nyuuzyou/wikis huggingface dataset:
Orthogonality of Legendre Polynomials/Inequality

== Theorem ==
Let $\map {P_n} x$ denote the Legendre polynomial of order $n$.
:$\ds \int_{-1}^1 \map {P_m} x \map {P_n} x \rd x = 0$
when $m \ne n$.
== Proof ==
== Sources ==
* : $\S 25$: Legendre Functions: Orthogonality of Legendre Polynomials: $25.25$
* : $\S 28$: Legendre and Associated Legendre Functions: Orthogonality of Legendre Polynomials: $28.25.$
Category:Orthogonality of Legendre Polynomials
Orthogonality of Resolution of the Identity evaluated at Disjoint Sets

== Theorem ==
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 $\EE : \map \BB X \to \map B {\HH}$ be a resolution of the identity.
Let $A, B \in \map \BB X$ be such that $A \cap B = \O$.
Then:
:$\innerprod {\map \EE A x} {\map \EE B y} = 0$ 
for each $x, y \in \HH$.
== Proof ==
From $(2)$ in the definition of a resolution of the identity, we have:
:$\map \EE A \map \EE B = \map \EE {A \cap B} = \map \EE \O$
From $(1)$ in the definition of a resolution of the identity, we then have:
:$\map \EE A \map \EE B = 0$
From Hilbert Space Projections with Zero Product are Pointwise Orthogonal, we therefore havE:
:$\innerprod {\map \EE A x} {\map \EE B y} = 0$ 
for each $x, y \in \HH$.
Category:Resolutions of the Identity
Orthogonality of Solutions to the Sturm-Liouville Equation with Distinct Eigenvalues

== Theorem == 
Let $\map f x$ and $\map g x$ be solutions of the Sturm-Liouville equation:
:$(1): \quad -\map {\dfrac \d {\d x} } {\map p x \dfrac {\d y} {\d x} } + \map q x y = \lambda \map w x y$
where $y$ is a function of the free variable $x$.
The functions $\map p x$, $\map q x$ and $\map w x$ are specified.
In the simplest cases they are continuous on the closed interval $\closedint a b$.
In addition:
:$(1a): \quad \map p x > 0$ has a continuous derivative
:$(1b): \quad \map w x > 0$
:$(1c): \quad y$ is typically required to satisfy some boundary conditions at $a$ and $b$.
Assume that the Sturm-Liouville problem is regular, that is, $\map p x^{-1} > 0$, $\map q x$, and $\map w x > 0$ are real-valued integrable functions over the closed interval $\closedint a b$, with separated boundary conditions of the form: 
:$(2): \quad \map y a \cos \alpha - \map p a \map {y'} a \sin \alpha = 0$
:$(3): \quad \map y b \cos \beta - \map p b \map {y'} b \sin \beta = 0$
where $\alpha, \beta \in \hointr 0 \pi$.
Then:
:$\ds \innerprod f g = \int_a^b \overline {\map f x} \map q x \map w x \rd x = 0$
where $\map f x$ and $\map g x$ are solutions to the Sturm-Liouville equation corresponding to distinct eigenvalues and $\map w x$ is the "weight" or "density" function.
== Proof ==
Multiply the equation for $\map g x$ by $\overline {\map f x}$ (the complex conjugate of $\map f x$) to get:
:$-\overline {\map f x} \dfrac {\map \d {\map p x \map {\dfrac {\d g} {\d x} } x} } {\d x} + \overline {\map f x} \map q x \map g x = \mu \overline {\map f x} \map w x \map g x$
Only $\map f x$, $\map g x$, $\lambda$ and $\mu $ may be complex.
All other quantities are real.
Complex conjugate this equation, exchange $\map f x$ and $\map g x$, and subtract the new equation from the original:
Integrate this between the limits $x = a$ and $x = b$:
:$\ds \paren {\mu - \bar \lambda} \int_a^b \overline {\map f x} \map g x \map w x \rd x = \map p b \paren {\map g b \map {\frac {\d \bar f} {\d x} } b - \overline {\map f b} \map {\frac {\d g} {\d x} } b} - \map p a \paren {\map g a \map {\frac {\d \bar f} {\d x} } a - \overline {\map f a} \map {\frac {\d g} {\d x} } a}$
The right side of this equation vanishes because of the boundary conditions, which are either:
:periodic boundary conditions, that is, that $\map f x$, $\map g x$, and their first derivatives (as well as $\map p x$) have the same values at $x = b$ as at $x = a$
or:
:that independently at $x = a$ and at $x = b$ either:
::the condition cited in equation $(2)$ or $(3)$ holds
:or:
::$\map p x = 0$.
So:
:$\ds \paren {\mu - \bar \lambda} \int_a^b \overline {\map f x} \map g x \map w x \rd x = 0$
If we set $f = g$, so that the integral surely is non-zero, then it follows that $\bar \lambda = \lambda$.
That is, the eigenvalues are real, making the differential operator in the Sturm-Liouville equation self-adjoint (hermitian).
So:
:$\ds \paren {\mu - \lambda} \int_a^b \overline {\map f x} \map g x \map w x \rd x = 0$
It follows that, if $f$ and $g$ have distinct eigenvalues, then they are orthogonal.
Category:Sturm-Liouville Theory
Orthonormal Sequence in Hilbert Space Converges Weakly to Zero

