nyuuzyou/wikis huggingface dataset:
Characterization of Convergent Net in Metric Space

== Theorem ==
Let $\struct {X, d}$ be a metric space.
Let $\struct {\Lambda, \preceq}$ be a directed set.
Let $\family {x_\lambda}_{\lambda \mathop \in \Lambda}$ be a net in $X$. 
Let $x \in X$. 
Then $\family {x_\lambda}_{\lambda \mathop \in \Lambda}$ converges in $\struct {X, d}$ :
:for each $\epsilon > 0$ there exists $\lambda_0 \in \Lambda$ such that $\map d {x_\lambda, x} < \epsilon$ for $\lambda \in \Lambda$ with $\lambda_0 \preceq \lambda$. 
== Proof ==
=== Necessary Condition ===
Suppose that $\family {x_\lambda}_{\lambda \mathop \in \Lambda}$ converges to $x$ in $\struct {X, d}$.
Let $\epsilon > 0$. 
From Open Ball is Open Set, the open ball $\map {B_\epsilon} x$ with radius $\epsilon$ and center $x$ is open in $\struct {X, d}$.
Since $\family {x_\lambda}_{\lambda \mathop \in \Lambda}$ converges to $x$, there exists $\lambda_0 \in \Lambda$ such that: 
:$x_\lambda \in \map {B_\epsilon} x$ for $\lambda \in \Lambda$ with $\lambda_0 \preceq \lambda$. 
That is: 
:$\map d {x_\lambda, x} < \epsilon$ for $\lambda \in \Lambda$ with $\lambda_0 \preceq \lambda$.
=== Sufficient Condition ===
Suppose that: 
:for each $\epsilon > 0$ there exists $\lambda_0 \in \Lambda$ such that $\map d {x_\lambda, x} < \epsilon$ for $\lambda \in \Lambda$ with $\lambda_0 \preceq \lambda$.
Let $U$ be an open neighborhood of $x$ in $\struct {X, \tau}$.
Since $U$ is open, there exists $\epsilon > 0$ such that: 
:$\map {B_\epsilon} x \subseteq U$
By hypothesis, there exists $\lambda_0 \in \Lambda$ such that $\map d {x_\lambda, x} < \epsilon$ for $\lambda \in \Lambda$ with $\lambda_0 \preceq \lambda$.
Then for $\lambda_0 \preceq \lambda$ we have $x_\lambda \in \map {B_\epsilon} x \subseteq U$.
Category:Metric Spaces
Category:Nets (Set Theory)
Characterization of Convergent Net in Weak-* Topology

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \tau}$ be a topological vector space.
Let $X^\ast$ be the topological dual of $X$. 
Let $w^\ast$ be the weak-$\ast$ topology on $X$.
Let $f \in X^\ast$. 
Let $\struct {\Lambda, \preceq}$ be a directed set.
Let $\family {f_\lambda}_{\lambda \in \Lambda}$ be a net. 
Then:
:$\family {f_\lambda}_{\lambda \in \Lambda}$ converges to $f$ in $\struct {X^\ast, w^\ast}$
:
:for each $x \in X$, the net $\family {\map {f_\lambda} x}_{\lambda \in \Lambda}$ converges to $\map f x$ in $\GF$. 
== Proof ==
=== Necessary Condition ===
Suppose that: 
:$\family {f_\lambda}_{\lambda \in \Lambda}$ converges to $f$ in $\struct {X^\ast, w^\ast}$
Let $x \in X$. 
From Characterization of Continuity of Linear Functional in Weak-* Topology, $x^\wedge : \struct {X^\ast, w^\ast} \to \GF$ is continuous.
From Characterization of Continuity in terms of Nets, we have that: 
:for each $f \in X^\ast$, the net $\family {\map {x^\wedge} {f_\lambda} }_{\lambda \in \Lambda} = \family {\map {f_\lambda} x}_{\lambda \in \Lambda}$ converges to $\map f x$.
=== Sufficient Condition ===
Suppose that: 
:for each $f \in X^\ast$, the net $\family {\map f {x_\lambda} }_{\lambda \in \Lambda}$ converges to $\map f x$.
Let $U$ be an open neighborhood of $f$ in $\struct {X^\ast, w^\ast}$.
From Open Sets in Weak-* Topology of Topological Vector Space, there exists $x_1, \ldots, x_n \in X$ and $\epsilon > 0$ such that: 
:$\set {g \in X^\ast : \cmod {\map f {x_i} - \map g {x_i} } < \epsilon \text { for each } 1 \le i \le n} \subseteq U$
Then for each $1 \le i \le n$, we have that: 
:$\family {\map {f_\lambda} {x_i} }_{\lambda \in \Lambda}$ converges to $\map f {x_i}$.
So for each $1 \le i \le n$ there exists $\lambda_i \in \Lambda$ such that for all $\lambda \in \Lambda$ with $\lambda_i \preceq \lambda$ we have:
:$\cmod {\map {f_\lambda} {x_i} - \map f {x_i} } < \epsilon$
from Characterization of Convergent Net in Metric Space.
From Existence of Upper Bound of Finite Subset of Directed Set, there exists $\lambda_\ast \in \Lambda$ such that $\lambda_i \preceq \lambda_\ast$ for each $1 \le i \le n$. 
Then, for $\lambda \in \Lambda$ with $\lambda_\ast \preceq \lambda$, we have $\lambda_i \preceq \lambda$ for each $1 \le i \le n$ by transitivity. 
Then: 
:$\cmod {\map {f_\lambda} {x_i} - \map f {x_i} } < \epsilon$
for each $1 \le i \le n$ and $\lambda_\ast \preceq \lambda$. 
So $f_\lambda \in U$ for $\lambda_\ast \preceq \lambda$. 
So $\family {f_\lambda}_{\lambda \in \Lambda}$ converges to $f$ in $\struct {X^\ast, w^\ast}$.
Category:Nets (Set Theory)
Category:Weak-* Topologies
Characterization of Convergent Net in Weak Topology

