nyuuzyou/wikis huggingface dataset:
Stone Space of Boolean Lattice is Topological Space

== Theorem ==
Let $\struct {B, \preceq, \wedge, \vee}$ be a Boolean lattice.
Let $\struct {U, \tau}$ be the Stone space of $B$.
Then $\struct {U, \tau}$ is a topological space.
== Proof ==
The topology of the Stone space is defined as the topology generated by the basis $Q$ consisting of all sets of the form
:$\set {x \in U: b \in x}$
for some $b \in B$, where $U$ is the set of all ultrafilters on $B$.
By Union from Synthetic Basis is Topology, it suffices to show that $Q$ is a synthetic basis.
First, we must show that $Q$ is a cover of $U$.
Let $x_0 \in U$ be arbitrary.
By filter axiom $\paren 1$, $x_0 \ne \O$.
Thus, there is some $b_0 \in x_0$.
Then, $x_0 \in \set {x \in U: b_0 \in x} \in Q$.
As $x_0$ was arbitrary, it follows that $Q$ is a cover of $U$.
Hence, synthetic basis axiom $\paren {\text B 1}$ holds for $Q$.
Let $P, Q \in Q$ be arbitrary.
Then for some $b, c \in B$:
:$P = \set {x \in U: b \in x}$
:$Q = \set {y \in U: c \in y}$
Let $R = \set {z \in U : b \wedge c \in z}$
We will show that:
:$P \cap Q = R$
Let $z \in P \cap Q$ be arbitrary.
Then by the definitions of $P$ and $Q$:
:$b \in z$
:$c \in z$
By definition of lattice, $\wedge$ denotes the meet operator.
Therefore:
:$b \wedge c \preceq b$
:$b \wedge c \preceq c$
:$\forall d: (d \preceq b) \land (d \preceq c) \implies (d \preceq b \wedge c)$
Therefore, by filter axioms $\paren 2$ and $\paren 3$, $b \wedge c \in z$.
As $z \in P \cap Q$ was arbitrary, it follows that $P \cap Q \subseteq R$.
Now, let $v \in R$ be arbitrary.
Then $b \wedge c \in v$.
As stated above:
:$b \wedge c \preceq b$
:$b \wedge c \preceq c$
Therefore, by filter axiom $\paren 3$:
:$b \in v$
:$c \in v$
Thus $v \in P \cap Q$.
Since this holds for each $v \in R$, $R \subseteq P \cap Q$.
From the above, it follows that $P \cap Q = R$ by the definition of set equality.
Therefore, synthetic basis axiom $\paren {\text B 2}$ holds for $Q$ by taking $\AA = \set {P \cap Q}$.
Category:Stone Spaces
Stopped Clock Paradox

== Paradox ==
Of two clocks, one is better than the other if it shows the absolutely correct time more often than the other.
Consider:
:Clock $A$, which loses $1$ minute every day
:Clock $B$, which is stopped.
Clock $A$ is $1$ minute slow after day $1$, and $2$ minutes slow after day $2$, and so on.
There are $720$ minutes in $12$ hours.
Hence it will be $720$ days, or nearly $2$ years, before it shows the right time again.
On the other hand, clock $B$ shows the correct time every $12$ hours.
Hence clock $B$, a stopped clock, is better than clock $A$, which loses but a minute a day.
== Resolution ==
First thing to note is that a clock that loses $1$ minute a day is a poor clock, and on an practical level not all that much better than a clock that does not run at all.
But let us modify our absolutely correct to correct within $30$ seconds, which is reasonable for an old dial clock.
One can assume that an adequate clock can be adjusted so that it will lose or gain no more than $30$ seconds over the course of a number of months.
On the old-fashioned mechanical clocks, there is usually a control on the back which one can use to minutely control the speed of a clock so that it can be made to be as accurate as your patience can achieve.
If I had a clock which was worse than that, I would throw it out and get a better one.
== Historical Note ==
== Sources ==
* 
* : Chapter Two: Paradoxes for Everyone
Category:Veridical Paradoxes
Stopped Martingale is Martingale

