nyuuzyou/wikis huggingface dataset:
Pointwise Exponentiation preserves A.E. Equality

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space and let $p \in \hointr 0 \infty$. 
Let $f, g : X \to \R$ be real-valued $\Sigma$-measurable functions such that: 
:$\size f = \size g$ $\mu$-almost everywhere. 
Then: 
:$\size f^p = \size g^p$ $\mu$-almost everywhere.
== Proof ==
Since: 
:$\size f = \size g$ $\mu$-almost everywhere
there exists a $\mu$-null set $N$ such that: 
:if $\size {\map f x} \ne \size {\map g x}$ then $x \in N$. 
If $\size {\map f x} = \size {\map g x}$, then $\size {\map f x}^p = \size {\map g x}^p$.
So by the Rule of Transposition, we have if $\size {\map f x}^p \ne \size {\map g x}^p$ then $\size {\map f x} \ne \size {\map g x}$.
So, if $x \in X$ has $\size {\map f x}^p \ne \size {\map g x}^p$ then $x \in N$. 
So:
:$\size f^p = \size g^p$ $\mu$-almost everywhere.
Category:Almost-Everywhere Equality Relation
Pointwise Infimum of Measurable Functions is Measurable

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $I$ be a countable set.
Let $\family {f_i}_{i \mathop \in I}$, $f_i: X \to \overline \R$ be an $I$-indexed family of $\Sigma$-measurable functions.
Then the pointwise infimum $\ds \inf_{i \mathop \in I} f_i: X \to \overline \R$ is also $\Sigma$-measurable.
== Proof ==
From Infimum as Supremum, we have the Equality of Mappings:
:$\ds \inf_{i \mathop \in I} f_i = -\paren {\sup_{i \mathop \in I} \paren {-f_i} }$
Now, from Negative of Measurable Function is Measurable and Pointwise Supremum of Measurable Functions is Measurable, it follows that:
:$\ds - \paren {\sup_{i \mathop \in I} \paren {-f_i} }$
is a measurable function.
Hence the result.
== Sources ==
* : $8.9$
Category:Measurable Functions
Pointwise Infimum of Stopping Times is Stopping Time

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {T_n}_{n \in \N}$ be a sequence of stopping times with respect to $\sequence {\FF_n}_{n \ge 0}$.
Let:
:$\ds T = \inf_{n \in \N} T_n$
be the pointwise infimum of the $\sequence {T_n}_{n \in \N}$.
Then $T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
== Proof ==
We have, for $t \in \Z_{\ge 0}$ and $\omega \in \Omega$: 
:$\map T \omega \le t$ $\map {T_n} \omega \le t$ for some $n \in \N$.
That is: 
:$\ds \set {\omega \in \Omega : \map T \omega \le t} = \bigcup_{n \in \N} \set {\omega \in \Omega : \map {T_n} \omega \le t}$
for each $t \in \Z_{\ge 0}$.
Now fix $t \in \Z_{\ge 0}$. 
Since each $T_n$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$, we have: 
:$\set {\omega \in \Omega : \map {T_n} \omega \le t} \in \FF_t$
for each $n \in \N$.
Then, since $\FF_t$ is a $\sigma$-algebra and $\sigma$-algebras are closed under countable union, we have: 
:$\ds \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
So $T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Category:Stopping Times
Pointwise Inverse in Induced Structure

== Theorem ==
Let $\struct {G, \oplus}$ be a group whose identity is $e_G$.
Let $S$ be a set.
Let $\struct {G^S, \oplus}$ be the structure on $G^S$ induced by $\oplus$.
Let $f \in G^S$.
Let $f^* \in G^S$ be the pointwise inverse of $f$:
:$\forall x \in S: \map {f^*} x = \paren {\map f x}^{-1}$
Then $f^*$ is the inverse of $f$ for the pointwise operation induced on $G^S$ by $\oplus$.
== Proof ==
Let $f \in G^S$.
Similarly for $\map {\paren {f^* \oplus f} } x$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Theorem $13.6: \ 3^\circ$
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.4$
Category:Inverse Elements
Category:Pointwise Operations
Pointwise Limit Inferior of Stopping Times is Stopping Time