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $\sequence {e_n}_{n \mathop \in \N}$ be a sequence in $\HH$ such that: 
:$\innerprod {e_n} {e_m} = 0$ if $n \ne m$
and:
:$\norm {e_n} = 1$ for each $n \in \N$. 
Then: 
:$e_n \weakconv 0$
where $\rightharpoonup$ denotes weak convergence.
== Proof ==
From Bessel's Inequality, we have: 
:$\ds \sum_{n \mathop = 1}^\infty \size {\innerprod y {e_n} }^2$ converges for each $y \in \HH$.
So from Terms in Convergent Series Converge to Zero, we have: 
:$\cmod {\innerprod y {e_n} }^2 \to 0$ for each $y \in \HH$. 
We then have, from Complex Sequence is Null iff Positive Integer Powers of Sequence are Null: 
:$\cmod {\innerprod y {e_n} } \to 0$ for each $y \in \HH$
and so, from Complex Sequence is Null iff Modulus of Sequence is Null: 
:$\innerprod y {e_n} \to 0$ for each $y \in \HH$.
From Weak Convergence in Hilbert Space: Corollary, we therefore have:
:$e_n \weakconv 0$
== Sources ==
* $27.1$: Weak Convergence
Category:Weak Convergence (Normed Vector Spaces)
Category:Hilbert Spaces
Orthonormal Subset/Examples/L-2 Space over Interval of Zero to Two Pi

== Example of Orthonormal Subset ==
Let $L^2_\C \closedint 0 {2 \pi}$ be the complex $L^2$ space over the closed interval $\closedint 0 {2 \pi}$.
Let $\innerprod \cdot \cdot$ be the $L^2$ inner product.
For $n \in \Z$, let $e_n: \closedint 0 {2 \pi} \to \C$ be defined by:
:$\map {e_n} t = \paren{ 2 \pi }^{-1/2} \map \exp {i n t}$
Then $\set{ e_n : n \in \Z}$ is an orthonormal subset of $L^2_\C \closedint 0 {2 \pi}$.
== Proof == 
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 4.$ Orthonormal Sets of Vectors and Bases: Example $4.3$
Category:Examples of Orthonormal Subsets
Category:Lp Spaces
Orthonormal Subset of Hilbert Space Extends to Basis

== Theorem ==
Let $H$ be a Hilbert space, and let $S$ be an orthonormal subset of $H$.
Then there exists a basis for $H$ that contains $S$ as a subset.
== Proof ==
Consider the set $\OO$ of orthonormal subsets $S'$ of $H$ that contain $S$:
:$\OO := \set{ S' \subseteq H: S \subseteq S', \text{$S'$ is orthonormal} }$
In particular, $S \in \OO$ so that $\OO$ is non-empty.
Give $\OO$ the subset ordering.
For a chain $\CC \subseteq \OO$, we assert that $\bigcup \CC \in \OO$.
For each $c \in \bigcup \CC$, there is a $C \in \CC$ such that $c \in C$.
Hence $\norm c = 1$.
Moreover, for $c, c' \in \bigcup \CC$, there is a $C \in \CC$ such that $c, c' \in C$ because $\CC$ is a chain.
Then since $C \in \OO$, it follows that $c \perp c'$.
Hence $\bigcup \CC \in \OO$.
Therefore, the conditions of Zorn's Lemma apply to $\OO$.
Let $O$ be a maximal element of $\OO$.
Then $O$ is a maximal orthonormal subset of $H$.
That is, $O$ is a basis containing $S$.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 4.$ Orthonormal Sets of Vectors and Bases: Proposition $4.2$
Category:Hilbert Spaces
Oscillation on Subset