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \tau}$ be a topological vector space over $\GF$ admitting a weak topology $w$. 
Let $X^\ast$ be the topological dual of $X$. 
Let $x \in X$.
Let $\struct {\Lambda, \preceq}$ be a directed set.
Let $\family {x_\lambda}_{\lambda \in \Lambda}$ be a net in $X$. 
Then: 
:$\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$ in $\struct {X, w}$
:
:for each $f \in X^\ast$, the net $\family {\map f {x_\lambda} }_{\lambda \in \Lambda}$ converges to $\map f x$.
== Proof ==
=== Necessary Condition ===
Suppose that: 
:$\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$ in $\struct {X, w}$.
Let $f \in X^\ast$. 
From Characterization of Continuity of Linear Functional in Weak Topology, $f : \struct {X, w} \to \GF$ is continuous.
From Characterization of Continuity in terms of Nets, we have that: 
:for each $f \in X^\ast$, the net $\family {\map f {x_\lambda} }_{\lambda \in \Lambda}$ converges to $\map f x$.
=== Sufficient Condition ===
Suppose that: 
:for each $f \in X^\ast$, the net $\family {\map f {x_\lambda} }_{\lambda \in \Lambda}$ converges to $\map f x$.
Let $U$ be an open neighborhood of $x$ in $\struct {X, w}$.
From Open Sets in Weak Topology of Topological Vector Space, there exists $f_1, f_2, \ldots, f_n \in X^\ast$ and $\epsilon > 0$ such that: 
:$\set {y \in X : \cmod {\map {f_i} {y - x} } < \epsilon \text { for each } 1 \le i \le n} \subseteq U$
Then for each $1 \le i \le n$, we have that: 
:$\family {\map {f_i} {x_\lambda} }_{\lambda \in \Lambda}$ converges to $\map {f_i} x$.
So for each $1 \le i \le n$ there exists $\lambda_i \in \Lambda$ such that for all $\lambda \in \Lambda$ with $\lambda_i \preceq \lambda$ we have:
:$\cmod {\map {f_i} {x - x_\lambda} } < \epsilon$
from Characterization of Convergent Net in Metric Space.
From Existence of Upper Bound of Finite Subset of Directed Set, there exists $\lambda_\ast \in \Lambda$ such that $\lambda_i \preceq \lambda_\ast$ for each $1 \le i \le n$. 
Then, for $\lambda \in \Lambda$ with $\lambda_\ast \preceq \lambda$, we have $\lambda_i \preceq \lambda$ for each $1 \le i \le n$ by transitivity. 
Then: 
:$\cmod {\map {f_i} {x - x_\lambda} } < \epsilon$
for each $1 \le i \le n$ and $\lambda_\ast \preceq \lambda$. 
So $x_\lambda \in U$ for $\lambda_\ast \preceq \lambda$. 
So $\family {x_\lambda}_{\lambda \in \Lambda}$ converges to $x$ in $\struct {X, w}$.
Category:Weak Topologies on Topological Vector Spaces
Category:Nets (Set Theory)
Characterization of Cosine Integral Function