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {T_n}_{n \in \N}$ be a sequence of stopping times with respect to $\sequence {\FF_n}_{n \ge 0}$.
Let:
:$\ds T = \liminf_{n \mathop \to \infty} T_n$
be the pointwise limit inferior of the $\sequence {T_n}_{n \in \N}$.
Then $T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
== Proof ==
From the definition of the limit inferior we have:
:$\ds \liminf_{n \mathop \to \infty} T_n = \sup_{n \in \N} \paren {\inf_{m \ge n} T_m}$
Writing:
:$\ds \inf_{m \ge n} T_m = \inf_{m \in \N} T_{m + n - 1}$
From Pointwise Infimum of Stopping Times is Stopping Time, we have:
:$\ds \inf_{m \ge n} T_m$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Then, from Pointwise Supremum of Stopping Times is Stopping Time:
:$\ds \sup_{n \in \N} \paren {\inf_{m \ge n} T_m}$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
That is:
:$T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Category:Stopping Times
Pointwise Limit Superior of Stopping Times is Stopping Time

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {T_n}_{n \in \N}$ be a sequence of stopping times with respect to $\sequence {\FF_n}_{n \ge 0}$.
Let:
:$\ds T = \limsup_{n \mathop \to \infty} T_n$
be the pointwise limit superior of the $\sequence {T_n}_{n \in \N}$.
Then $T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
== Proof ==
From the definition of the limit superior we have:
:$\ds \limsup_{n \mathop \to \infty} T_n = \inf_{n \in \N} \paren {\sup_{m \ge n} T_m}$
Writing:
:$\ds \sup_{m \ge n} T_m = \sup_{m \in \N} T_{m + n - 1}$
From Pointwise Supremum of Stopping Times is Stopping Time, we have:
:$\ds \sup_{m \ge n} T_m$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Then, from Pointwise Infimum of Stopping Times is Stopping Time:
:$\ds \inf_{n \in \N} \paren {\sup_{m \ge n} T_m}$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
That is:
:$T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
Category:Stopping Times
Pointwise Limit of Increasing Functions is Increasing

== Theorem ==
Let $X \subseteq \R$.
Let $\sequence {f_n}$ be a sequence of real functions $X \to \R$ converging pointwise to a function $f : X \to \R$. 
Let $f_n$ be a increasing function for each $n$.
Then $f$ is increasing.
== Proof ==
Suppose that $f$ is not increasing.
That is:
:there exists $x, y \in X$ such that $x < y$ and $\map f x > \map f y$. 
Let: 
:$r = \map f x - \map f y > 0$
Since $\sequence {f_n}$ converges to $f$ converging pointwise, the sequence $\sequence {\map {f_n} x}$ converges to $\map f x$. 
That is, there exists $N_1 \in \N$ such that: 
:$\size {\map {f_n} x - \map f x} < \dfrac r 3$
for $n > N_1$.
Similarly, $\sequence {\map {f_n} y}$ converges to $\map f y$.
That is, there exists $N_2 \in \N$ such that: 
:$\size {\map {f_n} y - \map f y} < \dfrac r 3$
for $n > N_2$.
Take $n > \max \set {N_1, N_2}$.
Then, we have: 
:$-\dfrac r 3 < \map {f_n} x - \map f x < \dfrac r 3$
and:
:$-\dfrac r 3 < \map {f_n} y - \map f y < \dfrac r 3$
so that: 
:$-\dfrac {2 r} 3 < \paren {\map {f_n} x - \map f x} - \paren {\map {f_n} y - \map f y} < \dfrac {2 r} 3$
Note that: 
Then, we have: 
:$\map {f_n} x - \map {f_n} y > \dfrac r 3 > 0$
But $x < y$, so this contradicts that $f_n$ is increasing.
So $f$ is increasing.
Category:Real Analysis
Pointwise Limit of Measurable Functions is Measurable

