[!INFO] As you can see, the rendering in this is a bit janky (this is what I get for deciding not to use
$\LaTeX{}$ for this). I had higher hopes for<details/>blocks, but alas. You might find more success rendering the markdown locally (this is at least true for me).Should be okay, since nobody is going to read this anyway π€·
A
$\sigma$ -algebra over a set$X$ is a subset$\Sigma\subseteq 2^X$ (that is, a set of subsets of$X$ ), satisfying:
- If
$I$ is some countable set ($0\leq|I|\leq\omega$ ), and$U_i\in\Sigma$ for$i\in I$ , then$\bigcup_{i\in I}U_i \in \Sigma$ .- If
$U\in\Sigma$ , then$X\setminus U\in\Sigma$ .A measurable space is a pair
$(X, \Sigma)$ , where$X$ is some set, and$\Sigma$ is a$\sigma$ -algebra over$X$ .In such a situation, a subset
$S\subseteq X$ is called measurable iff$S\in\Sigma$ .
Note, in particular, that every
Fix a measurable space
$(X, \Sigma)$ . A measure on$(X, \Sigma)$ is a function$\mu : \Sigma \to [0, \infty]$ such that:
$\mu(\varnothing) = 0$ .- If
$I$ is some countable set, and$U_i\in\Sigma$ for$i\in I$ such that$U_i\cap U_j=\varnothing$ for$i\neq j$ , then$$\mu\left(\bigcup_{i\in I}U_i\right) = \sum_{i\in I}\mu(U_i)$$ Naturally, a measure space is a triple
$(X, \Sigma, \mu)$ , where$(X, \Sigma)$ is a measurable space, and$\mu$ is a measure on$(X, \Sigma)$ .
A Boolean algebra is a sextuple
$(A, \land, \lor, \lnot, \top, \bot)$ satisfying:
$(A, \land, \top)$ is a commutative monoid.$(A, \lor, \bot)$ is a commutative monoid.- (Absorption) For all
$a, b, c\in A$ , we have$a \land (a\lor b) = a\lor(a\land b) = a$ .- (Distributivity) For
$a, b, c\in A$ , we have$$\begin{aligned} a\land(b\lor c) &= (a\land b)\lor(a\land c) \\\ a\lor(b\land c) &= (a\lor b)\land(a\lor c) \end{aligned}$$
- (Annihilation) For all
$a\in A$ , we have$a\land\bot=\bot$ , and$a\lor\top=\top$ .- (Complementation) For all
$a\in A$ , we have$a\land\lnot a=\bot$ and$a\lor\lnot a=\top$ .
What is a commutative monoid?
A monoid is a triple
- Multiplication is associative: for
$a,b,c\in M$ , we have$a\circ(b\circ c) = (a\circ b)\circ c$ . -
$1$ is a neutral element: for$a\in M$ , we have$1\circ a = a = a\circ1$ .
A monoid is further called commutative if
Every Boolean algebra
The partial ordering completely determines the Boolean algebra, through suprema and infima (and the fact that axiom 6 completely determines
Proof (of unique complements).
Suppose
b = b\land\top = b\land(a\lor b') = (b\land a)\lor(b\land b') = \bot\lor(b\land b') = b\land b'
Analogously,
Note. Since conjunction and disjunction are symmetric, the definition of
$\lnot a$ also implies$\lnot\lnot a = a$ (i.e., double negation).
Proof (of partial ordering being well-defined).
If
It remains to show that this indeed defines a partial ordering.
- (Reflexivity)
$a\preceq a$ for all$a\in A$ , since$a\land a = a\land(a\lor\bot) = a$ . - (Antisymmetry) If
$a\preceq b$ (i.e.,$a\land b=a$ ) and$b\preceq a$ (i.e.,$b\land a=b$ ), then$a = a\land b=b\land a=b$ . - (Transitivity) If
$a\preceq b$ (i.e.,$a\land b=a$ ) and$b\preceq c$ (i.e.,$b\land c=b$ ), then$a\land c=(a\land b)\land c=a\land(b\land c)=a\land b=a$ ; that is,$a\preceq c$ .
This completes the proof.