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {X_n}_{n \ge 0}$ be an $\sequence {\FF_n}_{n \ge 0}$-martingale.
Let $T$ be a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Let $\sequence {X_n^T}_{n \ge 0}$ be the stopped process.
Then $\sequence {X_n^T}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \ge 0}$-martingale.
=== Corollary ===
== Proof ==
From Adapted Stochastic Process is Martingale iff Supermartingale and Submartingale:
:$\sequence {X_n}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \ge 0}$-supermartingale and $\sequence {\FF_n}_{n \ge 0}$-submartingale.
From Stopped Supermartingale is Supermartingale and Stopped Submartingale is Submartingale:
:$\sequence {X_n^T}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \ge 0}$-supermartingale and $\sequence {\FF_n}_{n \ge 0}$-submartingale.
From Adapted Stochastic Process is Martingale iff Supermartingale and Submartingale:
:$\sequence {X_n^T}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \ge 0}$-martingale.
Category:Stopped Processes
Category:Martingales
Category:Stopped Martingale is Martingale
Stopped Martingale is Martingale/Corollary

== Corollary ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {X_n}_{n \ge 0}$ be an $\sequence {\FF_n}_{n \ge 0}$-martingale.
Let $T$ be a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Let $\sequence {X_n^T}_{n \ge 0}$ be the stopped process.
Then we have: 
:$\expect {X_n^T} = \expect {X_0}$ for each $n \in \Z_{\ge 0}$.
== Proof ==
From Stopped Martingale is Martingale:
:$\sequence {X_n^T}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \ge 0}$-martingale.
From Expected Value of Martingale is Constant in Time, we have:
:$\expect {X_n^T} = \expect {X_0^T}$ 
We have that $\map T \omega \ge 0$ for each $\omega \in \Omega$, so $0 \wedge T = 0$.
So, we have:
:$X_0^T = X_0$
and so:
:$\expect {X_0^T} = \expect {X_0}$
giving the result.
Category:Stopped Martingale is Martingale
Stopped Process is Adapted Stochastic Process

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {X_n}_{n \ge 0}$ be an $\sequence {\FF_n}_{n \ge 0}$-adapted stochastic process.
Let $T$ be a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$. 
Let $\sequence {X_n}_{n \ge 0}$ be a $\sequence {\FF_n}_{n \ge 0}$-adapted process.
Let $\sequence {X_n^T}_{n \ge 0}$ be the stopped process.
Then $\sequence {X_n^T}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \ge 0}$-adapted process.
== Proof ==
Let $n \in \Z_{\ge 0}$. 
From Constant Function is Stopping Time, $n$ is a stopping time.
From Pointwise Minimum of Stopping Times is Stopping Time, $n \wedge T$ is a stopping time, where $\wedge$ is the pointwise minimum.
From Adapted Stochastic Process at Stopping Time is Measurable with respect to Stopped Sigma-Algebra, we have: 
:$X_{n \wedge T}$ is $\FF_{n \wedge T}$-measurable
where $\FF_{n \wedge T}$ is the stopped $\sigma$-algebra associated with $n \wedge T$. 
We have by the definition of pointwise minimum: 
:$n \wedge T \le n$
From Stopped Sigma-Algebra preserves Inequality between Stopping Times and Stopped Sigma-Algebra of Constant Stopping Time coincides with Filtration, we have:
:$\FF_{n \wedge T} \subseteq \FF_n$ 
So:
:$X_{n \wedge T}$ is $\FF_n$-measurable.
So:
:$\sequence {X_n^T}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \ge 0}$-adapted process.
Category:Stopped Processes
Category:Adapted Stochastic Processes
Stopped Sigma-Algebra is Sigma-Algebra