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\sequence {f_n}_{n \mathop \in \N}$, $f_n: X \to \overline \R$ be a sequence of $\Sigma$-measurable functions.
Then the pointwise limit $\ds \lim_{n \mathop \to \infty} f_n: X \to \overline \R$ is also $\Sigma$-measurable.
== Proof ==
From Convergence of Limsup and Liminf, it follows that:
:$\ds \lim_{n \mathop \to \infty} f_n = \limsup_{n \mathop \to \infty} f_n$
We have Pointwise Upper Limit of Measurable Functions is Measurable.
Hence the result.
== Sources ==
* : $8.9$
Category:Measurable Functions
Pointwise Limit of Sequence of Bounded Linear Transformations from Banach Space to Normed Vector Space is Bounded Linear Transformation

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm \cdot_X}$ be a Banach space over $\Bbb F$.
Let $\struct {Y, \norm \cdot_Y}$ be a normed vector space over $\Bbb F$.
Let $T : X \to Y$ be a function.
Let $\sequence {T_n}_{n \mathop \in \N}$ be a sequence of bounded linear transformations such that: 
:$\ds T x = \lim_{n \mathop \to \infty} T_n x$
for each $x \in X$.
Then $T$ is a bounded linear transformation.
== Proof ==
We first show that $T$ is a linear transformation.
Let $\alpha, \beta \in \Bbb F$ and $x, y \in X$. 
Then, we have: 
So $T$ is linear
Now we show that $T$ is bounded.
We use the Banach-Steinhaus Theorem.
From Modulus of Limit: Normed Vector Space, we have: 
:$\ds \norm {T x}_Y = \lim_{n \mathop \to \infty} \norm {T_n x}_Y$
From Convergent Real Sequence is Bounded, we have: 
:$\norm {T_n x}_Y$ is bounded in $n$ for each $x \in X$.
That is: 
:$\ds \sup_{n \in \N} \norm {T_n x}_Y$ is finite for each $x \in X$.
So, from the Banach-Steinhaus Theorem, we have: 
:$\ds \sup_{n \in \N} \norm {T_n}$ is finite, where $\norm \cdot$ is the norm on bounded linear transformation.
Pick $M > 0$ such that: 
:$\norm {T_n} \le M$
for each $n \in \N$.
Then: 
:$\norm {T_n x}_Y \le M \norm x_X$
for each $x \in X$ and $n \in \N$ from the definition of the norm on bounded linear transformation. 
Then for each $x \in X$ we have: 
So $T$ is bounded.
== Sources ==
* $22.2$: The Principle of Uniform Boundedness
Category:Bounded Linear Transformations
Category:Linear Transformations on Banach Spaces
Pointwise Lower Limit of Measurable Functions is Measurable

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\sequence {f_n}_{n \mathop \in \N}$, $f_n: X \to \overline \R$ be a sequence of $\Sigma$-measurable functions.
Then the pointwise lower limit:
:$\ds \liminf_{n \mathop \to \infty} f_n: X \to \overline \R$
is also $\Sigma$-measurable.
== Proof ==
By definition of limit inferior, we have:
:$\ds \liminf_{n \mathop \to \infty} f_n = \sup_{m \mathop \in \N} \ \inf_{n \mathop \ge m} f_n$
The result follows from combining:
:Pointwise Infimum of Measurable Functions is Measurable
:Pointwise Supremum of Measurable Functions is Measurable
== Sources ==
* : $8.9$
Category:Measurable Functions
Pointwise Maximum of Finite Family of Seminorms is Seminorm

== Theorem ==
Let $\struct {K, \norm {\,\cdot\,}_K}$ be a normed division ring.
Let $X$ be a vector space over $K$. 
Let $\II$ be a set of seminorms on $X$.
Define: 
:$\ds q =: \max_{p \mathop \in \II} p$
where $\max$ denotes the pointwise maximum over $\II$. 
== Proof ==
=== Proof of $(\text N 2)$ ===
Let $\lambda \in K$ and $x \in X$. 
Then: 
proving $(\text N 2)$.
=== Proof of $(\text N 3)$ ===
Let $x, y \in X$. 
Then for each $p \in \II$ we have: 
Taking pointwise maximums over $p \in \II$ we have: 
:$\map q {x + y} \le \map q x + \map q y$
Category:Seminorms
Pointwise Maximum of Integrable Functions is Integrable Function