Given a partially-ordered set
$(P, \preceq)$ , and a subset$S\subseteq P$ , an upper bound of$S$ is any$u\in P$ such that$s\preceq u$ for every$s\in S$ . If there is an upper bound $u^\in P$ of $S$ such that, for any other upper bound $u\in P$ of $S$, we have $u^\preceq u$, then $u^$ is called the supremum of $S$, denoted $u^ =: \sup S$.Entirely dually, a lower bound of
$S$ is any$\ell\in P$ such that$\ell\preceq s$ for every$s\in S$ . If there is a lower bound $\ell^\in P$ of $S$ such that, for any other lower bound $\ell\in P$ of $S$, we have $\ell\preceq\ell^$, then $ \ell^$ is called the infimum of $S$, denoted $\ell^ =: \inf S$.
In a Boolean algebra, one can show that
Proof.
If
The argument for
A Boolean algebra
$A$ is complete if every subset$S\subseteq A$ admits a supremum and an infimum.For a family
$(a_i)_{i\in I}$ of elements of$A$ , the supremum and infimum may be denoted$$\bigvee_{i\in I}a_i := \sup_{i\in I}a_i, \qquad \bigwedge_{i\in I}a_i := \inf_{i\in I}a_i$$
Let
$A$ be a Boolean algebra, and$I$ some indexing set. Then,$A$ has suprema of all$I$ -indexed families iff it has infima of all$I$ -indexed families. In such a situation, if$S = (a_i)_{i\in I}$ is a family of elements of$A$ , then$$\bigvee_{i\in I}a_i = \lnot\bigwedge_{i\in I}\lnot a_i, \qquad \bigwedge_{i\in I}a_i = \lnot\bigvee_{i\in I}\lnot a_i$$ In particular, a Boolean algebra is complete iff it has arbitrary suprema.
Proof.
Let
Since
Suppose
By the universal property of
As
As
The set of Boolean truth values
$\Omega = {\top, \bot}$ defines a complete Boolean algebra.
Since Boolean algebras form an algebraic theory, they admit arbitrary products.
This employs us to construct larger Boolean algebras from
Given a set
$X$ , the power set$2^X$ can be identified with the product of$|X|$ -many copies of (the underlying set of)$\Omega$ , and so admits a canonical Boolean algebra structure.Under this identification,
$S\land T := S\cap T$ $S\lor T := S\cup T$ $\lnot S := X\setminus S$ $\top := X$ $\bot := \varnothing$ .
Every
$\sigma$ -algebra is a Boolean algebra.
Proof.
Since a
Given a Boolean algebra
$A$ , an ideal of$A$ is a subset$I\subseteq A$ such that
$\bot \in I$ ,- If
$x, y\in I$ , then$x\lor y\in I$ ,- If
$a\in A$ and$z\in I$ , then$a\land z\in I$ .
Let
$(X, \Sigma, \mu)$ be a measure space. Say that$S\in\Sigma$ is a null set if$\mu(S) = 0$ . Then, the null sets define an ideal of$\Sigma$ .
Proof.
The other axioms follow from more general properties of a measure: for
-
$\mu(U\cup V)\leq\mu(U)+\mu(V)$ (which implies axiom 2). -
$\mu(U\cap V)\leq\mu(U)$ (which implies axiom 3).
These both follow from monotonicity: if
To see this, note that if
Given an ideal
Proof that this defines an equivalence relation.
We need to show that the equivalence relation is reflexive, symmetric, and transitive.
Reflexivity is clear: for any
Symmetry is by design:
For transitivity, we will show that
To see that
We already have that
To this end, notice that the right hand side of
\begin{aligned}
(a\oplus b)\oplus c &= ((a\land\lnot b)\lor(b\land\lnot a))\oplus c \\
&= (((a\land\lnot b)\lor(b\land\lnot a))\land\lnot c)\lor(c\land \lnot((a\land\lnot b)\lor(b\land\lnot a))) \\
&= ((a\land\lnot b\land\lnot c)\lor(\lnot a\land b\land\lnot c))\lor(c\land((\lnot a\lor b)\land(\lnot b\lor a))) \\
&= ((a\land\lnot b\land\lnot c)\lor(\lnot a\land b\land\lnot c))\lor(c\land((\lnot a\land\lnot b)\lor(\lnot a\land a)\lor(b\land\lnot b)\lor(b\land a))) \\
&= ((a\land\lnot b\land\lnot c)\lor(\lnot a\land b\land\lnot c))\lor(c\land((\lnot a\land\lnot b)\lor(a\land b))) \\
&= ((a\land\lnot b\land\lnot c)\lor(\lnot a\land b\land\lnot c))\lor((\lnot a\land\lnot b\land c)\lor(a\land b\land c)) \\
&= (a\land\lnot b\land\lnot c)\lor(\lnot a\land b\land\lnot c)\lor(\lnot a\land\lnot b\land c)\lor(a\land b\land c)
\end{aligned}
is invariant under permutations of
The quotient of
If
$A$ is a Boolean algebra, and$I\subseteq A$ is an ideal, then the quotient$A/I$ inherits the structure of a Boolean algebra.
