avewright/proofwiki-math huggingface dataset:
Theorem: Closure of Densely-Defined Bounded Linear Operator

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$. 
Let $\tuple {\map D T, T}$ be a densely-defined linear operator that is bounded.
Let $\widetilde T : \HH \to \HH$ be the bounded linear operator extending $T$, given by Bounded Linear Transformation to Banach Space has Unique Extension to Closure of Domain.
Then $\tuple {\map D T, T}$ is closable with closure $\tuple {\HH, \widetilde T}$.

Proof:
From Characterization of Closable Densely-Defined Linear Operators in terms of Closure of Graph and Closable Densely-Defined Linear Operator has Smallest Closed Extension, to show that the closure of $\tuple {\map D T, T}$ is $\tuple {\HH, \widetilde T}$, it is enough to show that:
:$\map \cl {\map \GG T} = \map \GG {\widetilde T}$
where the closure is understood in the direct product norm and $\GG$ denotes the graph of a linear operator.
Let $\tuple {x, y} \in \map \cl {\map \GG T}$.
Then there exists a convergent sequence $\sequence {\tuple {x_n, T x_n} }_{y \mathop \in \N}$ valued in $\map \GG T$ such that:
:$\tuple {x_n, T x_n} \to \tuple {x, y}$
From Convergence in Direct Product Norm, we have:
:$x_n \to x$ as $n \to \infty$
and:
:$\widetilde T x_n = T x_n \to y$ as $n \to \infty$.
We cannot conclude that $T x_n \to T x$ since we do not necessarily have $x \in \map D T$. 
However, from the boundedness and global definition of $\widetilde T$ we do obtain $\widetilde T x_n \to \widetilde T x$ as $n \to \infty$.
From Convergent Sequence in Normed Vector Space has Unique Limit, we have $\widetilde T x = y$. 
Hence we have:
:$\tuple {x, y} = \tuple {x, \widetilde T x} \in \map \GG {\widetilde T}$
So:
:$\map \cl {\map \GG T} \subseteq \map \GG {\widetilde T}$. 
Conversely let $\tuple {x, \widetilde T x} \in \map \GG {\widetilde T}$.
Since $\map D T$ is everywhere dense in $\HH$, there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\map D T$ such that $x_n \to x$.
Then we have $T x_n = \widetilde T x_n \to \widetilde T x$. 
Hence from Convergence in Direct Product Norm we have:
:$\tuple {x_n, T x_n} \to \tuple {x, \widetilde T x}$
Hence $\tuple {x, \widetilde T x}$ is the limit of a sequence in $\map \GG T$.
Hence:
:$\map \GG {\widetilde T} \subseteq \map \cl {\map \GG T}$
We conclude:
:$\map \cl {\map \GG T} = \map \GG {\widetilde T}$
Hence by Characterization of Closable Densely-Defined Linear Operators in terms of Closure of Graph and Closable Densely-Defined Linear Operator has Smallest Closed Extension, $\tuple {\map D T, T}$ is closable with closure $\tuple {\HH, \widetilde T}$.
{{qed}}
Category:Closable Densely-Defined Linear Operators
h9xfzkrmmwhryml33b5kbfebdkk34ul
Theorem: Measure is Monotone/Resolution of the Identity

Let $X$ be a topological space.
Let $\map \BB X$ be the Borel $\sigma$-algebra of $X$.
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\map B \HH$ be the space of bounded linear transformations on $\HH$.
Let $\le_{\map B \HH}$ be the canonical preordering of $\map B \HH$.
Let $\EE : \map \BB X \to \map B {\HH}$ be a resolution of the identity.
Let $E, F \in \map \BB X$ be such that $E \subseteq F$.
Then:
:$\map \EE E \le_{\map B \HH} \map \EE F$

Proof:
Write:
:$F = E \cup \paren {F \setminus E}$
From $(4)$ in the definition of a resolution of the identity, we have:
:$\map \EE F = \map \EE E + \map \EE {F \setminus E}$ 
From Bounds on Projection in Unital C*-Algebra, we have:
:${\mathbf 0}_{\map B \HH} \le_{\map B \HH} \map \EE {F \setminus E}$ 
From $(1)$ in the definition of a preordered vector space, we have:
:$\map \EE E + \map \EE {F \setminus E} \ge_{\map B \HH} \map \EE E$ 
Hence:
:$\map \EE E \le_{\map B \HH} \map \EE F$ 
{{qed}}
Category:Resolutions of the Identity
Category:Measure is Monotone
n50n1p42zl7r2f098wv6h29mnp6pzt8
Theorem: Bounds on Projection in Unital C*-Algebra

Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a unital $\text C^\ast$-algebra.
Let $\le_A$ be the canonical preordering of $A$.
Let $p$ be a projection on $A$.
Then we have:
:${\mathbf 0}_A \le_A p \le_A {\mathbf 1}_A$

Proof:
Since $p$ is a projection, it is Hermitian.
From Spectrum of Projection in *-Algebra: Corollary, we have:
:$\map {\sigma_A} p \subseteq \set {0, 1}$
From Bounds on Hermitian Element of Unital C*-Algebra in terms of Bounds on Spectrum, we have:
:${\mathbf 0}_A \le_A p \le_A {\mathbf 1}_A$
{{qed}}
Category:Projections (*-Algebras)
Category:Canonical Preorderings of C*-Algebras
iesydpzhm46ia4ma4r2hcqwgq7lq1tt
Theorem: Characterization of Canonical Preordering of Projections on Hilbert Space

Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$. 
Let $\map B \HH$ be the space of bounded linear operators on $\HH$ understood as a $\text C^\ast$-algebra.
Let $\le_{\map B \HH}$ be the canonical preordering of $\map B \HH$.
Let $P, Q$ be Hilbert space projections on $\HH$.
{{TFAE}}
:$(1) \quad$ $P \le_{\map B \HH} Q$
:$(2) \quad$ $P Q = P$
:$(3) \quad$ $Q P = P$
:$(4) \quad$ $\Img P \subseteq \Img Q$
:$(5) \quad$ for all $x \in \HH$ we have $\norm {P x} \le \norm {Q x}$
:$(6) \quad$ $Q - P$ is a Hilbert space projection

Proof:
We first note that from Characterization of Projections, $P$ is Hermitian, and that $P, Q$ is are orthogonal projections onto $\Img P$ and $\Img Q$ respectively.
Theorem: Everywhere Dense Subset of Countable Hilbert Space Direct Sum in terms of Everywhere Dense Subsets of Summands

Let $\family {\tuple {\HH_n, \innerprod \cdot \cdot_n} }_{n \mathop \in \N}$ be a sequence of Hilbert spaces over $\C$.
For each $n \in \N$, let $\family {e^{(n)}_\alpha}_{\alpha \in \map J n}$ be an everywhere dense subset of $\HH_n$.
Let $\tuple {\HH, \innerprod \cdot \cdot}$ be the Hilbert space direct sum of $\family {\tuple {\HH_n, \innerprod \cdot \cdot_n} }_{n \mathop \in \N}$. 
Let $\SS$ be the set of $f \in \HH$ such that:
:there exists $N \in \N$ such that:
::$\map f n = e^{(n)}_{\map \alpha n}$ for some $\map \alpha n \in \map J n$ and for all $n \le N$
::$\map f n = {\mathbf 0}_{\HH_n}$ for $n > N$. 
Then $\SS$ is everywhere dense in $\HH$.

Proof:
Let $f \in \HH$ and $\epsilon > 0$. 
We need to show that there exists $g \in \SS$ with $\norm {f - g} < \epsilon$.
Then by definition of the Hilbert space direct sum and Net Convergence Equivalent to Absolute Convergence:
:$\ds \sum_{n \mathop = 1}^\infty \norm {\map f n}_{\HH_n}^2 < \infty$
Hence there exists $N \in \N$ such that:
:$\ds \sum_{n \mathop = N + 1}^\infty \norm {\map f n}_{\HH_n}^2 < \frac {\epsilon^2} 2$
For $n \le N$, pick $\map \alpha n \in \map J n$ such that:
:$\norm {\map f n - e_{\map \alpha n}^{(n)} }_{\HH_n}^2 < \dfrac {\epsilon^2} {2 N}$
Define:
:$\map g n = e_{\map \alpha n}^{(n)} \in \HH_n$ for $n \le N$
:$\map g n = {\mathbf 0}_{\HH_n}$ for $n > N$.
We have:
:$\ds \sum_{n \mathop = 1}^\infty \norm {\map g n}_{\HH_n}^2 = \sum_{n \mathop = 1}^N \norm {\map g n}_{\HH_n}^2 < \infty$
since the sum on the {{RHS}} is finite.
Hence $g \in \HH$.
We then have:
{{begin-eqn}}
{{eqn	| l = \norm {f - g}_\HH^2
		| r = \sum_{n \mathop = 1}^\infty \norm {\map f n - \map g n}_{\HH_n}^2
}}
{{eqn	| r = \sum_{n \mathop = 1}^N \norm {\map f n - \map g n}_{\HH_n}^2 + \sum_{n \mathop = N + 1}^\infty \norm {\map f n}_{\HH_n}^2 
}}
{{eqn	| o = < 
		| r = \sum_{n \mathop = 1}^N \frac {\epsilon^2} {2 N} + \frac {\epsilon^2} 2 
}}
{{eqn	| r = \frac {\epsilon^2} 2 + \frac {\epsilon^2} 2 
}}
{{eqn	| r = \epsilon^2
}}
{{end-eqn}}
hence:
:$\norm {f - g}_\HH < \epsilon$
as required.
{{qed}}
Category:Hilbert Space Direct Sums
ql85fajxk5ucnqxihf1kudb5ul3sf1i
Theorem: Direct Sum of Densely-Defined Linear Operators is Densely-Defined Linear Operator