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g: X \to \overline \R$ be $\mu$-integrable functions.
Then $\map \max {f, g}$, the pointwise maximum of $f$ and $g$, is also a $\mu$-integrable function.
That is, the space of $\mu$-integrable functions $\LL^1_{\overline \R}$ is closed under pointwise maximum.
== Proof ==
== Sources ==
* : $10.4 \, \text{(iii)}$
Category:Integrable Functions (Measure Space)‎
Pointwise Maximum of Measurable Functions is Measurable

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f, g: X \to \overline \R$ be $\Sigma$-measurable functions.
Then the pointwise maximum $\max \set {f, g}: X \to \overline \R$ is also $\Sigma$-measurable.
== Proof ==
For all $x \in X$ and $a \in \R$, we have by Max Operation Yields Supremum of Parameters that:
:$\max \set {\map f x, \map g x} \le a$
 both $\map f x \le a$ and $\map g x \le a$.
That is, for all $a \in \R$:
:$\set {x \in X: \max \set {\map f x, \map g x} \le a} = \set {x \in X: \map f x \le a} \cap \set {x \in X: \map g x \le a}$
By Characterization of Measurable Functions: $(1) \implies (2)$, the two sets on the RHS are elements of $\Sigma$, i.e. measurable.
By Sigma-Algebra Closed under Intersection, it follows that:
:$\set {x \in X: \max \set {\map f x, \map g x} \le a} \in \Sigma$
Hence $\max \set {f, g}$ is measurable, by Characterization of Measurable Functions: $(2) \implies (1)$.
== Sources ==
* : $8.10$
Category:Measurable Functions
Pointwise Maximum of Simple Functions is Simple

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f, g : X \to \R$ be simple functions. 
Then the pointwise maximum $\max \set {f, g}: X \to \R$ is also simple function.
 
== Proof 1 ==
 
== Proof 2 ==
Category:Simple Functions
Category:Pointwise Maximum of Simple Functions is Simple
Pointwise Maximum of Simple Functions is Simple/Proof 1

== Theorem ==
== Proof ==
From Pointwise Sum of Simple Functions is Simple Function:
:$f + g$ is simple.
From Scalar Multiple of Simple Function is Simple Function:
:$-g$ is simple.
Then, from Pointwise Sum of Simple Functions is Simple Function, we have: 
:$f - g$ is simple.
From Absolute Value of Simple Function is Simple Function:
:$\size {f - g}$ is simple.
From Pointwise Sum of Simple Functions is Simple Function, we then have: 
:$\paren {f + g} + \size {f - g}$ is simple. 
Finally, from Scalar Multiple of Simple Function is Simple Function, we have: 
:$\dfrac 1 2 \paren {\paren {f + g} + \size {f - g} }$ is simple.
By Maximum Function in terms of Absolute Value, we have: 
:$\ds \max \set {f, g} = \frac 1 2 \paren {\paren {f + g} + \size {f - g} }$
so:
:$\max \set {f, g}$ is simple.
Category:Pointwise Maximum of Simple Functions is Simple
Pointwise Maximum of Simple Functions is Simple/Proof 2