== Theorem ==
Let $f: D \to \R$ be a real function where $D \subseteq \R$.
Let $x$ be a point in $D$.
Let $S_x$ be a set of real sets that contain (as an element) $x$.
Let $I$ be an element of $S_x$.
Let $\map {\omega_f} {I \cap D}$ be the oscillation of $f$ on $I \cap D$:
:$\map {\omega_f} {I \cap D} = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
Let $I \in S_x$.
Let $\map {\omega_f} {I \cap D} \in \R$.
Let $J \in S_x$ be a subset of $I$.
Then:
:$\map {\omega_f} {J \cap D} \in \R$
:$\map {\omega_f} {J \cap D} \le \map {\omega_f} {I \cap D}$
== Proof ==
Let:
:$I, J \in S_x$
:$J \subset I$
:$\map {\omega_f} {I \cap D} \in \R$
where:
:$\map {\omega_f} {I \cap D} = \sup \set {\size {\map f y - \map f z}: y, z \in I \cap D}$
We need to prove:
:$\map {\omega_f} {J \cap D} \in \R$
:$\map {\omega_f} {J \cap D} \le \map {\omega_f} {I \cap D}$
We intend to prove that $\map {\omega_f} {J \cap D} \in \R$.
We start by proving that $\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is bounded above and non-empty.
We have that $J$ is a subset of $I$.
Therefore:
:$\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is a subset of $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$.
The statement $\map {\omega_f} {I \cap D} \in \R$ means that $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ admits a supremum.
Therefore, $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ is bounded above.
Accordingly:
:$\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is bounded above as $\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is a subset of $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$
We observe that $\size {\map f y - \map f z} = 0$ for $y = z = x$.
Therefore, $0 \in \set {\size {\map f y - \map f z}: y, z \in J \cap D}$ as $x \in J \cap D$.
Accordingly:
:$\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is non-empty
We have that $\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is a real set as $\size {\map f y - \map f z} \in \R$ for every $y, z \in D$.
We have shown that $\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is non-empty and bounded above.
Therefore, $\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ admits a supremum by the Continuum Property.
In other words:
We finished proving that $\map {\omega_f} {J \cap D} \in \R$.
It remains to prove that $\map {\omega_f} {J \cap D} \le \map {\omega_f} {I \cap D}$.
We have:
:$\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ is a subset of $\set {\size {\map f y - \map f z}: y, z \in I \cap D}$
:$\set {\size {\map f y - \map f z}: y, z \in I \cap D}$ admits a supremum
:$\set {\size {\map f y - \map f z}: y, z \in J \cap D}$ admits a supremum
Then:
Category:Real Analysis
Ostrowski's Theorem

== Theorem ==
Every non-trivial norm on the rational numbers $\Q$ is equivalent to either:
:the $p$-adic norm $\norm {\, \cdot \,}_p$ for some prime $p$
or:
:the absolute value, $\size {\, \cdot \,}$.
== Proof ==
Let $\norm {\, \cdot \,}$ be a non-trivial norm on the rational numbers $\Q$.
=== Archimedean Norm Case ===
Let $\norm {\, \cdot \,}$ be an Archimedean norm.
=== Non-Archimedean Norm Case ===
Let $\norm {\, \cdot \,}$ be a non-Archimedean Norm.
== Also see ==
* Complete Archimedean Valued Field is Real or Complex Numbers, also sometimes called Ostrowski's theorem.
== Historical Note ==
== Sources ==
* : $\S 3.1$ Absolute Values on $\Q$: Theorem $3.1.3$
* : $\S 1.9$ Metrics and norms on the rational numbers. Ostrowski’s Theorem: Theorem $1.50$
Category:Normed Division Rings
Category:P-adic Number Theory
Category:Ostrowski's Theorem
Ostrowski's Theorem/Archimedean Norm