Let $\family {\tuple {\HH_n, \innerprod \cdot \cdot_n} }_{n \mathop \in \N}$ be a sequence of Hilbert spaces over $\C$.
Let $\tuple {\HH, \innerprod \cdot \cdot}$ be the Hilbert space direct sum of $\family {\tuple {\HH_n, \innerprod \cdot \cdot_n} }_{n \mathop \in \N}$.
For each $n \in \N$, let $\tuple {\map D {T_n}, T_n}$ be a densely-defined linear operator. 
Define:
:$\ds \map D T = \set {\sequence {\phi_n}_{n \mathop \in \N} \in \HH : \sum_{n \mathop = 1}^\infty \norm {T_n \phi_n}^2_{\HH_n} < \infty}$
Define $T : \map D T \to \HH$ by:
:$\map T {\sequence {\phi_n}_{n \mathop \in \N} } = T_\phi$
for each $\phi = \sequence {\phi_n}_{n \mathop \in \N} \in \HH$, where $T_\phi \in \HH$ is such that:
:$\map {T_\phi} n = T_n \phi_n$
for each $n \in \N$.
Then $\tuple {\map D T, T}$ is also a densely-defined linear operator.

Proof:
Let $\SS$ be the set of $f \in \HH$ such that:
:there exists $N \in \N$ such that:
::$\map f n = {\mathbf 0}_{\HH_n}$ for $n > N$.
For $\phi = \sequence {\phi_n}_{n \mathop \in \N} \in \SS$ and $N \in \N$ picked as a witness to the above condition, we have:
:$\ds \sum_{n \mathop = 1}^\infty \norm {T_n \phi_n}_{\HH_n}^2 = \sum_{n \mathop = 1}^N \norm {T_n \phi_n}_{\HH_n}^2 < \infty$ 
Hence $\phi \in \map D T$.
So $\SS \subseteq \map D T$.
From Everywhere Dense Subset of Countable Hilbert Space Direct Sum in terms of Everywhere Dense Subsets of Summands, $\SS$ is everywhere dense in $\HH$.
Hence from Set Closure Preserves Set Inclusion, $\map D T$ is everywhere dense in $\HH$.
It remains to show that $T$ is linear.
Let $\phi = \sequence {\phi_n}_{n \mathop \in \N} \in \map D T$, $\psi = \sequence {\psi_n}_{n \mathop \in \N} \in \map D T$ and $\lambda \in \C$.
Then, for each $n \in \N$, we have:
{{begin-eqn}}
{{eqn	| l = \map {T_{\phi + \lambda \psi} } n
		| r = \map {T_n} {\phi_n + \lambda \psi_n}
}}
{{eqn	| r = T_n \phi_n + \lambda T_n \psi_n 
		| c = $T_n$ is linear
}}
{{eqn	| r = \map {T_\phi} n + \lambda \map {T_\psi} n
}}
{{end-eqn}}
Hence we have:
:$T_{\phi + \lambda \psi} = T_\phi + \lambda T_\psi$
Hence:
:$\map T {\phi + \lambda \psi} = T \phi + \lambda T \psi$
So $T$ is linear.
{{qed}}
Category:Direct Sums of Densely-Defined Linear Operators
rrmim3txumdyhd9719oqq0j02e46s4a
Theorem: Characterization of Completely Prime Filter in Complete Lattice/Necessary Condition