== Theorem ==
== Proof ==
From Simple Function is Measurable, we have that: 
:$f$ and $g$ are $\Sigma$-measurable.
For brevity let:
:$h = \max \set {f, g}$
From Pointwise Maximum of Measurable Functions is Measurable, we have that: 
:$h$ is $\Sigma$-measurable.
From Measurable Function is Simple Function iff Finite Image Set, we aim to show that: 
:$\map h X$ is a finite set. 
From Measurable Function is Simple Function iff Finite Image Set, we have: 
:$\map f X$ and $\map g X$ are finite sets.
Let $x \in X$. 
If $\map f x < \map g x$, then: 
:$\map h x = \map g x$
so that: 
:$\map h x \in \map g X$
If $\map g x \le \map f x$, then: 
:$\map h x = \map f x$
so that: 
:$\map h x \in \map f X$
Since for $x \in X$ we have either $\map g x \le \map f x$ or $\map f x < \map g x$, we obtain:
:$\map h X \subseteq \map f X \cup \map g X$
From Union of Finite Sets is Finite, we have that: 
:$\map f X \cup \map g X$ is finite.
Then, from Subset of Finite Set is Finite: 
:$\map h X$ is finite.
So:
:$h$ is simple.
Category:Pointwise Maximum of Simple Functions is Simple
Pointwise Maximum of Stopping Times is Stopping Time

== 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 $\sequence {\FF_n}_{n \ge 0}$.
Let $S \vee T$ be the pointwise maximum of $S$ and $T$.
Then $S \vee T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
== Proof ==
We have, for $t \in \Z_{\ge 0}$ and $\omega \in \Omega$: 
:$\map {\paren {S \vee T} } \omega \le t$ $\map S \omega \le t$ and $\map T \omega \le t$
That is: 
:$\set {\omega \in \Omega : \map {\paren {S \vee T} } \omega \le t} = \set {\omega \in \Omega : \map S \omega \le t} \cap \set {\omega \in \Omega : \map T \omega \le t}$ 
Since $S$ and $T$ are stopping times with respect to $\sequence {\FF_n}_{n \ge 0}$, we have:
:$\set {\omega \in \Omega : \map S \omega \le t} \in \FF_t$
and:
:$\set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
So, since $\FF_t$ is closed under finite union, we have: 
:$\set {\omega \in \Omega : \map {\paren {S \vee T} } \omega \le t} \in \FF_t$
Category:Stopping Times
Pointwise Minimum of Integrable Functions is Integrable Function

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g: X \to \overline \R$ be $\mu$-integrable functions.
Then $\map \min {f, g}$, the pointwise minimum of $f$ and $g$, is also a $\mu$-integrable function.
That is, the space of $\mu$-integrable functions $\LL^1_{\overline \R}$ is closed under pointwise minimum.
== Proof ==
== Sources ==
* : $10.4 \, \text{(iii)}$
Category:Integrable Functions (Measure Space)‎
Pointwise Minimum of Measurable Functions is Measurable

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f, g: X \to \overline \R$ be $\Sigma$-measurable functions.
Then the pointwise minimum $\min \set {f, g}: X \to \overline \R$ is also $\Sigma$-measurable.
== Proof ==
For all $x \in X$ and $a \in \R$, we have by Min Operation Yields Infimum of Parameters that:
:$a \le \min \set {\map f x, \map g x}$
 both $a \le \map f x$ and $a \le \map g x$.
That is, for all $a \in \R$:
:$\set {x \in X: \min \set {\map f x, \map g x} \ge a} = \set {x \in X: \map f x \ge a} \cap \set {x \in X: \map g x \ge a}$
By Characterization of Measurable Functions: $(1) \implies (4)$, the two sets on the RHS are elements of $\Sigma$, that is, measurable.
By Sigma-Algebra Closed under Intersection:
:$\set {x \in X: \min \set {\map f x, \map g x} \ge a} \in \Sigma$
Hence $\min \set {f, g}$ is measurable, by Characterization of Measurable Functions: $(4) \implies (1)$.
== Sources ==
* : $8.10$
Category:Measurable Functions
Pointwise Minimum of Metric and Positive Real Number is Topologically Equivalent Metric