== Definition ==
Let $\Ci: \R_{>0}: \R$ denote the cosine integral function:
:$\map \Ci x = \ds \int_{t \mathop = x}^{t \mathop \to +\infty} \frac {\cos t} t \rd t$
Then:
:$\map \Ci x = -\gamma - \ln x + \ds \int_{t \mathop \to 0}^{t \mathop = x} \frac {1 - \cos t} t \rd t$
where $\gamma$ is the Euler-Macheroni constant.
== Proof ==
== Sources ==
* : $\S 35$: Miscellaneous Special Functions: Cosine Integral: $35.14$
* 
Category:Cosine Integral Function
Characterization of Derivative by Local Basis

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A$ be a subset of $S$.
Let $x$ be a point of $T$.
Let $\BB \subseteq \tau$ be a local basis at $x$.
Then
:$x \in A'$
:
:for every $U \in \BB$, there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \ne y$
where:
:$A'$ denotes the derivative of $A$.
== Proof ==
=== Sufficient Condition ===
Let $x \in A'$.
By Characterization of Derivative by Open Sets:
For every open set $U$ of $T$:
:if $x \in U$
:then there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \ne y$
As the elements of $\BB$ are all open sets, it follows that:
For every open set $U \in \BB$:
:if $x \in U$
:then there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \ne y$
=== Necessary Condition ===
Let $x$ be such that:
$(1): \quad$ for every subset $U \in \BB$, there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \ne y$.
By Characterization of Derivative by Open Sets, to prove that $x \in A'$ it is enough to prove:
:for every open set $U$ of $T$:
::if $x \in U$
::then there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \ne y$.
Let $U$ be an open set of $T$.
Let that $x \in U$.
Then by definition of local basis:
:there exists $V \in \BB$ such that:
::$x \in V \subseteq U$
By $(1)$:
:there exists a point $y$ of $T$ such that $y \in A \cap V$ and $x \ne y$.
By the corollary to Set Intersection Preserves Subsets:
:$A \cap V \subseteq A \cap U$
and so:
:$y \in A \cap V \implies y \in A \cap U$
and so $y$ fulfils the conditions of the hypothesis.
== Sources ==
* 
* 
Category:Set Derivatives
Characterization of Derivative by Open Sets

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A$ be a subset of $T$.
Let $x$ be a point of $T$.
Then
:$x \in A'$
:
:for every open set $U$ of $T$:
::if $x \in U$
::then there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \ne y$
where
:$A'$ denotes the derivative of $A$.
== Proof ==
=== Sufficient Condition ===
Let $x \in A'$.
Then $x$ is an accumulation point of $A$ by Definition:Set Derivative.
Then by definition of accumulation point:
:$(1): \quad x \in \paren {A \setminus \set x}^-$
where $A^-$ denotes the closure of $A$.
Let $U$ be an open set of $T$.
Let $x \in U$.
Then by $(1)$ and Condition for Point being in Closure:
:$\paren {A \setminus \set x} \cap U \ne \O$
Then there exists $y$ being a point such that:
:$(2): \quad y \in A \setminus \set x$ and $y \in U$.
Then $y \in A$ and $y \notin \set x$ by definition of set difference.
Hence by $(2)$ and the definitions of set intersection and singleton:
:$y \in A \cap U$ and $x \ne y$
=== Necessary Condition ===
Let $x$ be such that:
:$(3): \quad$ for every open set $U$ of $T$ if $x \in U$, then there exists a point $y$ of $T$ such that $y \in A \cap U$ and $x \ne y$.
It will be proved that:
:for every open set $G$ of $T$:
::if $x \in G$
::then $\paren {A \setminus \set x} \cap U \ne \O$.
Let $G$ be an open set of $T$ such that $x \in G$.
Then by $(3)$ there exists a point $y$ of $T$ such that:
:$(4): \quad y \in A \cap G$ and $x \ne y$
Then by the definitions of set intersection and singleton:
:$y \in A$ and $y \notin \set x$
By definition of set difference:
:$y \in A \setminus \set x$
By $(4)$ and the definition of set intersection:
:$y \in G$
Hence:
: $\paren {A \setminus \set x} \cap U \ne \O$
Then by Condition for Point being in Closure:
:$x \in \paren {A \setminus \set x}^-$
Then, by definition, $x$ is an accumulation point of $A$.
Hence by definition of set derivative:
:$x \in A'$
== Sources ==
* 
* 
Category:Set Derivatives
Characterization of Derivative of Test Function