Proof.
For
$[a]\land[b] := [a\land b]$ $[a]\lor[b] := [a\lor b]$ $\lnot[a] := [\lnot a]$
That these operations satisfy the axioms of a Boolean algebra is automatic. What isn't automatic is that these operations are well-defined.
Suppose
$a\land b\sim_Ia'\land b$ $a\lor b\sim_Ia'\lor b$ $\lnot a\sim_I\lnot a'$
(Commutativity saves us from having to check more than this.)
These follow from generally useful facts about the symmetric difference operator.
a + b = c
Claim 3.4.i.
Conjunction distributes over symmetric difference:
$(a\oplus a')\land b = (a\land b)\oplus(a'\land b)$ .
Proof.
\begin{align*}
(a\land b)\oplus(a'\land b) &= ((a\land b)\land\lnot(a'\land b))\lor(\lnot(a\land b)\land(a'\land b)) \tag{definition} \\
&= (a\land b\land(\lnot a\lor\lnot b))\lor((\lnot a\lor\lnot b)\land a'\land b) \tag{de Morgan} \\
&= ((a\land b\land\lnot a')\lor\underbrace{(a\land b\land\lnot b)}_{=\bot})\lor((\lnot a\land a'\land b)\lor\underbrace{(\lnot b\land a'\land b)}_{=\bot}) \tag{distributivity} \\
&= (a\land b\land\lnot a')\lor(\lnot a\land a'\land b) \tag{$\bot$ is $\lor$ -neutral} \\
&= ((a\land\lnot a')\lor(\lnot a\land a'))\land b \tag{distributivity} \\
&= (a\oplus a')\land b \tag{definition}
\end{align*}
This completes the proof.
In particular, if
(a\land b)\oplus(a'\land b) = \underbrace{(a\oplus a')}_{\in I}\land b \in I
shows that conjunction in
Claim 3.4.ii.
Disjunction anti-distributes over symmetric difference:
$(a\oplus a')\land\lnot b = (a\lor b)\oplus(a'\lor b)$ .
Proof.
\begin{align*}
(a\lor b)\oplus(a'\lor b) &= ((a\lor b)\land\lnot(a'\lor b))\lor(\lnot(a\lor b)\land(a'\lor b)) \tag{definition} \\
&= ((a\lor b)\land\lnot a'\land\lnot b)\lor(\lnot a\land\lnot b\land(a'\lor b)) \tag{de Morgan} \\
&= (a\land\lnot a'\land\lnot b)\lor\underbrace{(b\land\lnot a'\land\lnot b)}_{=\bot}\lor(\lnot a\land\lnot b\land a')\lor\underbrace{(\lnot a\land\lnot b\land b)}_{=\bot} \tag{distributivity} \\
&= (a\land\lnot a'\land\lnot b)\lor(\lnot a\land\lnot b\land a') \tag{$\bot$ is $\lor$ -neutral} \\
&= ((a\land\lnot a')\lor(\lnot a\land a'))\land\lnot b \tag{distributivity} \\
&= (a\oplus a')\land\lnot b \tag{definition}
\end{align*}
This completes the proof.
In particular, if
(a\lor b)\oplus(a'\lor b) = \underbrace{(a\oplus a')}_{\in I}\land\lnot b\in I
shows that disjunction in
Claim 3.4.iii
Negation transfers over symmetric difference:
$(\lnot a)\oplus a' = a\oplus(\lnot a')$ .
Proof.
\begin{align*}
(\lnot a)\oplus a' &= (\lnot a\land\lnot a')\lor(a'\land\lnot(\lnot a)) \tag{definition} \\
&= (\lnot a\land\lnot a')\lor(a\land a') \tag{double negation}
\end{align*}
By commutativity, the last expression is invariant under swapping
In particular, if
(\lnot a)\oplus(\lnot a') = a\oplus(\lnot\lnot a') = a\oplus a' \in I
This concludes the proof.