Let $\struct{L, \vee, \wedge, \preceq}$ be a complete lattice.
Let $F \subseteq L$ be a completely prime filter.
Then:
:$(1)\quad\forall A \subseteq L : \bigvee A \in F \iff \paren{\exists a \in A : a \in F}$
:$(2)\quad\forall $ finite $A \subseteq L : \bigwedge A \in F \iff \paren{\forall a \in A : a \in F}$
where:
:* $\bigvee A$ denotes the supremum of $A$ in $L$
:* $\bigwedge A$ denotes the infimum of $A$ in $L$

Proof:
==== $F$ satisfies Statement $(1)$ ====
Let $A \subseteq L$.
Let $\bigvee A \in F$
By definition of completely prime filter:
:$A \cap F \ne \O$
Let $x \in A \cap F$
By definition of set intersection:
:$x \in A$ and $x \in F$
Let $x \in A : x \in F$
By definition of supremum:
:$x \preceq \bigvee A$
By definition of filter:
:$\bigvee A \in F$
Hence:
:$(1)\quad\forall A \subseteq L : \bigvee A \in F \iff \paren{\exists a \in A : a \in F}$
{{qed|lemma}}
==== $F$ satisfies Statement $(2)$ ====
Let $A \subseteq L$ be finite.
Let $\bigwedge A \in F$
By definition of infimum:
:$\forall x \in A : \bigwedge A \preceq x$
By definition of filter:
:$\forall x \in A : x \in F$
Let:
:$\forall x \in A : x \in F$
Let $A \ne \O$.
By definition of filter:
:$F$ is sublattice of $L$
From Existence of Non-Empty Finite Infima in Meet Semilattice:
:$\bigwedge A \in F$
Let $A = \O$.
From Infimum of Empty Set is Greatest Element:
:$\top = \bigwedge A$
From Filter Contains Greatest Element:
:$\top \in F$
Hence:
:$(2)\quad\forall $ finite $A \subseteq L : \bigwedge A \in F \iff \paren{\forall a \in A : a \in F}$
{{qed|lemma}}
Theorem: Characterization of Completely Prime Filter in Complete Lattice/Sufficient Condition

Let $\struct{L, \vee, \wedge, \preceq}$ be a complete lattice.
Let $F \subseteq L$ satisfy:
:$(1)\quad\forall A \subseteq L : \bigvee A \in F \iff \paren{\exists a \in A : a \in F}$
:$(2)\quad\forall $ finite $A \subseteq L : \bigwedge A \in F \iff \paren{\forall a \in A : a \in F}$
where:
:* $\bigvee A$ denotes the supremum of $A$ in $L$
:* $\bigwedge A$ denotes the infimum of $A$ in $L$
Then:
:$F$ is a completely prime filter

Proof:
==== $F$ is a Proper Subset ====
From Supremum of Empty Set is Smallest Element:
:$\bot = \bigvee \O$
By definition of empty set:
:$\forall x \in F : x \notin \O$
By the contrapositive statement of $(1)$:
:$\bot = \bigvee \O \notin F$
It follows that $F \neq L$
{{qed|lemma}}
==== $F$ is not Empty ====
From Infimum of Empty Set is Greatest Element:
:$\top = \bigwedge \O$
By definition of empty set, $\O$ vacuously satisfies:
:$\forall x \in \O : x \in F$
By $(2)$:
:$\top = \bigwedge \O \in F$
It follows that $F \neq \O$
Hence $F$ satisfies filter axiom (1).
{{qed|lemma}}
==== $F$ is a Down-Directed Subset of $L$ ====
Let $x, y \in F$.
From $(2)$:
:$x \wedge y \in F$
From Meet Precedes Operands:
:$x \wedge y \preceq x$ and $x \wedge y \preceq y$
Hence $F$ satisfies filter axiom (2).
{{qed|lemma}}
==== $F$ is an Upper Section of $L$ ====
Let $x \in F$.
Let $x \preceq y$.
From Successor is Supremum:
:$x \vee y = y$
From $(1)$:
:$y \in F$
Hence $F$ satisfies filter axiom (3).
{{qed|lemma}}
It follows that $F$ is a filter.
==== $F$ is Completely Prime ====
Let $A \subseteq L$.
Let $\bigvee A \in F$.
From $(1)$:
:$\exists a \in A : a \in F$
By definition of subset:
:$a \in A \cap F$
Hence $A \cap F \ne \O$.
{{qed|lemma}}
Hence $F$ is a completely prime filter by definition.
{{qed}}
Theorem: Eigenstates of Hamiltonian have Well Defined Symmetry only if Potential is Even