== Theorem ==
Let $\struct {X, d}$ be a metric space.
Let $c > 0$ be a real number. 
For each $x, y \in X$, define: 
:$\map {d'} {x, y} = \min \set {\map d {x, y}, c}$ 
Then $d'$ is a metric that is topologically equivalent to $d$. 
== Proof ==
=== ===
Suppose that $x, y \in X$ are such that: 
:$\map {d'} {x, y} = 0$
Since $c > 0$, this implies that: 
:$\map d {x, y} = 0$
Since $d$ is a metric, we have $x = y$ by for $d$. 
Hence is fulfilled.
=== ===
Let $x, y, z \in X$. 
We go casewise. 
Suppose that $\map d {x, y} > c$ and $\map d {y, z} > c$. 
Then $\map {d'} {x, y} = c$ and $\map {d'} {y, z} = c$, so that: 
:$\map {d'} {x, y} + \map {d'} {y, z} = 2 c$ 
By definition, we also have: 
:$\map {d'} {x, z} \le c \le 2 c$ 
so that: 
:$\map {d'} {x, z} \le 2 c = \map {d'} {x, y} + \map {d'} {y, z}$
Now suppose that $\map d {x, y} \le c$ or $\map d {y, z} \le c$.
Then $\map {d'} {x, y} = c$ or $\map {d'} {y, z} = c$ so that: 
:$\map {d'} {x, y} + \map {d'} {y, z} \ge c \ge \map {d'} {x, y}$ 
Finally, suppose that $\map d {x, y} < c$ and $\map d {y, z} < c$. 
Then, we have $\map {d'} {x, y} = \map d {x, y}$ and $\map {d'} {y, z} = \map d {y, z}$. 
By for $d$, we have: 
:$\map d {x, z} \le \map d {x, y} + \map d {y, z} = \map {d'} {x, y} + \map {d'} {y, z}$
while: 
:$\map {d'} {x, z} \le \map d {x, z}$ 
so we get: 
:$\map {d'} {x, z} \le \map {d'} {x, y} + \map {d'} {y, z}$ 
Hence is fulfilled.
=== ===
For each $x, y \in X$ we have: 
Hence is fulfilled.
=== ===
Let $x, y \in X$ be such that $x \ne y$. 
Then $\map d {x, y} \ne 0$ by for $d$. 
If $0 < \map d {x, y} < c$, then: 
:$\map {d'} {x, y} = \map d {x, y} > 0$
If $\map d {x, y} \ge c$, then: 
:$\map {d'} {x, y} = c > 0$ 
So $\map {d'} {x, y} > 0$ if $x \ne y$.
Hence is fulfilled.
=== Topological Equivalence ===
We show that $U$ is $d$-open it is $d'$-open.
First note that if $r < c$, then for $x, y \in X$ we have: 
:$\map {d'} {x, y} < c$ 
:
:$\map d {x, y} < c$ 
So, for each $x \in X$ and $r < c$ we have: 
:$\map {B_r} {x, d} = \map {B_r} {x, d'}$
Let $U$ be $d$-open. 
Then, by Set is Open iff Union of Open Balls, for each $x \in U$, there exists $r_x > 0$ such that: 
:$\ds U = \bigcup_{x \mathop \in U} \map {B_{r_x} } {x, d}$ 
Setting: 
:$\delta_x = \min \set {r_x, \dfrac c 2}$ 
we have: 
:$\map {B_{\delta_x} } {x, d} \subseteq \map {B_{r_x} } {x, d}$ 
so that: 
:$\ds \bigcup_{x \mathop \in U} \map {B_{\delta_x} } {x, d} \subseteq \bigcup_{x \mathop \in U} \map {B_{r_x} } {x, d} \subseteq U$ 
Conversely, $x \in \map {B_{\delta_x} } {x, d}$ for each $x \in U$, so that:
:$\ds U \subseteq \bigcup_{x \mathop \in U} \map {B_{\delta_x} } {x, d}$
giving:
:$\ds U = \bigcup_{x \mathop \in U} \map {B_{\delta_x} } {x, d}$
Since $\delta_x < c$, we have: 
:$\ds U = \bigcup_{x \mathop \in U} \map {B_{\delta_x} } {x, d'}$
So $U$ is $d'$-open.
Swapping $d$ and $d'$ in the above proof, we get that if $U$ is $d'$-open, it is $d$-open.
So $d'$ is a metric that is topologically equivalent to $d$.
Category:Metric Spaces
Category:Topologically Equivalent Metrics
Pointwise Minimum of Simple Functions is Simple

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f, g : X \to \R$ be simple functions. 
Then the pointwise minimum $\min \set {f, g}: X \to \R$ is also simple function.
 