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $T$ be a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$. 
Let $\FF_T$ be the stopped $\sigma$-algebra associated with $T$.
Then $\FF_T$ is a $\sigma$-algebra.
== Proof ==
We show that $\Omega \in \FF_T$, that $\FF_T$ is closed under countable intersection, and relative complement. 
For each $t \in \Z_{\ge 0}$ we have: 
:$\Omega \cap \set {\omega \in \Omega : \map T \omega \le t} = \set {\omega \in \Omega : \map T \omega \le t}$
Since $T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$ we have:
:$\Omega \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
So $\Omega \in \FF_T$.
Now let $\sequence {A_n}_{n \mathop \in \N}$ be a sequence of sets in $\FF_T$.
Then for each $t \in \Z_{\ge 0}$, we have: 
:$\ds \paren {\bigcap_{n \in \N} A_n} \cap \set {\omega \in \Omega : \map T \omega \le t} = \bigcap_{n \in \N} \paren {A_n \cap \set {\omega \in \Omega : \map T \omega \le t} }$
from Intersection is Associative.
Since $A_n \in \FF_T$ for each $n \in \N$, we have:
:$A_n \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$ 
for each $n \in \N$.
Since $\FF_t$ is closed under countable intersection, we have: 
:$\ds \paren {\bigcap_{n \in \N} A_n} \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
for each $t \in \Z_{\ge 0}$.
So:
:$\ds \paren {\bigcap_{n \in \N} A_n} \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_T$
Finally we look to prove that if $A \in \FF_T$ then $A^c \in \FF_T$.
Let $t \in \Z_{\ge 0}$. 
Note that:
From De Morgan's Laws (Set Theory): Set Difference: Family of Sets: Difference with Intersection:
:$A^c \cup \set {\omega \in \Omega : \map T \omega > t} = \paren {A \cap \set {\omega \in \Omega : \map T \omega \le t} }^c$
Since $A \in \FF_T$ we have:
:$A \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
Since $\FF_t$ is closed under relative complement we have:
:$\paren {A \cap \set {\omega \in \Omega : \map T \omega \le t} }^c \in \FF_t$
Since $T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$, we have:
:$\set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
Since $\FF_t$ is closed under finite intersection, we have:
:$\set {\omega \in \Omega : \map T \omega \le t} \cap \paren {A^c \cup \set {\omega \in \Omega : \map T \omega > t} } \in \FF_t$
so that:
:$A^c \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
Since this holds for all $t \in \Z_{\ge 0}$, we have $A^c \in \FF_T$.
Category:Stopped Sigma-Algebras
Category:Sigma-Algebras
Category:Examples of Sigma-Algebras
Stopped Sigma-Algebra of Constant Stopping Time coincides with Filtration

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {X_n}_{n \ge 0}$ be an $\sequence {\FF_n}_{n \ge 0}$-adapted stochastic process.
Let $n \in \Z_{\ge 0}$.
Let $T$ be a stopping time such that: 
:$\map T \omega = n$
for each $\omega \in \Omega$.
Let $\FF_T$ be the stopped $\sigma$-algebra associated with $T$.
Then:
:$\FF_T = \FF_n$
== Proof ==
We first show $\FF_n \subseteq \FF_T$.
Let $A \in \FF_n$. 
Note that for $t \in \Z_{\ge 0}$, we have: 
:$\set {\omega \in \Omega : \map T \omega \le t} = \O$
if $t < n$.
That is:
:$A \cap \set {\omega \in \Omega : \map T \omega \le t} = \O$ 
if $t < n$.
So certainly:
:$A \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
if $t < n$. 
If $t \ge n$ we have:
:$\set {\omega \in \Omega : \map T \omega \le t} = \Omega$
so that:
:$A \cap \set {\omega \in \Omega : \map T \omega \le t} = A \in \FF_n$
So:
:$A \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_n$
for $t \ge n$. 
Since $\sequence {\FF_n}_{n \ge 0}$ is a filtration and $t \ge n$, we have: 
:$\FF_n \subseteq \FF_t$
and so:
:$A \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
for $t \ge n$ as well.
So:
:$A \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
for all $t \in \Z_{\ge 0}$.
So $A \in \FF_T$.
So $\FF_n \subseteq \FF_T$.
Now we show $\FF_T \subseteq \FF_n$. 
Let $A \in \FF_T$. 
Then we have: 
:$A \cap \set {\omega \in \Omega : \map T \omega \le n} \in \FF_n$ 
But we have:
:$\set {\omega \in \Omega : \map T \omega \le n} = \Omega$
so:
:$A \cap \set {\omega \in \Omega : \map T \omega \le n} = A$
giving $A \in \FF_n$. 
So $\FF_T \subseteq \FF_n$.
We conclude $\FF_T = \FF_n$.
Category:Stopped Sigma-Algebras
Stopped Sigma-Algebra of Pointwise Minimum of Stopping Times

