu(A_{m,n} \triangle A_{m,n+1}) < 1/N$.

Since have $E_{m+1} \subset E_m$, we can define

$B_{1,n} := A_{1,n}$
$B_{m+1,n} := B_{m,n} \cap A_{m,n}$

and it will still be the case that $((B_{m,n}))$ measures $(E_m)$ at the same uniform rate of convergence [details omitted].

Define $C_{N,m} := \bigcap_{n=N}^m B_{m,n}$ to be the intersection of $(B_{m,n})_{n \geq 1}$ from index $N$ to $m$. Then we have $C_{N,m+1} \subset C_{N,m}$ due to $\bigcap_{n=N}^{m+1} B_{m+1,n} \subset \bigcap_{n=N}^{m+1} B_{m,n} \subset \bigcap_{n=N}^m B_{m,n}$, and in addition, this sequence must decrease to the empty set because

\begin{align*} \bigcap_{m \geq 1} \bigcap_{n=N}^m B_{m,n} &= \bigcap_{M \geq 1} \bigcap_{m \geq M} \bigcap_{n=N}^m B_{m,n} \\ &\subset \bigcap_{M \geq 1} \bigcap_{m \geq M} \bigcap_{n=N}^m B_{M,n} \\ &= \bigcap_{M \geq 1} \bigcap_{n \geq N} B_{M,n} \\ &\subset \bigcap_{M \geq 1} \liminf_{n \to \infty} B_{M,n} \\ &\subset \bigcap_{M \geq 1} E_M \\ &= \varnothing \end{align*}

Therefore for any $N \geq 1$ we have $\lim_{m \to \infty} \mu(C_{N,m}) = 0$ by the continuity of $\mu$ in ${\mathcal R}$. As a result, we can calculate $\lim_{m \to \infty} \mu(E_m)$ via:

\begin{align*} \lim_{m \to \infty} \mu(E_n) &= \lim_{m \to \infty} \lim_{n \to \infty} \mu(B_{m,n}) \\ &\leq \inf_{N \geq 1}\lim_{m \to \infty} \sup_{n \geq N} \mu(B_{m,n}) \\ &\leq \inf_{N \geq 1}\left(\lim_{m \to \infty} \inf_{N_1 \geq 1} \left(\mu\left(\bigcap_{n=N}^{N_1} B_{m,n}\right) + 1/N\right)\right) \\ &\leq \inf_{N \geq 1}\left(\lim_{m \to \infty} \mu\left(\bigcap_{n=N}^{m} B_{m,n}\right) + 1/N\right) \\ &= \inf_{N \geq 1}\left(\lim_{m \to \infty} \mu(C_{N,m}) + 1/N\right) \\ &= \inf_{N \geq 1} (0 + 1/N) \\ &= 0 \end{align*}

$\square$

We have therefore proved that $(\mathrm{Measurable}({\mathcal R}, \mu), \mu)$ is a nearly continuously measured ring, so long as we modify the definition of continuous to only include measurable sequences, rather than arbitrary shrinking sequences.

However, it still remains to prove that $(\mathrm{Measurable}({\mathcal R}, \mu), \mu)$ is complete, i.e., that every measurable absolutely convergent sequence has a limit. I am not sure how to do this.

It might not be possible.