Given a certain Hamiltonian describing a quantum system, its eigenstates will have a well-defined symmetry in space (even or odd) only if the potential said system is subjected to is an even function of space

Proof:
Given some Hamiltonian $H$, we can find its eigenfunctions in order to determine the eigenstates of the system, thus forming an orthogonal (or orthonormal, if normalized) base of the corresponding Hilbert space. If these eigenfunctions are to have a well-defined symmetry, then they must also be eigenfunctions of the parity operator.
The only way the eigenstates can be eigenfunctions of both the Hamiltonian and the parity operator simultaneously is if their commutator is zero. Therefore, the proof of this lemma is reduced to finding the minimal set of conditions necessary for $[P, H]=0$ to be true. Let us examine how this commutator acts on some wave function $\phi(x)$ (this proof is carried out in one dimension in space for simplicity, but it can be extrapolated to an arbitrary dimension trivially).
<blockquote>$[P, H]\phi = (PH-HP)\phi$
$= P\left(\frac{p²}{2m}\phi(x)+V(x)\phi(x)\right)-\left(\frac{p²}{2m}P\phi(x)+V(x)P\phi(x)\right)$</blockquote>Taking into account the parity operator acts as $P\phi(x) = \phi(-x)$ and that the kinetic term is even (because $p²=-\hbar²\frac{d²}{dx²}$, so trivially $p²$ will remain unchanged under the parity operator).
<blockquote>$= \frac{p²}{2m}\phi(-x)+V(-x)\phi(-x)-\frac{p²}{2m}\phi(-x)-V(x)\phi(-x)$
$= \left[V(-x)-V(x)\right]\phi(-x)=0$</blockquote>Therefore, the condition we must impose for the commutator of the Hamiltonian and the parity operator to be zero $\forall x \in Dom(\phi)$ is that $V(x)=V(-x)$, that is to say, for $V(x)$ to be an even function of space $\blacksquare$
jxlp8syq8r7qb9lhwbwh9opmmxulwh8
Theorem: Complex-Valued Function is Measurable iff Real and Imaginary Part Measurable

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\struct {\C, \map \BB \C}$ be the complex numbers made into a measurable space with its Borel $\sigma$-algebra.
Let $\struct {\R, \map \BB \R}$ be the real numbers made into a measurable space with its Borel $\sigma$-algebra.
Let $f : X \to \C$ be a function.
Let $\map \Re f : X \to \R$ and $\map \Im f : X \to \R$ be the pointwise real part and imaginary part of $f$ respectively.
Then $f$ is $\Sigma/\map \BB \C$-measurable {{iff}} $\map \Re f$ and $\map \Im f$ are $\Sigma/\map \BB \R$-measurable.

Proof:
=== Necessary Condition ===
Suppose that $f$ is $\struct {X, \Sigma}$/$\struct {\C, \map \BB \C}$-measurable.
From Real and Imaginary Part Projections are Continuous and Continuous Mapping is Measurable, the real part and imaginary part are $\map \BB \C/\map \BB \R$-measurable as mappings $\C \to \R$.
Hence from Composition of Measurable Mappings is Measurable, $\map \Re f$ and $\map \Im f$ are $\Sigma/\map \BB \R$-measurable.
{{qed|lemma}}
Theorem: Measurability in Trace Sigma-Algebra

Let $\struct {A, \Sigma_A}$ and $\struct {C, \Sigma_C}$ be measurable spaces.
Let $B \subseteq C$. 
Let $\Sigma_B$ be the trace $\sigma$-algebra on $B$ induced by $\Sigma_C$. 
Let $f : A \to B$ be $\Sigma_A/\Sigma_B$-measurable.
Then $f$ is $\Sigma_A/\Sigma_C$-measurable.