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a discrete-time filtered probability space.
Let $T$ and $S$ be stopping times with respect to $\sequence {\FF_n}_{n \ge 0}$.
Let $S \wedge T$ be the pointwise minimum of $S$ and $T$.
Then: 
:$\FF_{S \wedge T} = \FF_S \cap \FF_T$
where $\FF_{\paren \cdot}$ denotes the stopped $\sigma$-algebra.
== Proof ==
From the definition of pointwise minimum, we have:
:$S \wedge T \le S$
and:
:$S \wedge T \le T$ 
Then from Stopped Sigma-Algebra preserves Inequality between Stopping Times, we have:
:$\FF_{S \wedge T} \subseteq \FF_S$
and:
:$\FF_{S \wedge T} \subseteq \FF_T$
so that:
:$\FF_{S \wedge T} \subseteq \FF_S \cap \FF_T$
Now let $A \in \FF_S \cap \FF_T$ 
We have, for $t \in \Z_{\ge 0}$ and $\omega \in \Omega$: 
:$\map {\paren {S \wedge T} } \omega \le t$ $\map S \omega \le t$ or $\map T \omega \le t$
That is: 
:$\set {\omega \in \Omega : \map {\paren {S \wedge T} } \omega \le t} = \set {\omega \in \Omega : \map S \omega \le t} \cup \set {\omega \in \Omega : \map T \omega \le t}$ 
Then, from Intersection Distributes over Union:
:$A \cap \set {\omega \in \Omega : \map {\paren {S \wedge T} } \omega \le t} = \paren {A \cap \set {\omega \in \Omega : \map S \omega \le t} } \cup \paren {A \cap \set {\omega \in \Omega : \map T \omega \le t} }$
Since $A \in \FF_S$, we have: 
:$A \cap \set {\omega \in \Omega : \map S \omega \le t} \in \FF_t$
Since $A \in \FF_T$, we have: 
:$A \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
So, since $\FF_t$ is closed under finite union, we have:
:$\paren {A \cap \set {\omega \in \Omega : \map S \omega \le t} } \cup \paren {A \cap \set {\omega \in \Omega : \map T \omega \le t} } \in \FF_t$
so:
:$A \cap \set {\omega \in \Omega : \map {\paren {S \wedge T} } \omega \le t} \in \FF_t$
Category:Stopped Sigma-Algebras
Stopped Sigma-Algebra preserves Inequality between Stopping Times

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $T$ and $S$ be stopping times with respect to $\sequence {\FF_n}_{n \ge 0}$ such that: 
:$\map S \omega \le \map T \omega$
for each $\omega \in \Omega$.
Let $\FF_S$ and $\FF_T$ be the stopped $\sigma$-algebras associated with $S$ and $T$ respectively.
Then:
:$\FF_S \subseteq \FF_T$
== Proof ==
Let $A \in \FF_S$ and $t \in \Z_{\ge 0}$.
If $\omega \in \Omega$ is such that: 
:$\map T \omega \le t$
we have:
:$\map S \omega \le t$ 
So:
:$\set {\omega \in \Omega : \map T \omega \le t} \subseteq \set {\omega \in \Omega : \map S \omega \le t}$ 
for each $t \in \Z_{\ge 0}$.
So, from Intersection with Subset is Subset, we have:
:$\set {\omega \in \Omega : \map T \omega \le t} \cap \set {\omega \in \Omega : \map S \omega \le t} = \set {\omega \in \Omega : \map T \omega \le t}$
So, we have from Intersection is Associative:
:$A \cap \set {\omega \in \Omega : \map T \omega \le t} = \paren {A \cap \set {\omega \in \Omega : \map S \omega \le t} } \cap \set {\omega \in \Omega : \map T \omega \le t}$
Since $T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$ we have:
:$\set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
and since $A \in \FF_S$ we have:
:$A \cap \set {\omega \in \Omega : \map S \omega \le t} \in \FF_t$
Since $\FF_t$ is closed under finite intersection, we have:
:$A \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
for all $t \in \Z_{\ge 0}$. 
So $A \in \FF_T$, giving $\FF_S \subseteq \FF_T$.
== Sources ==
* : Lemma $9.21$
Category:Stopped Sigma-Algebras
Stopped Submartingale is Submartingale

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {X_n}_{n \ge 0}$ be an $\sequence {\FF_n}_{n \ge 0}$-submartingale.
Let $T$ be a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Let $\sequence {X_n^T}_{n \ge 0}$ be the stopped process.
Then $\sequence {X_n^T}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \ge 0}$-submartingale.
=== Corollary ===
== Proof ==
From Adapted Stochastic Process is Supermartingale iff Negative is Submartingale, we have:
:$\sequence {-X_n}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \ge 0}$-supermartingale.
From Stopped Supermartingale is Supermartingale, we have:
:$\sequence {-X_n^T}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \ge 0}$-supermartingale.
From Adapted Stochastic Process is Supermartingale iff Negative is Submartingale, we have:
:$\sequence {X_n^T}$ is a $\sequence {\FF_n}_{n \ge 0}$-submartingale.
Category:Submartingales
Category:Stopped Processes
Category:Stopped Submartingale is Submartingale
Stopped Submartingale is Submartingale/Corollary