== Theorem ==
Let $\Phi, \phi \in \map \DD \R$ be test functions.
Let:
:$\ds X = \set {\Phi' : \Phi \in \map \DD \R} $
:$\ds Y = \set {\phi \in \map \DD \R : \int_{- \infty}^\infty \map \phi x \rd x = 0}$
Then $X = Y$.
Furthermore, for every $\phi \in Y$ there exists only one $\Phi \in \map \DD \R$ such that $\Phi' = \phi$.
== Proof ==
=== $X$ is a subset of $Y$ ===
By definition of test functions:
:$\paren {\Phi \in \map \DD \R} \implies \paren {\Phi' \in \map \DD \R}$
Moreover, a test function has compact support:
:$\ds \int_{-\infty}^\infty \map {\Phi'} x \rd x = \map \Phi \infty - \map \Phi {-\infty} = 0$
Hence:
:$X \subseteq Y$
=== $Y$ is a subset of $X$ ===
Suppose $\phi \in \map \DD \R$ is such that:
:$\ds \int_{-\infty}^\infty \map \phi x \rd x = 0$
Let $\Phi$ be such that:
:$\ds \map \Phi x = \int_{-\infty}^x \map \phi \xi \rd \xi$
Then there exists $\Phi'$ such that:
:$\map {\Phi'} x = \map \phi x$
Since $\phi$ is a test function:
:$\Phi' \in \map {C^\infty} \R$.
Let $\phi$ be supported on $\closedint {-a} a$ with $a \in \R_{> 0}$.
Then:
:$\forall x \in \R : x < -a \implies \map \Phi x = 0$
On the other hand:
:$\ds \forall x \in \R : x > a \implies \map \Phi x = \int_{-\infty}^x \map \phi \xi \rd \xi = \int_{-\infty}^\infty \map \phi \xi \rd \xi = 0$
Hence, $\Phi$ is also supported on $\closedint {-a} a$, and:
:$Y \subseteq X$
=== Uniqueness ===
Let $\phi \in Y$.
Suppose, for $\Phi'_1, \Phi'_2 \in \map \DD \R$ we have that $\Phi'_1 = \phi = \Phi'_2$.
Then $\paren {\Phi_1 - \Phi_2}' = 0$ and $\Phi_1 - \Phi_2 = C$ with $C \in \R$ being a constant.
However, both $\Phi_1$ and $\Phi_2$ have compact supports inherited from $\phi$, say $K \subset \R$.
Then:
:$\forall x \notin K : \map {\Phi_1} x - \map {\Phi_2} x = 0 - 0 = 0$
$\Phi_1$ and $\Phi_2$ are test functions and thus smooth.
Therefore, $C = 0$ holds for all $x \in \R$.
== Sources ==
* : Chapter $\S 6.1$: A glimpse of distribution theory. Test functions, distributions, and examples
Category:Test Functions
Characterization of Differentiability