== Theorem ==
Let $\norm {\, \cdot \,}$ be a non-trivial Archimedean norm on the rational numbers $\Q$.
Then $\norm {\, \cdot \,}$ is equivalent to the absolute value $\size {\, \cdot \,}$.
== Proof ==
By Characterisation of Non-Archimedean Division Ring Norms then:
:$\exists n \in \N$ such that $\norm n > 1$
Let $n_0 = \min \set {n \in \N : \norm n > 1}$
By Norm of Unity then:
:$n_0 > 1$
Let $\alpha = \dfrac {\log \norm {n_0} } {\log n_0}$
Since $n_0, \norm n_0 > 1$ then:
:$\alpha > 0$
==== Lemma 1.1 ====
==== Lemma 1.2 ====
Hence:
:$\forall n \in \N: \norm n = n^\alpha = \size n^\alpha$
By Equivalent Norms on Rational Numbers then $\norm {\, \cdot \,}$ is equivalent to the absolute value $\size {\, \cdot \,}$.
== Sources ==
* : $\S 3.1$ Absolute Values on $\Q$, Theorem $3.1.3$
* : $\S 1.9$ Metrics and norms on the rational numbers. Ostrowski’s Theorem, Theorem $1.50$
Category:Ostrowski's Theorem
Ostrowski's Theorem/Archimedean Norm/Lemma 1.1

== Theorem ==
Let $\norm {\, \cdot \,}$ be a non-trivial Archimedean norm on the rational numbers $\Q$.
Let $n_0 = \min \set {n \in \N : \norm n > 1}$
Let $\alpha = \dfrac {\log \norm {n_0} } {\log n_0 }$
Then:
:$\forall n \in N: \norm n \le n^\alpha$
== Proof ==
By the definition of $\alpha$ then:
:$\norm {n_0} = {n_0}^\alpha$
By the definition of $n_0$ then:
:${n_0}^\alpha > 1$
Let $n \in \N$.
By Basis Representation Theorem then $n$ can be written:
:$n = a_0 + a_1 n_0 + a_2 {n_0}^2 + \cdots + a_s {n_0}^s$
where $0 \le a_i < n_0$ and $a_s \ne 0$
Since all of the $a_i < n_0$, by choice of $n_0$ then:
:$\forall a_i: \norm {a_i} \le 1$
Then:
Let $C = \paren {\dfrac { {n_0}^\alpha} { {n_0}^\alpha - 1} }$
Hence:
:$\norm n \le C n^\alpha$
As $n \in \N$ was arbitrary:
:$\forall n \in N: \norm n \le C n^\alpha$
Let $n, N \in N$
Then:
:$\norm {n^N} \le C \paren {n^N}^\alpha$
Now:
By Limit of Root of Positive Real Number:
:$\sqrt [N] C \to 1$ as $N \to \infty$
By the Multiple Rule for Real Sequences:
:$\sqrt [N] C n^\alpha \to n^\alpha$ as $N \to \infty$
By Inequality Rule for Real Sequences, letting $N \to \infty$ for fixed $n$:
:$\norm n \le n^\alpha$
The result follows.
== Sources ==
* : $\S 3.1$ Absolute Values on $\Q$, Theorem $3.1.3$
* : $\S 1.9$ Metrics and norms on the rational numbers. Ostrowski’s Theorem, Theorem $1.50$
Category:Ostrowski's Theorem
Ostrowski's Theorem/Archimedean Norm/Lemma 1.2

== Theorem ==
Let $\norm {\, \cdot \,}$ be a non-trivial Archimedean norm on the rational numbers $\Q$.
Let $n_0 = \min \set {n \in \N : \norm n > 1}$
Let $\alpha = \dfrac {\log \norm {n_0} } {\log n_0}$
Then:
:$\forall n \in N: \norm n \ge n^\alpha$
== Proof ==
By the definition of $\alpha$:
:$\norm {n_0} = {n_0}^\alpha$
By the definition of $n_0$:
:${n_0}^\alpha > 1$
Let $n \in \N$.
By Basis Representation Theorem then $n$ can be written:
:$n = a_0 + a_1 n_0 + a_2 {n_0}^2 + \cdots + a_s {n_0}^s$
where $0 \le a_i < n_0$ and $a_s \ne 0$
By Bounds for Integer Expressed in Base k:
:${n_0}^{s + 1} > n \ge {n_0}^s$
By Lemma 1.1:
:$\norm { {n_0}^{s + 1} - n} \le \paren { {n_0}^{s + 1} - n}^\alpha$
Hence:
Let $C' = \paren{1 - \paren{1 - \frac 1 {n_0} }^\alpha}$.
Then:
:$\norm n \ge C' n^\alpha$
As $n \in \N$ was arbitrary:
:$\forall n \in N: \norm n \ge C' n^\alpha$
Let $n, N \in N$.
Then:
:$\norm {n^N} \ge C' \paren {n^N}^\alpha$
Now:
By Limit of Root of Positive Real Number:
:$\sqrt [N] {C'} \to 1$ as $N \to \infty$
By the Multiple Rule for Real Sequences:
:$\sqrt [N] {C'} n^\alpha \to n^\alpha$ as $N \to \infty$
By Inequality Rule for Real Sequences, letting $N \to \infty$ for fixed $n$:
:$\norm n \ge n^\alpha$
The result follows.
== Sources ==
* : $\S 3.1$ Absolute Values on $\Q$, Theorem $3.1.3$
* : $\S 1.9$ Metrics and norms on the rational numbers. Ostrowski’s Theorem, Theorem $1.50$
Category:Ostrowski's Theorem
Ostrowski's Theorem/Non-Archimedean Norm