== Corollary ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {X_n}_{n \ge 0}$ be an $\sequence {\FF_n}_{n \ge 0}$-submartingale.
Let $T$ be a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Let $\sequence {X_n^T}_{n \ge 0}$ be the stopped process.
Then we have:
:$\expect {X_n^T} \ge \expect {X_0}$ for each $n \in \Z_{\ge 0}$. 
== Proof ==
From Stopped Submartingale is Submartingale:
:$\sequence {X_n^T}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \ge 0}$-submartingale.
From Expected Value of Submartingale Greater Than or Equal To Initial Expected Value, we have:
:$\expect {X_n^T} \ge \expect {X_0^T}$ 
We have that $\map T \omega \ge 0$ for each $\omega \in \Omega$, so $0 \wedge T = 0$.
So, we have:
:$X_0^T = X_0$
and so:
:$\expect {X_0^T} = \expect {X_0}$
giving the result.
Category:Stopped Submartingale is Submartingale
Stopped Supermartingale is Supermartingale

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {X_n}_{n \ge 0}$ be an $\sequence {\FF_n}_{n \ge 0}$-supermartingale.
Let $T$ be a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Let $\sequence {X_n^T}_{n \ge 0}$ be the stopped process.
Then $\sequence {X_n^T}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \ge 0}$-supermartingale.
=== Corollary ===
== Proof ==
By Stopped Process is Adapted Stochastic Process, $\sequence {X_n^T}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \ge 0}$-adapted stochastic process.
From Integrable Adapted Stochastic Process at Stopping Time is Integrable:
:$X_n^T$ is integrable for each $n \in \Z_{\ge 0}$.
Note that by definition we have for $\omega \in \Omega$ and $n \in \Z_{\ge 0}$:
:$\map {X_n^T} \omega = \map {X_n} \omega$ if $n \le \map T \omega$
and:
:$\map {X_n^T} \omega = \map {X_t} \omega$ if $n > \map T \omega = t$
So we can write: 
:$\ds \map {X_{n + 1}^T} \omega = \sum_{k \mathop = 0}^n \map {X_k} \omega \map {\chi_{\set {\omega \in \Omega : \map T \omega = k} } } \omega + \map {X_{n + 1} } \omega \map {\chi_{\set {\omega \in \Omega : \map T \omega \ge n + 1} } } \omega$
for each $\omega \in \Omega$.
Since $T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$ we have:
:$\set {\omega \in \Omega : \map T \omega = k} \in \FF_k$
for each $k \in \Z_{\ge 0}$ with $0 \le k \le n$.
So, since $\sequence {\FF_n}_{n \ge 0}$ is a filtration, we have: 
:$\set {\omega \in \Omega : \map T \omega = k} \in \FF_n$
Then, from Characteristic Function Measurable iff Set Measurable:
:$\chi_{\set {\omega \in \Omega : \map T \omega = k} }$ is $\FF_n$-measurable. 
Since $\sequence {X_n}_{n \ge 0}$ is $\sequence {\FF_n}_{n \ge 0}$-adapted we have that:
:$X_k$ is $\FF_k$-measurable
so:
:$X_k$ is $\FF_n$-measurable.
Then, from Pointwise Product of Measurable Functions is Measurable, we have:
:$X_k \chi_{\set {\omega \in \Omega : \map T \omega = k} }$ is $\FF_n$-measurable.
So, from Pointwise Sum of Measurable Functions is Measurable: General Result:
:$\ds \sum_{k \mathop = 0}^n X_k \chi_{\set {\omega \in \Omega : \map T \omega = k} }$ is $\FF_n$-measurable.
Finally, note that since $T$ is a stopping time with respect to $\sequence {\FF_n}$ and: 
:$\set {\omega \in \Omega : \map T \omega \ge n + 1}^c = \set {\omega \in \Omega : \map T \omega \le n}$
we have that:
:$\set {\omega \in \Omega : \map T \omega \ge n + 1} \in \FF_n$
since $\sigma$-algebras are closed under relative complement. 
From Characteristic Function Measurable iff Set Measurable, we have: 
:$\chi_{\set {\omega \in \Omega : \map T \omega \ge n + 1} }$ is $\FF_n$-measurable.
We can now calculate:
almost surely.
So $\sequence {X_n^T}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \ge 0}$-supermartingale.
Category:Supermartingales
Category:Stopped Processes
Category:Stopped Supermartingale is Supermartingale
Stopped Supermartingale is Supermartingale/Corollary