== Proof 1 ==
== Proof 2 ==
Category:Pointwise Minimum of Simple Functions is Simple
Category:Simple Functions
Pointwise Minimum of Simple Functions is Simple/Proof 1

== Theorem ==
== Proof ==
From Pointwise Sum of Simple Functions is Simple Function:
:$f + g$ is simple.
From Scalar Multiple of Simple Function is Simple Function:
:$-g$ is simple.
Then, from Pointwise Sum of Simple Functions is Simple Function we have: 
:$f - g$ is simple.
From Absolute Value of Simple Function is Simple Function:
:$\size {f - g}$ is simple.
So, from Scalar Multiple of Simple Function is Simple Function, we have: 
:$-\size {f - g}$ is simple.
From Pointwise Sum of Simple Functions is Simple Function, we then have: 
:$\paren {f + g} - \size {f - g}$ is simple. 
Finally, from Scalar Multiple of Simple Function is Simple Function, we have: 
:$\dfrac 1 2 \paren {\paren {f + g} - \size {f - g} }$ is simple.
By Minimum Function in terms of Absolute Value, we have: 
:$\ds \min \set {f, g} = \frac 1 2 \paren {\paren {f + g} - \size {f - g} }$
so:
:$\min \set {f, g}$ is simple.
Category:Pointwise Minimum of Simple Functions is Simple
Pointwise Minimum of Simple Functions is Simple/Proof 2

== Theorem ==
== Proof ==
From Simple Function is Measurable, we have that: 
:$f$ and $g$ are $\Sigma$-measurable.
For brevity let:
:$h = \min \set {f, g}$
From Pointwise Minimum of Measurable Functions is Measurable, we have that: 
:$h$ is $\Sigma$-measurable.
From Measurable Function is Simple Function iff Finite Image Set, we aim to show that: 
:$\map h X$ is a finite set. 
From Measurable Function is Simple Function iff Finite Image Set, we have: 
:$\map f X$ and $\map g X$ are finite sets.
Let $x \in X$. 
If $\map f x < \map g x$, then: 
:$\map h x = \map f x$
so that: 
:$\map h x \in \map f X$
If $\map g x \le \map f x$, then: 
:$\map h x = \map g x$
so that: 
:$\map h x \in \map g X$
Since for $x \in X$ we have either $\map g x \le \map f x$ or $\map f x < \map g x$, we obtain:
:$\map h X \subseteq \map f X \cup \map g X$
From Union of Finite Sets is Finite, we have that: 
:$\map f X \cup \map g X$ is finite.
Then, from Subset of Finite Set is Finite: 
:$\map h X$ is finite.
So:
:$h$ is simple.
Category:Pointwise Minimum of Simple Functions is Simple
Pointwise Minimum of Stopping Times is Stopping Time

== 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 $\sequence {\FF_n}_{n \ge 0}$.
Let $S \wedge T$ be the pointwise minimum of $S$ and $T$.
Then $S \wedge T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.
== Proof ==
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: 
:$\set {\omega \in \Omega : \map {\paren {S \wedge T} } \omega \le t} = \set {\omega \in \Omega : \map S \omega \le t} \cup \set {\omega \in \Omega : \map T \omega \le t}$ 
Since $S$ and $T$ are stopping times with respect to $\sequence {\FF_n}_{n \ge 0}$, we have:
:$\set {\omega \in \Omega : \map S \omega \le t} \in \FF_t$
and:
:$\set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
So, since $\FF_t$ is closed under finite union, we have: 
:$\set {\omega \in \Omega : \map {\paren {S \wedge T} } \omega \le t} \in \FF_t$
Category:Stopping Times