Let
Let
$(X, \Sigma, \mu)$ be a measure space such that$X = \bigcup_{n=0}^\infty U_n$ , where$U_n\in\Sigma$ and$\mu(U_n)\lt\infty$ . (Such a measure space is called$\sigma$ -finite).Then,
$L(X)$ is a complete Boolean algebra.
Proof.
We need to show that
For organisation purposes, we prove this theorem through a sequence of claims.
Claim 4.1.i.
$L(X)$ has suprema for all countable chains.
Proof.
[!INFO] In any partially-ordered set
$(P,\preceq)$ , a chain in$P$ is a totally-ordered subset$C\subseteq P$ , meaning for all$c_1,c_2\in C$ , either$c_1\preceq c_2$ or$c_2\preceq c_1$ .
Suppose
[V_0]\preceq[V_1]\preceq[V_2]\preceq\cdots
is a countable chain in
Then, let
Suppose
\begin{align*}
\mu(V\setminus U) &= \mu\left(\left(\bigcup_{n\geq0}V_n\right)\setminus U\right) \\
&= \mu\left(\bigcup_{n\geq0}(V_n\setminus U)\right) \\
&\leq \sum_{n\geq0}\underbrace{\mu(V_n\setminus U)}_{=0} \\
&= 0
\end{align*}
showing that
Claim 4.1.ii.
$L(X)$ has suprema for all well-ordered chains.
Proof.
Since
Let
Let
Since each
\xi_m^n := \inf\left\{0\leq\xi\lt\kappa ~\middle|~\mu(V_\xi^n)\geq\mu^n-\frac1m\right\}
Then, the
If
On the other hand, if
This means
\begin{align*}
\mu^n = \mu(V_\xi^n) &= \mu(V_\xi^n\setminus V^n) + \mu(V_\xi^n\cap V^n) \\
&= \mu(V_\xi^n\setminus V^n) + \mu(V^n) \tag{since $[V^n]\preceq[V_\xi^n]$} \\
&\geq \mu(V_\xi^n\setminus V^n) + \mu(V_{\xi_m^n}^n) \quad \forall m\geq1 \\
&\geq \mu(V_\xi^n\setminus V^n) + \mu^n - \frac1m \quad \forall m\geq1 \tag{definition of $\xi_m^n$} \\
\implies\qquad \mu(V_\xi^n\setminus V^n) &\leq \frac1m \quad \forall m\geq 1 \\
\implies\qquad \mu(V_\xi^n\setminus V^n) &= 0
\end{align*}
and therefore that
This completes the proof that
Now, we have a countable chain of suprema
Let
Claim 4.1.iii.
$L(X)$ admits arbitrary suprema.
Proof.
[!WARNING] I hope you like the axiom of choice! π
Let
Construct a chain
-
$[\tilde V_0] := [\varnothing]$ . - Given
$[\tilde V_\xi]$ , define$[\tilde V_{\xi+1}] := [\tilde V_\xi]\lor [V_\xi]$ . - For a limit ordinal
$\lambda\leq\kappa$ , if we have$[\tilde V_\xi]$ for every$\xi\lt\lambda$ , then define$[\tilde V_\lambda] := \sup_{\xi\lt\lambda}[\tilde V_\xi]$ to be the supremum of the chain constructed thus far, by Claim 4.1.ii.
An invariant of this construction is that
In particular,
Combining Claim 4.1.iii with de Morgan's laws, this concludes the proof that
Given a complete Boolean algebra
$A$ , an atom is a minimal non-bottom element of$A$ ; that is,$a\in A$ is an atom if$a\neq\bot$ , and whenever$a'\preceq a$ , either$a'=\bot$ or$a'=a$ .A complete Boolean algebra
$A$ is called atomic if every element of$A$ is the supremum of some set of atoms.
We are finally in a position to provide an example of a complete Boolean algebra that is not atomic.
Suppose
$(X, \Sigma, \mu)$ is a$\sigma$ -finite measure space such that for every$U\in\Sigma$ , there exists some measurable subset$V\subseteq U$ with$\mu(V) = \frac{\mu(U)}2$ .Then,
$L(X)$ has no atoms.
[!INFO] An example of such a measure is the Lebesgue measure on
$\mathbb{R}^n$ .
You thought nobody's gonna read it, but here I am, having already found mistake.$\mu:\Sigma\to[0,\infty]$ , not $\mu:X\to[0,\infty]$ .
In Definition 1.2, a measure is a function
I wonder if I find more, will you then have to pay penance for your penance? π€