== Theorem ==
Let $\norm {\, \cdot \,}$ be a non-trivial non-Archimedean norm on the rational numbers $\Q$.
Then $\norm {\, \cdot \,}$ is equivalent to the $p$-adic norm $\norm {\, \cdot \,}_p$ for some prime $p$.
== Proof ==
From Characterisation of Non-Archimedean Division Ring Norms then:
:$\forall n \in \N: \norm n \le 1$
==== Lemma 2.1 ====
Let $n_0 = \min \set {n \in N : \norm n < 1}$.
==== Lemma 2.2 ====
Let $p = n_0$.
Let $\alpha = - \dfrac {\log \norm p } {\log p}$ then:
:$\norm p = p^{-\alpha} = \paren {p^{-1}}^\alpha = \norm p_p^\alpha$
Let $b \in N$
=== Case 1: $p \nmid b$ ===
Let $p \nmid b$.
From Prime not Divisor implies Coprime:
:$p$ and $b$ are coprime, that is, $p \perp b$
From Corollary 5 of Three Points in Ultrametric Space have Two Equal Distances:
:$\norm b = 1$
By the definition of the $p$-adic norm:
:$\norm b_p = 1$ 
Hence:
:$\norm b = 1 = 1^\alpha = \norm b_p^\alpha$
=== Case 2: $p \divides b$ ===
Let $p \divides b$.
Let $\nu = \map {\nu_p} b$ where $\nu_p$ is the $p$-adic valuation on $\Z$.
Then:
:$b = p^\nu c$
where $p \nmid c$
From #Case 1:
:$\norm c = 1$
Hence:
In either case:
:$\norm b = \norm b_p^\alpha$
Since $b$ was arbitrary, it has been shown:
:$\forall b \in \N: \norm b = \norm b_p^\alpha$
From Equivalent Norms on Rational Numbers:
:$\norm {\, \cdot \,}$ is equivalent to the $p$-adic norm $\norm {\, \cdot \,}_p$.
== Sources ==
* : $\S 3.1$ Absolute Values on $\Q$, Theorem $3.1.3$
* : $\S 1.9$ Metrics and norms on the rational numbers. Ostrowski’s Theorem, Theorem $1.50$
Category:Ostrowski's Theorem
Ostrowski's Theorem/Non-Archimedean Norm/Lemma 2.1

== Theorem ==
Let $\norm {\, \cdot \,}$ be a non-trivial non-Archimedean norm on the rational numbers $\Q$.
Then:
:$\exists n \in \N: 0 < \norm n < 1$.
== Proof ==
Because $\norm {\, \cdot \,}$ is non-trivial:
:$\exists \dfrac a b \in \Q : 0 < \norm {\dfrac a b} \mbox { and } \norm {\dfrac a b} \ne 1$
By Norm of Inverse then:
:$\norm {\dfrac a b} > 1 \implies \norm {\dfrac b a} < 1$
Hence either $\norm {\dfrac a b} < 1$ or $\norm {\dfrac b a} < 1$.
 assume $\norm {\dfrac a b} < 1$.
By Norm of Quotient then:
:$\dfrac {\norm a} {\norm b} < 1$
Hence:
:${\norm a} < \norm b$
Let $n = \size a$ and $m = \size b$ where $\size {\,\cdot\,}$ is the absolute value on $\Q$.
Then $n, m \in \N$
By Norm of Negative then:
:$\norm n = \norm a$
:$\norm m = \norm b$
Hence:
:$\norm n < \norm m$
Since $\norm {\, \cdot \,}$ is non-Archimedean then:
:$\norm m \le 1$
Hence:
:$\norm n < \norm m \le 1$
== Sources ==
* : $\S 3.1$ Absolute Values on $\Q$, Theorem $3.1.3$
* : $\S 1.9$ Metrics and norms on the rational numbers. Ostrowski’s Theorem, Theorem $1.50$
Category:Ostrowski's Theorem
Ostrowski's Theorem/Non-Archimedean Norm/Lemma 2.2