Proof:
Let $S \in \Sigma_C$.
By Image is Subset of Codomain/Corollary 2, we have:
:$f \sqbrk A \subseteq B$
we have:
:$f^{-1} \sqbrk S = f^{-1} \sqbrk {S \cap B}$
From the definition of the trace $\sigma$-algebra, we have:
:$S \cap B \in \Sigma_B$
Since $f$ is $\Sigma_A/\Sigma_B$-measurable, we have:
:$f^{-1} \sqbrk {S \cap B} \in \Sigma_A$
Thus:
:$f^{-1} \sqbrk S \in \Sigma_A$.
Therefore we have shown that:
:$\forall S \in \Sigma_C$: $f^{-1} \sqbrk S \in \Sigma_A$
Hence $f$ is $\Sigma_A/\Sigma_C$-measurable. 
{{qed}}
Category:Measurable Functions
Category:Trace Sigma-Algebras
s3bq77owvzu130suglxte7npc7qksru
Theorem: Linear Combination of Measurable Functions valued in Topological Vector Space is Measurable

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\struct {K, +_K, \circ_K, \tau_K}$ be a topological field be a topological field.
Let $\struct {Y, \tau}$ be a topological vector space over $K$.
Let $\map \BB Y$ be the Borel $\sigma$-algebra of $\struct {Y, \tau}$.
Let $f : X \to Y$ and $g : X \to Y$ be $\map \BB X/\map \BB Y$-measurable mappings.
Let $\lambda \in K$.
Then $f + \lambda g : X \to Y$ is $\map \BB X/\map \BB Y$-measurable.

Proof:
Define $m_\lambda : Y \to Y$ by:
:$\map {m_\lambda} y = \lambda y$
for each $y \in Y$. 
By the definition of a topological vector space, $m_\lambda$ is continuous.
Hence from Continuous Mapping is Measurable, $m_\lambda$ is $\map \BB Y/\map \BB Y$-measurable.
Equip $Y^2$ with the product topology.
Define also $s : Y^2 \to Y$ by:
:$\map s {x, y} = x + y$
for each $\tuple {x, y} \in Y^2$.
Again by the definition of a topological vector space, $s$ is continuous.
Hence from Continuous Mapping is Measurable, $m_\lambda$ is $\map \BB {Y^2}/\map \BB Y$-measurable.
Define $F : Y^2 \to Y$ by:
:$\map F {x, y} = \tuple {x, \lambda y}$ 
for each $\tuple {x, y} \in Y^2$. 
Let $\pr_1: Y^2 \to Y$ and $\pr_2: Y^2 \to Y$ be the first and second projections, respectively.
Hence we have:
:$\map {\pr_1 \circ F} {x, y} = x$
and:
:$\map {\pr_2 \circ F} {x, y} = \lambda y = \map {m_\lambda} y$
From Identity Mapping is Continuous, Continuous Mapping is Measurable and Measurable Mappings from Product Measurable Space, we have that $\pr_1 \circ F$ is $\map \BB {Y^2}/\map \BB Y$-measurable.
Similarly from Measurable Mappings from Product Measurable Space we have that $\pr_2 \circ F$ is $\map \BB {Y^2}/\map \BB Y$-measurable.
Hence from Measurable Mappings to Product Measurable Space, we obtain that $F$ is $\map \BB {Y^2}/\map \BB {Y^2}$-measurable.
Hence from Composition of Measurable Mappings is Measurable, $s \circ F$ is $\map \BB {Y^2}/\map \BB Y$-measurable.
Finally, define $G : X \to Y^2$ by:
:$\map G x = \tuple {\map f x, \map g x}$
for each $x \in X$.
Since $f$ and $g$ are $\map \BB X/\map \BB Y$-measurable mappings, we have that $G$ is $\map \BB X/\map \BB {Y^2}$-measurable from Measurable Mappings to Product Measurable Space.
Finally from Composition of Measurable Mappings is Measurable, $s \circ F \circ G$ is $\map \BB X/\map \BB Y$-measurable with:
{{begin-eqn}}
{{eqn	| l = \map {\paren {s \circ F \circ G} } x
		| r = \map {\paren {s \circ F} } {\map f x, \map g x}
        | c = {{Defof|Composition of Mappings}}
}}
{{eqn	| r = \map s {\map F {\map f x, \map g x} }
        | c = {{Defof|Composition of Mappings}}
}}
{{eqn	| r = \map s {\map f x, \lambda \map g x}
        | c = Definition of $F$
}}
{{eqn	| r =  \map f x + \lambda \map g x
        | c = Definition of $s$
}}
{{end-eqn}}
for each $x \in X$.
Hence $f + \lambda g : X \to Y$ is $\map \BB X/\map \BB Y$-measurable.
{{qed}}
Category:Measurable Functions
mq45ztg58zwyj63ugu5dzfzy8z9hesu
Theorem: Content of Subdivided Rectangle

Let $R$ be a closed $n$-rectangle.
Let $P$ be a subdivision of $R$.
Then:
:$\ds \map V R = \sum_{r \mathop \in P^*} \map V r$
where:
:$\map V R$ denotes the content of the rectangle $R$
:$P^*$ denotes the set of subrectangles of $P$.