== Theorem ==
Let $\mathbb X$ be an open rectangle of $\R^n$. 
Let $f: \mathbb X \to \R, \mathbf x \mapsto \map f {\mathbf x}$ be a real-valued function.
Let $\mathbf x = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix} \in \R^n$.
Let $\map {\Delta f} {\mathbf x} = \map f {\mathbf x + \Delta \mathbf x} - \map f {\mathbf x}$ .
Let $\dfrac {\partial f} {\partial x_j}$ be the partial derivative of $f$ $x_j$.
Then $f$ is differentiable there exists some $\map {\Delta f} {\mathbf x}$ such that:
:$\ds \map {\Delta f} {\mathbf x} = \sum_{i \mathop = 1}^n \frac {\partial \map f {\mathbf x} } {\partial x_i} \Delta x_i + \sum_{i \mathop = 1}^n \varepsilon_i \Delta x_i$
where $\forall i: 1 \le i \le n: \varepsilon_i \to 0$ as $\Delta x_i \to 0$.
== Proof ==
Suppose $f: \R^1 \to \R$.
Define:
:$\map f x = y$
:$\Delta y = \map f {x + \Delta x} - \Delta x$
From the definition of the derivative of a real function, we can say that $f$ is differentiable :
:$\dfrac {\Delta y} {\Delta x} \to \dfrac {\d y} {\d x}$
as $\Delta x \to 0$.
Clearly, this is equivalent to saying that $f$ is differentiable :
:$\dfrac {\Delta y} {\Delta x} - \dfrac {\d y} {\d x} = \varepsilon$
as $\varepsilon \to 0$, where $\varepsilon \in \R$ is some real number.
Solving this equation for $\Delta y$:
:$\Delta y = \dfrac {\d y} {\d x} \Delta x + \varepsilon \Delta x$
That is, the real function is differentiable $\varepsilon \to 0$ as $\Delta x \to 0$.
Now consider $f: \R^n \to \R$, $n > 1$.
From the definition of differentiability of a real-valued function, $f$ is differentiable :
:$\map {\Delta f} {\mathbf x} = \map {\nabla f} {\mathbf x} \cdot \Delta \mathbf x + \begin{bmatrix} \\ \varepsilon_1 \\ \varepsilon_2 \\ \vdots \\ \varepsilon_n \end {bmatrix} \cdot \Delta \mathbf x$
such that $\begin{bmatrix} \\ \varepsilon_1 \\ \varepsilon_2 \\ \vdots \\ \varepsilon_n \end{bmatrix} \to \mathbf 0$ as $\Delta \mathbf x \to \mathbf 0$.
 
Observe that:
where $\begin {bmatrix} \\ \varepsilon_1 \\ \varepsilon_2 \\ \vdots \\ \varepsilon_n \end {bmatrix} \to \begin {bmatrix} \\ 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix}$ as $\begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix} \to \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}$.
== Sources ==
* : $13.4$
Category:Differential Calculus
Characterization of Dual Operator

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ and $Y$ be normed vector spaces over $\GF$.
Let $X^\ast$ and $Y^\ast$ be the normed dual spaces of $X$ and $Y$ respectively.
Let $X^{\ast \ast}$ and $Y^{\ast \ast}$ be the second normed duals of $X$ and $Y$ respectively. 
Let $T : Y^\ast \to X^\ast$ be a bounded linear transformation.
Let $\iota_X : X \to X^{\ast \ast}$ be the evaluation linear transformation for $X$. 
Let $\iota_Y : Y \to Y^{\ast \ast}$ be the evaluation linear transformation for $Y$.
:$(1) \quad$ $T$ is $\struct {w^\ast, w^\ast}$-continuous
:$(2) \quad$ there exists a bounded linear transformation $S : X \to Y$ such that $S^\ast = T$, where $S^\ast$ denotes the dual operator of $S$
:$(3) \quad$ $T^\ast \sqbrk {\iota_X X} \subseteq \iota_Y Y$
== Proof ==
=== $(1)$ implies $(2)$ ===
Suppose that $T$ is $\struct {w^\ast, w^\ast}$-continuous.
From Characterization of Continuity of Linear Functional in Weak-* Topology ‎, we have that: 
:$x^\wedge : \struct {X^\ast, w^\ast} \to \GF$ is continuous.
From Composite of Continuous Mappings is Continuous:
:$x^\wedge \circ T : \struct {Y^\ast, w^\ast} \to \GF$ is continuous.
Applying Characterization of Continuity of Linear Functional in Weak-* Topology , there exists $S x \in Y$ such that: 
:$x^\wedge \circ T = \paren {S x}^\wedge$
We first show that $S : X \to Y$ is linear. 
We have:
We now show that $S$ is bounded. 
We have, for each $x \in X$: 
So $S$ is bounded. 
Finally, for $f \in Y^\ast$ and $x \in X$ we have: 
That is: 
:$f \circ S = T f$ 
for each $f \in Y^\ast$. 
From the definition of the dual operator we have: 
:$S^\ast f = T f$
for each $f \in Y^\ast$.
So $S^\ast = T$. 
=== $(2)$ implies $(3)$ ===
Suppose that: 
:there exists a bounded linear transformation $S : X \to Y$ such that $S^\ast = T$, where $S^\ast$ denotes the dual operator of $S$.
We have $S^{\ast \ast} = T^\ast$, where $S^{\ast \ast}$ denotes the second dual operator.
Let $x \in X$ and $f \in Y^\ast$. 
Then:
That is: 
:$\map {T^\ast} {\iota_X x} = \map {\iota_Y} {S x}$
=== $(3)$ implies $(1)$ ===
Suppose that $T^\ast \sqbrk {\iota_X X} \subseteq \iota_Y Y$.
For each $x^\wedge \in \iota_X X$, we then have that: 
:$\map {T^\ast} {x^\wedge} \in \iota_Y Y$
That is, there exists $y \in Y$ such that: 
:$x^\wedge \circ T = y^\wedge$
From Characterization of Continuity of Linear Functional in Weak-* Topology, we then have: 
:$x^\wedge \circ T : \struct {Y^\ast, w^\ast} \to \GF$ is continuous
for each $x \in X$. 
Conversely from Characterization of Continuity of Linear Functional in Weak-* Topology, we have that every continuous linear functional $\Phi : \struct {Y^\ast, w^\ast} \to \GF$ has the form $\Phi = x^\wedge$. 
So from Continuity in Initial Topology, we have that: 
:$T : \struct {Y^\ast, w^\ast} \to \struct {X^\ast, w^\ast}$ is continuous.
Category:Dual Operators 
Category:Weak-* Topologies
Characterization of Ergodicity in terms of Koopman Operator

== Theorem ==
Let $\struct {X, \BB, \mu}$ be a probability space.
Let $T: X \to X$ be a measure-preserving transformation.
Let $\map \MM {X, \R}$ be the set of $\BB$-measurable functions.
Let $\map {\LL^2} \mu$ denote the Lebesgue $2$-space.
Let $U_T : \map \MM {X, \R} \to \map \MM {X, \R}$ be the Koopman operator:
:$U_T : f \mapsto f \circ T$
Then the following are equivalent:
:$(1):$ $T$ is ergodic
:$(2):$ For all $f \in \map \MM {X, \R}$:
::if $\map {U_T} f = f$, then $f$ is constant $\mu$-a.e.
:$(3):$ For all $f \in \map \MM {X, \R}$:
::if $\map {U_T} f = f$ for $\mu$-a.e., then $f$ is constant $\mu$-a.e.
:$(4):$ For all $f \in \map {\LL ^2} \mu$:
::if $\map {U_T} f = f$, then $f$ is constant $\mu$-a.e.
:$(5):$ For all $f \in \map {\LL ^2} \mu$:
::if $\map {U_T} f = f$ for $\mu$-a.e., then $f$ is constant $\mu$-a.e.
== Proof ==
=== $(1) \implies (3)$ ===
This is clear, since $(3)$ is exactly the definition of ergodicity.
=== $(3) \implies (2)$ and $(5) \implies (4)$ ===
These are a direct consequence of .
If $\map {U_T} f = f$, the same holds especially $\mu$-almost everywhere, since:
=== $(2) \implies (4)$ and $(3) \implies (5)$ ===
In view of :
:$\map {\LL ^2} \mu \subseteq \map \MM {X, \R}$
The claims follow from this.
=== $(4) \implies (1)$ ===
Let $A \in \BB$ be such that $T^{-1} \sqbrk A = A$.
Let $\chi_A : X \to \set {0, 1}$ be the characteristic function of $A$.
Note that $\chi_A^2 = \chi_A$, as $0^2=0$ and $1^2=1$.
In particular, $\chi_A \in \map {\LL^2} \mu$, since:
:$\ds \int \chi_A^2 \rd \mu = \map \mu A < + \infty$
Moreover:
:$\chi_A \circ T = \chi_A$
since for all $x \in X$:
That is, $\map {U_T} f = f$ by .
Therefore , $\chi_A$ is constant $\mu$-almost everywhere.
The claim follows from this, since by :
:$A = \set {x \in X : \map {\chi_A} x = 1}$
and:
:$X \setminus A = \set {x \in X : \map {\chi_A} x = 0}$
== Sources ==
* $1.5$: Ergodicity
Category:Ergodic Measure-Preserving Transformations
Characterization of Essentially Bounded Functions