== Theorem ==
Let $\norm {\, \cdot \,}$ be a non-trivial non-Archimedean norm on the rational numbers $\Q$.
Let $n_0 = \min \set {n \in N : \norm n < 1}$.
Then:
:$n_0$ is a prime number.
== Proof ==
 $n_0$ is a composite number. 
Let $n_1, n_2 \in \N$ such that $n_1, n_2 < n_0$ and $n_0 = n_1 n_2$. 
By the definition of $n_0$ then:
:$\norm {n_1} = 1$ 
:$\norm {n_2} = 1$ 
By :
:$\norm {n_0} = \norm {n_1 n_2} = \norm {n_1} \norm {n_2} = 1$
This contradicts the assumption that $\norm {n_0} < 1$. 
So $n_0$ must be a prime number.
== Sources ==
* : $\S 3.1$ Absolute Values on $\Q$, Theorem $3.1.3$
* : $\S 1.9$ Metrics and norms on the rational numbers. Ostrowski’s Theorem, Theorem $1.50$
Category:Ostrowski's Theorem
Outer Jordan Content Never Smaller than Inner Jordan Content

== Theorem ==
Let $M$ be a bounded subspace of Euclidean space.
Let $S$ be an orthotope enclosing $M$.
Let $\map {m^*} A$ denote the outer Jordan content of $A$.
Let $\map V A$ denote the content of $A$.
Then:
:$\map {m^*} M \ge \map V S - \map {m^*} {S \setminus M}$
== Proof ==
Let $P$ be a finite covering of $M$, and $Q$ a finite covering of $S \setminus M$.
If $Q \supsetneq S$, we can find another finite covering $Q' = \set {Q_i \cap S : Q_i \in Q} \subseteq S$, where $\map V {Q'} \le \map V Q$.
, assume that $Q \subseteq S$.
Then:
:$S \setminus Q \subseteq M \subseteq P$
It follows that:
:$\map V {S \setminus Q} \le \map V P$.
But because $Q \subseteq S$:
:$\map V {S \setminus Q} = \map V S - \map V Q$
Therefore:
:$\map V S - \map V Q \le \map V P$
Thus, by the definition of Outer Jordan Content:
:$\map V S - \map {m^*} {S \setminus M} \le \map {m^*} M$
Category:Measure Theory
Outer Jordan Content is Monotone

== Theorem ==
Let $A, B \subseteq \R^n$ be bounded subspaces of Euclidean $n$-space.
Suppose that $A \subseteq B$.
Then:
:$\map {m^*} A \le \map {m^*} B$
where $m^*$ denotes the outer Jordan content.
== Proof ==
Let $C$ be a finite covering of $B$ by closed rectangles.
Since $A \subseteq B$, by Cover is Cover of Subset:
:$C$ is a finite covering of $A$ by closed rectangles.
Therefore, by definition of the outer Jordan content:
:$\ds \map {m^*} A \le \sum_{R \mathop \in C} \map V R$
But, since $C$ was arbitrary:
:$\map {m^*} A$ is a lower bound of $\ds \set {\sum_{R \mathop \in C} \map V R : C}$
where $C$ ranges over all finite coverings of $B$ by closed rectangles.
Hence, by definition of outer Jordan content, and of infimum:
:$\map {m^*} A \le \map {m^*} B$
Category:Outer Jordan Content
Category:Measure Theory
Outer Jordan Content is Subadditive