Proof:
We will proceed by induction on $n$.
Theorem: Homogeneity of Space implies Conservation of Linear Momentum

Because space is homogeneous in an inertial frame of reference, linear momentum is conserved if the system is closed.

Proof:
In inertial frames of reference, space is homogeneous. Therefore, working exclusively with inertial frames of reference, in closed systems (those which do not interact in any ways with external agents and are therefore causally disconnected from anything external), we can make a translation of every particle in said system and nothing will change (broadly speaking, this means "no region of space is special or privileged"). In order to prove this, let us examine what effects it would have on the Lagrangian function describing a system of these characteristics if we were to make an inifinitesimally small translation of all the particles in it. For simplicity, the proof shall be carried out with a one-particle system, but it is trivial to extrapolate it to an arbitrary number of them.
Because the system is closed, the Lagrangian of the system (of the particle) does not depend on time, and it is a function exclusively of the particle's position and velocity:
:$\LL = \map \LL {\vec r_i, \dot {\vec r_i} }$
Therefore, an infinitesimal change in the Lagrangian function can be written in cartesian coordinates as follows:
:$\ds \delta \LL = \sum_i \frac {\partial \LL} {\partial x_i} \delta x_i + \sum_i \frac {\partial \LL} {\partial \dot x_i} \delta \dot x_i$
Now, taking into account we are just carrying out a (constant) translation of the position vector, this translation does not depend on time, and therefore:
:$\delta \dot x_i = \dfrac \d {\d t} \delta x_i = 0$
Hence, we are only left with the first term.
Since our hypothesis is the Lagrangian remains unchanged under translations in space, we must then impose $\delta \LL = 0$ and see what conditions must be met in order to guarantee this equality.
Let us consider what we have so far:
:$\ds \delta \LL = \sum_i \frac {\partial \LL} {\partial x_i} \delta x_i = 0$
Since each infinitesimal displacement is independent from the rest, the only way in which this summation can be null is if each partial derivative is null:
:$\ds \therefore \sum_i \frac {\partial \LL} {\partial x_i} = 0$
Now, remembering Euler-Lagrange equations, we can state:
:$\dfrac {\partial \LL} {\partial x_i} = - \dfrac \d {\d t} \dfrac {\partial \LL} {\partial \dot x_i}$.
We reach the conclusion that:
:$\dfrac \d {\d t} \dfrac {\partial \LL} {\partial \dot x_i} = 0$
Therefore, $\dfrac {\partial \LL} {\partial \dot x_i} = \text{constant}$, but since $\dfrac {\partial \LL} {\partial \dot x_i} = \dfrac {\partial \paren {T + U} } {\partial \dot x_i}$ with $U = \map U {x_i}$, then we have:
:$\ds T = \sum_i \frac 1 2 m \dot x_i^2$
It is clear that: $\dfrac {\partial \LL} {\partial \dot x_i} = p$, where $p$ is the particle's linear momentum.
Therefore, we have:
:$p = \text{constant}$
{{qed}}
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Theorem: Decimal Expansion of Number is Terminating or Periodic iff Rational

Let $x$ be a real number.
Then:
:the decimal expansion of $x$ is either terminating or recurring
{{iff}}
:$x$ is a rational number.

Proof:
=== Necessary Condition ===
Let $x$ be a rational number.
Let $x$ be expressed in the form $\dfrac a b$ where $a \in \Z$ and $b \in \Z_{>0}$.
By the division algorithm:
:$a = b q + r$
for some integers $q, r$ and $0 \le r < b$.
It follows that if $r = 0$ the decimal terminates.
Suppose $r \ne 0$.
We have that $r$ is an element of a finite set of integers.
By the pigeonhole principle, at some point $r$ must repeat.
Thus the decimal expansion of a rational number either terminates or recurs.
{{qed|lemma}}