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space. 
Let $f : X \to \R$ be a $\Sigma$-measurable function.
:$(1) \quad$ $f$ is essentially bounded.
:$(2) \quad$ There exists a bounded function $g : X \to \R$ such that $f = g$ $\mu$-almost everywhere.
== Proof ==
=== $(1)$ implies $(2)$ ===
Suppose that there exists $c \in \R$ such that: 
:$\map \mu {\set {x \in X : \size {\map f x} > c} } = 0$
Let:
:$A = \set {x \in X : \size {\map f x} \le c}$
Define a function $g : X \to \overline \R$ by: 
:$\map g x = \begin{cases}\map f x & x \in A \\ 0 & x \not \in A\end{cases}$
Then:
:$\size {\map g x} \le c$ for all $x \in X$. 
So:
:$g$ is bounded.
Then if $x \in X$ is such that $\map f x \ne \map g x$, we must have $x \in X \setminus A$.
But :
:$\map \mu {X \setminus A} = \map \mu {\set {x \in X : \size {\map f x} > c} } = 0$
So:
:$f = g$ $\mu$-almost everywhere.
=== $(2)$ implies $(1)$ ===
Suppose that: 
:there exists a bounded function $g : X \to \R$ such that $f = g$ $\mu$-almost everywhere.
That is, there exists a $\mu$-null set $N \subseteq X$ such that: 
:if $x \in X$ is such that $\map f x \ne \map g x$ then $x \in N$.
Let: 
:$\ds M = \sup_{x \mathop \in X} \size {\map g x} < \infty$
Since: 
:$g \le M$
if $x \in X$ is such that: 
:$\map f x > M$
then: 
:$\map f x \ne \map g x$
so:
:$x \in N$
So:
:$\set {x \in X : \size {\map f x} > M} \subseteq N$
From Absolute Value of Measurable Function is Measurable, we have that: 
:$\size f$ is $\Sigma$-measurable.
So, from Characterization of Measurable Functions we have: 
:$\set {x \in X : \size {\map f x} > M}$ is $\Sigma$-measurable.
Then from Null Sets Closed under Subset, since $\map \mu N = 0$, we have: 
:$\map \mu {\set {x \in X : \size {\map f x} > M} } = 0$
That is: 
:$f$ is essentially bounded.
Category:Measure Theory
Characterization of Euclidean Borel Sigma-Algebra

== Theorem ==
Let $\OO^n$, $\CC^n$ and $\KK^n$ be the collections of open, closed and compact subsets of the Euclidean space $\struct {\R^n, \tau}$, respectively.
Let $\JJ_{ho}^n$ be the collection of half-open rectangles in $\R^n$.
Let $\JJ^n_{ho, \text {rat} }$ be the collection of half-open rectangles in $\R^n$ with rational endpoints.
Then the Borel $\sigma$-algebra $\map \BB {\R^n}$ satisfies:
:$\map \BB {\R^n} = \map \sigma {\OO^n} = \map \sigma {\CC^n} = \map \sigma {\KK^n} = \map \sigma {\JJ_{ho}^n} = \map \sigma {\JJ^n_{ho, \text {rat} } }$
where $\sigma$ denotes generated $\sigma$-algebra.
== Proof ==
By definition of Borel $\sigma$-algebra, $\map \BB {\R^n} = \map \sigma {\OO^n}$.
The rest of the proof will be split in proving the following equalities:
:$(1): \quad \map \sigma {\OO^n} = \map \sigma {\CC^n}$
:$(2): \quad \map \sigma {\CC^n} = \map \sigma {\KK^n}$
:$(3): \quad \map \sigma {\OO^n} = \map \sigma {\JJ_{ho}^n}$
:$(4): \quad \map \sigma {\JJ_{ho}^n} = \map \sigma {\JJ^n_{ho, \text {rat} } }$
=== Proof of $(1)$ ===
=== Proof of $(2)$ ===
=== Proof of $(3)$ ===
=== Proof of $(4)$ ===
== Sources ==
* : $\S 1.2$
* : $3.7, 8$
Category:Sigma-Algebras
Category:Characterization of Euclidean Borel Sigma-Algebra