== Theorem ==
Let $A, B \subseteq \R^n$ be bounded subspaces of Euclidean $n$-space.
Then:
:$\map {m^*} {A \cup B} \le \map {m^*} A + \map {m^*} B$
where $m^*$ denotes the outer Jordan content.
== Proof ==
Let $\epsilon > 0$ be arbitrary.
By Characterizing Property of Infimum of Subset of Real Numbers, select:
:$C$ to be a finite covering of $A$ by closed $n$-rectangles such that:
::$\ds \sum_{R \mathop \in C} \map V R < \map {m^*} A + \frac \epsilon 2$
:$D$ to be a finite covering of $B$ by closed $n$-rectangles such that:
::$\ds \sum_{R \mathop \in D} \map V R < \map {m^*} B + \frac \epsilon 2$
Define $E = C \cup D$.
By Union of Finite Sets is Finite, $E$ is a finite set of closed $n$-rectangles.
So, by Union of Covers is Cover of Union:
:$C \cup D$ is a finite covering of $A \cup B$ by closed $n$-rectangles.
We have:
As $\epsilon > 0$ was arbitrary, it follows from Real Plus Epsilon that:
:$\map {m^*} {A \cup B} \le \map {m^*} A + \map {m^*} B$
Category:Outer Jordan Content
Outer Jordan Content of Dilation

== Theorem ==
Let $M \subseteq \R^n$ be a bounded subspace of Euclidean $n$-space.
Let $c_1, c_2, \dotsc, c_n \in \R_{\ge 0}$ be non-negative real numbers.
Let $M' \subseteq \R^n$ be defined as:
:$M' = \set {\tuple {c_1 x_1, c_2 x_2, \dotsc, c_n x_n} : \tuple {x_1, x_2, \dotsc, x_n} \in \R^n}$
Then:
:$\map {m^*} {M'} = c_1 c_2 \dotsm c_n \map {m^*} M$
where $m^*$ denotes the outer Jordan content.
== Proof ==
Let $\epsilon > 0$ be arbitrary.
By Characterizing Property of Infimum of Subset of Real Numbers, let $C$ be a finite covering of $M$ by closed $n$-rectangles such that:
:$\ds \sum_{R \mathop \in C} \map V C < \map {m^*} M + \frac \epsilon {c_1 c_2 \dotsm c_n + 1}$
Let $C'$ be defined as:
:$\ds C' = \set {\closedint {c_1 a_1} {c_1 b_1} \times \dotso \times \closedint {c_n a_n} {c_n b_n} : \closedint {a_1} {b_1} \times \dotso \times \closedint {a_n} {b_n} \in C}$
For any $\tuple {c_1 x_1, \dotsc, c_n x_n} \in M'$, there must be some:
:$R = \closedint {a_1} {b_1} \times \dotso \times \closedint {a_n} {b_n} \in C$
such that:
:$\tuple {x_1, \dotsc, x_n} \in R$
by definition of a covering.
That is:
Therefore, by Real Number Ordering is Compatible with Multiplication:
Or, in other words:
:$\tuple {c_1 x_1, \dotsc, c_n x_n} \in \closedint {c_1 a_1} {c_1 b_1} \times \dotso \times \closedint {c_n a_n} {c_n b_n} \in C'$
Hence:
:$C'$ is a finite covering of $M'$ by closed $n$-rectangles.
Therefore:
As $\epsilon > 0$ was arbitrary, it follows from Real Plus Epsilon that:
:$\map {m^*} {M'} \le c_1 c_2 \dotsm c_n \map {m^*} M$
Now, if every $c_i > 0$, we have that:
:$M = \set {\tuple {\frac 1 {c_1} x_1, \dotsc, \frac 1 {c_n} x_n} : \tuple {x_1, \dotsc, x_n} \in M'}$
Therefore, the above argument can be applied with the roles of $M$ and $M'$ interchanged to prove that:
:$\map {m^*} M \le \frac 1 {c_1 c_2 \dotsm c_n} \map {m^*} {M'}$
which is to say:
:$c_1 c_2 \dotsm c_n \map {m^*} M \le \map {m^*} {M'}$
If the above does not hold, then $c_i = 0$ for some $i \in \set {1, 2, \dotsc, n}$.
Then:
:$c_1 c_2 \dotsm c_n \map {m^*} M = 0 \le \map {m^*} {M'}$
In either case, we have:
:$c_1 c_2 \dotsm c_n \map {m^*} M \le \map {m^*} {M'}$
The result follows from the two inequalities.
Category:Outer Jordan Content