== Corollary to Stopped Supermartingale is Supermartingale ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {X_n}_{n \ge 0}$ be an $\sequence {\FF_n}_{n \ge 0}$-supermartingale.
Let $T$ be a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Let $\sequence {X_n^T}_{n \ge 0}$ be the stopped process.
Then we have that $X_n^T$ is integrable for each $n \in \N$ and:
:$\expect {X_n^T} \le \expect {X_0}$ for each $n \in \Z_{\ge 0}$. 
== Proof ==
From Stopped Supermartingale is Supermartingale:
:$\sequence {X_n^T}_{n \ge 0}$ is a $\sequence {\FF_n}_{n \ge 0}$-supermartingale
and so:
:$\sequence {X_n^T}_{n \ge 0}$ is integrable. 
From Expected Value of Supermartingale Less Than or Equal To Initial Expected Value, we have:
:$\expect {X_n^T} \le \expect {X_0^T}$ 
We have that $\map T \omega \ge 0$ for each $\omega \in \Omega$, so $0 \wedge T = 0$.
So, we have:
:$X_0^T = X_0$
and so:
:$\expect {X_0^T} = \expect {X_0}$
giving the result.
Category:Stopped Supermartingale is Supermartingale
Straight Line Commensurable with Apotome

== Theorem ==
== Proof ==
:
Let $AB$ be an apotome.
Let $CD$ be commensurable in length with $AB$.
It is to be demonstrated that:
:$CD$ is an apotome
and:
:the order of $CD$ is the same as the order of $AB$.
Let $BE$ be the annex of $CD$.
This can be constructed uniquely by Construction of Apotome is Unique.
By definition, $AE$ and $EB$ are rational straight lines which are commensurable in square only.
From , let it be contrived that:
:$BE : DF = AB : CD$
From :
:$AE : CF = AB : CD$
and so from :
:$AE : CF = BE : DF$
But $AB$ is commensurable in length with $CD$.
Therefore from :
:$AE$ is commensurable in length with $CF$
and:
:$BE$ is commensurable in length with $DF$.
We have that $AE$ and $EB$ are rational straight lines which are commensurable in square only.
Therefore from :
: $CF$ and $FD$ are rational straight lines which are commensurable in square only.
Thus $CD$ is an apotome.
It remains to be shown that the order of $CD$ is the same as the order of $AB$.
We have that:
:$AE : CF = BE : DF$
So from :
:$AE : EB = CF : FD$
We have that:
: $AE^2 = EB^2 + \lambda^2$
where either:
:$\lambda$ is commensurable in length with $AE$
or:
:$\lambda$ is incommensurable in length with $AE$.
Suppose $\lambda$ is commensurable in length with $AE$.
Then by :
:$CF^2 > FD^2 + \mu^2$
where $\mu$ is commensurable in length with $CF$.
Let $AE$ be commensurable in length with a rational straight line $\alpha$ set out.
Then by :
:$CD$ is commensurable in length with $\alpha$.
Similarly, let $BE$ be commensurable in length with $\alpha$.
Then by :
:$DF$ is commensurable in length with $\alpha$.
Suppose neither $AE$ nor $BE$ is commensurable in length with $\alpha$.
Then by :
:neither $CF$ nor $FD$ will be commensurable in length with $\alpha$.
Let $\lambda$ be incommensurable in length with $AE$.
Then by :
:$CF^2 > FD^2 + \mu^2$
where $\mu$ is incommensurable in length with $CF$.
Let $AE$ be commensurable in length with a rational straight line $\alpha$ set out.
Then by :
:$CD$ is commensurable in length with $\alpha$.
Similarly, let $BE$ be commensurable in length with $\alpha$.
Then by :
:$DF$ is commensurable in length with $\alpha$.
Suppose neither $AE$ nor $BE$ is commensurable in length with $\alpha$.
Then by :
:neither $CF$ nor $FD$ will be commensurable in length with $\alpha$.
It follows that $CD$ is an apotome of the same as the order as $AB$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory