avewright/proofwiki-math huggingface dataset:
Theorem: Null Space Closed under Vector Addition

Let:
:$\map {\mathrm N} {\mathbf A} = \set {\mathbf x \in \R^n : \mathbf A \mathbf x = \mathbf 0}$
be the null space of $\mathbf A$, where:
:<nowiki>$\mathbf A_{m \times n} = \begin {bmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m1} & a_{m2} & \cdots & a_{mn} \\
\end{bmatrix}$</nowiki>
:$\mathbf x_{n \times 1} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}$
:$\mathbf 0_{m \times 1} = \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix}$
are matrices
:the column matrix $\mathbf x_{n \times 1}$ is interpreted as a vector in the real Euclidean space $\R^n$.
Then $\map {\mathrm N} {\mathbf A}$ is closed under vector addition:
:$\forall \mathbf v, \mathbf w \in \map {\mathrm N} {\mathbf A}: \mathbf v + \mathbf w \in \map {\mathrm N} {\mathbf A}$

Proof:
Let $\mathbf v, \mathbf w \in \map {\mathrm N} {\mathbf A}$.
By the definition of null space:
{{begin-eqn}}
{{eqn | l = \mathbf A \mathbf v
      | r = \mathbf 0
}}
{{eqn | l = \mathbf A \mathbf w
      | r = \mathbf 0
}}
{{end-eqn}}
Next, observe that:
{{begin-eqn}}
{{eqn | l = \mathbf A \paren {\mathbf v + \mathbf w}
      | r = \mathbf A \mathbf v + \mathbf A \mathbf w
      | c = Matrix Multiplication Distributes over Matrix Addition
}}
{{eqn | r = \mathbf 0 + \mathbf 0
}}
{{eqn | r = \mathbf 0
}}
{{end-eqn}}
The order is correct, by hypothesis.
Hence the result, by the definition of null space.
{{qed}}
Theorem: Null Space Closed under Scalar Multiplication

Let:
:$\map {\mathrm N} {\mathbf A} = \set {\mathbf x \in \R^n : \mathbf {A x} = \mathbf 0}$
be the null space of $\mathbf A$, where:
:<nowiki>$\mathbf A_{m \times n} = \begin{bmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m1} & a_{m2} & \cdots & a_{mn} \\
\end {bmatrix}$</nowiki>
:$\mathbf x_{n \times 1} = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix}$
:$\mathbf 0_{m \times 1} = \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix}$
are matrices where each column is an element of a real vector space.
Then $\map {\mathrm N} {\mathbf A}$ is closed under scalar multiplication:
:$\forall \mathbf v \in \map {\mathrm N} {\mathbf A} ,\forall \lambda \in \R: \lambda \mathbf v \in \map {\mathrm N} {\mathbf A}$

Proof:
Let $\mathbf v \in \map {\mathrm N} {\mathbf A}$, $\lambda \in \R$.
By the definition of null space:
{{begin-eqn}}
{{eqn | l = \mathbf {A v}
      | r = \mathbf {0}
}}
{{end-eqn}}
Observe that:
{{begin-eqn}}
{{eqn | l = \mathbf A \paren {\lambda \mathbf v}
      | r = \lambda \paren {\mathbf {A v} }
      | c = Matrix Multiplication is Homogeneous of Degree $1$
}}
{{eqn | r = \lambda \mathbf 0
      | c = 
}}
{{eqn | r = \mathbf 0
}}
{{end-eqn}}
Hence the result, by the definition of null space.
{{qed}}
Theorem: Null Space is Subspace

Let:
:$\map {\mathrm N} {\mathbf A} = \set {\mathbf x \in \R^n: \mathbf {A x} = \mathbf 0}$
be the null space of $\mathbf A$, where:
:<nowiki>$\mathbf A_{m \times n} = \begin {bmatrix}
a_{11} & a_{12} & \cdots & a_{1n} \\
a_{21} & a_{22} & \cdots & a_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
a_{m1} & a_{m2} & \cdots & a_{mn} \\
\end {bmatrix}$</nowiki>
:$\mathbf x_{n \times 1} = \begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix}$
:$\mathbf 0_{m \times 1} = \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix}$
are matrices.
Then $\map {\mathrm N} {\mathbf A}$ is a linear subspace of $\R^n$.

Proof:
$\map {\mathrm N} {\mathbf A} \subseteq \R^n$, by construction.
We have:
:$\mathbf 0 \in \map {\mathrm N} {\mathbf A}$, from Null Space Contains Zero Vector
:$\forall \mathbf v, \mathbf w \in \map {\mathrm N} {\mathbf A}: \mathbf v + \mathbf w \in \map {\mathrm N} {\mathbf A}$, from Null Space Closed under Vector Addition
:$\forall \mathbf v \in \map {\mathrm N} {\mathbf A}, \lambda \in \R: \lambda \mathbf v \in \map {\mathrm N} {\mathbf A}$, from Null Space Closed under Scalar Multiplication
The result follows from Vector Subspace of Real Vector Space.
{{qed}}
Theorem: Norm is Continuous

Let $\struct {V, \norm {\,\cdot\,} }$ be a normed vector space.
Then the mapping $x \mapsto \norm x$ is continuous.
Here, the metric used is the metric $d$ induced by $\norm {\,\cdot\,}$.

Proof:
Since $\norm x = \map d {x, \mathbf 0}$, the result follows directly from Distance Function of Metric Space is Continuous.
{{qed}}
Theorem: P-adic Valuation of Rational Number is Well Defined

The $p$-adic valuation:
:$\nu_p: \Q \to \Z \cup \set {+\infty}$
is well defined.

Proof:
Let $\dfrac a b = \dfrac c d \in \Q$.
Thus:
:$a d = b c \in \Z$
By  definition of rational number:
:$b, d \ne 0$
By definition of restricted $p$-adic valuation:
:$\map {\nu_p^\Z} b, \map {\nu_p^\Z} d < +\infty$
Theorem: P-adic Valuation is Valuation

The $p$-adic valuation $\nu_p: \Q \to \Z \cup \set {+\infty}$ is a valuation on $\Q$.

Proof:
To prove that $\nu_p$ is a valuation it is necessary to demonstrate:
{{begin-axiom}}
{{axiom | n = \text V 1
        | q = \forall q, r \in \Q
        | ml= \map {\nu_p} {q r}
        | mo= =
        | mr= \map {\nu_p} q + \map {\nu_p} r
}}
{{axiom | n = \text V 2
        | q = \forall q \in \Q
        | ml= \map {\nu_p} q = +\infty
        | mo= \iff
        | mr= q = 0
}}
{{axiom | n = \text V 3
        | q = \forall q, r \in \Q
        | ml= \map {\nu_p} {q + r}
        | mo= \ge
        | mr= \min \set {\map {\nu_p} q, \map {\nu_p} r}
}}
{{end-axiom}}
Let $q := \dfrac a b, r := \dfrac c d \in \Q$.
Theorem: P-adic Norm not Complete on Rational Numbers

Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.
Then:
:the valued field $\struct {\Q, \norm {\,\cdot\,}_p}$ is not complete.
That is, there exists a Cauchy sequence in $\struct {\Q, \norm{\,\cdot\,}_p}$ which does not converge to a limit in $\Q$.

Proof:
=== Case: $p \gt 3$ ===
{{:P-adic Norm not Complete on Rational Numbers/Proof 1/Case 1}}{{qed|lemma}}
=== Case: $p = 2$ or $3$ ===
{{:P-adic Norm not Complete on Rational Numbers/Proof 1/Case 2}}{{qed}}
Theorem: P-adic Norm not Complete on Rational Numbers

Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.
Then:
:the valued field $\struct {\Q, \norm {\,\cdot\,}_p}$ is not complete.
That is, there exists a Cauchy sequence in $\struct {\Q, \norm{\,\cdot\,}_p}$ which does not converge to a limit in $\Q$.

Proof:
Hensel's Lemma is used to prove the existence of a Cauchy sequence that does not converge.
==== Lemma 1 ====
{{:P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 1}}{{qed|lemma}}
Let $x_1 \in \Z_{>0}: p \nmid x_1, x_1 \ge \dfrac {p + 1} 2$
Let $k$ be a positive integer such that $k \ge 2, p \nmid k$.
Let $a = x_1^k + p$.
==== Lemma 2 ====
{{:P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 2}}{{qed|lemma}}
Let $\map f X \in \Z \sqbrk X$ be the polynomial:
:$X^k - a$
==== Lemma 3 ====
{{:P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 3}}{{qed|lemma}}
Let $\map {f'} X \in \Z \sqbrk X$ be the formal derivative of $\map f X$.
==== Lemma 4 ====
{{:P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 4}}{{qed|lemma}}
From Hensel's Lemma there exists a sequence of integers $\sequence {x_n}$ such that:
:$(1) \quad \forall n: \map f {x_n} \equiv 0 \pmod {p^n}$
:$(2) \quad \forall n: x_{n + 1} \equiv x_n \pmod {p^n}$
==== Lemma 5 ====
{{:P-adic Norm not Complete on Rational Numbers/Proof 2/Lemma 5}}{{qed|lemma}}
From Sequence of Consecutive Integers Modulo Power of p is Cauchy in P-adic Norm then:
:$\sequence {x_n}$ is a Cauchy sequence in $\struct {\Q, \norm {\,\cdot\,}_p}$
{{AimForCont}} $\sequence {x_n}$ is a sequence such that for some $c \in \Q$:
:$\ds \lim_{n \mathop \to \infty} x_n = c$
in $\struct {\Q, \norm {\,\cdot\,}_p}$
From Product Rule for Sequences in Normed Division Ring then:
:$\ds \lim_{n \mathop \to \infty} x_n^k = c^k$
Hence:
:$c^k = a \in \Z$.
From Nth Root of Integer is Integer or Irrational then:
:$c \in \Z$
This contradicts Lemma 2.
So the sequence $\sequence {x_n}$ does not converge in $\struct {\Q, \norm{\,\cdot\,}_p}$.
The result follows.
{{qed}}
Theorem: P-adic Norm not Complete on Rational Numbers

Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for some prime $p$.
Then:
:the valued field $\struct {\Q, \norm {\,\cdot\,}_p}$ is not complete.
That is, there exists a Cauchy sequence in $\struct {\Q, \norm{\,\cdot\,}_p}$ which does not converge to a limit in $\Q$.

Proof:
By Rational Numbers are Countably Infinite, the set of rational numbers is countably infinite.
By $p$-adic Numbers are Uncountable, the set of $p$-adic numbers $\Q_p$ is uncountably infinite.
Let $\CC$ be the commutative ring of Cauchy sequences over $\struct {\Q, \norm {\,\cdot\,}_p}$.
Let $\NN$ be the set of null sequences in $\struct {\Q, \norm {\,\cdot\,}_p}$.
The $p$-adic numbers $\Q_p$ is the quotient ring $\CC \, \big / \NN$ by definition.
By Embedding Division Ring into Quotient Ring of Cauchy Sequences, the mapping $\phi: \Q \to \Q_p$ defined by:
:$\map \phi r = \sequence {r, r, r, \dotsc} + \NN$
where $\sequence {r, r, r, \dotsc} + \NN$ is the left coset in $\CC \, \big / \NN$ that contains the constant sequence $\sequence {r, r, r, \dotsc}$, is a distance-preserving monomorphism.
By {{Corollary|Surjection from Natural Numbers iff Countable|2}} then $\phi$ is not a surjection.
Hence:
:$\exists \sequence {x_n} \in \CC: \sequence {x_n} + \NN \not \in \map \phi \Q$
By Cauchy Sequence Converges Iff Equivalent to Constant Sequence then $\sequence {x_n}$ is not convergent in $\struct {\Q, \norm {\,\cdot\,}_p}$.
The result follows.
{{qed}}
Theorem: Restricted P-adic Valuation is Valuation

Let $\nu_p^\Z: \Z \to \Z \cup \set {+\infty}$ be the $p$-adic valuation restricted to the integers.
Then $\nu_p^\Z$ is a valuation.

Proof:
To prove that $\nu_p^\Z$ is a valuation it is necessary to demonstrate:
{{begin-axiom}}
{{axiom | n = \text V 1
        | q = \forall m, n \in \Z
        | ml= \map {\nu_p^\Z} {m n}
        | mo= =
        | mr= \map {\nu_p^\Z} m + \map {\nu_p^\Z} n
}}
{{axiom | n = \text V 2
        | q = \forall n \in \Z
        | ml= \map {\nu_p^\Z} n = +\infty
        | mo= \iff
        | mr= n = 0
}}
{{axiom | n = \text V 3
        | q = \forall m, n \in \Z
        | ml= \map {\nu_p^\Z} {m + n}
        | mo= \ge
        | mr= \min \set {\map {\nu_p^\Z} m, \map {\nu_p^\Z} n}
}}
{{end-axiom}}
Theorem: Null Measure is Measure

Let $\struct {X, \Sigma}$ be a measurable space.
Then the null measure $\mu$ on $\struct {X, \Sigma}$ is a measure.

Proof:
Let us verify the measure axioms $(1)$, $(2)$ and $(3')$ for $\mu$.
Theorem: Infinite Measure is Measure

Let $\struct {X, \Sigma}$ be a measurable space.
Then the infinite measure $\mu$ on $\struct {X, \Sigma}$ is a measure.

Proof:
Let us verify the measure axioms $(1)$, $(2)$ and $(3')$ for $\mu$.
Theorem: Linear Combination of Measures

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu, \nu$ be measures on $\struct {X, \Sigma}$.
Then for all positive real numbers $a, b \in \R_{\ge 0}$, the pointwise sum:
:$a \mu + b \nu: \Sigma \to \overline \R, \ \map {\paren {a \mu + b \nu} } E := a \map \mu E + b \map \nu E$
is also a measure on $\struct {X, \Sigma}$.
{{improve|Add corollary for a positive linear combination of finitely many measures}}

Proof:
Verifying the axioms $(1)$, $(2)$ and $(3')$ for a measure in turn:
Theorem: Series of Measures is Measure

Let $\struct {X, \Sigma}$ be a measurable space.
Let $\sequence {\mu_n}_{n \mathop \in \N}$ be a sequence of measures on $\left({X, \Sigma}\right)$.
Let $\sequence {a_n}_{n \mathop \in \N} \subseteq \R_{\ge 0}$ be a sequence of positive real numbers.
Then the series of measures $\mu: \Sigma \to \overline \R$, defined by:
:$\ds \map \mu E := \sum_{n \mathop \in \N} a_n \map {\mu_n} E$
is also a measure on $\struct {X, \Sigma}$.

Proof:
Let us verify the conditions for a measure in turn.
Let $E \in \Sigma$.
Then:
:$\forall n \in \N: \map {\mu_n} E \ge 0$
and so:
:$a_n \map {\mu_n} E \ge 0$
Therefore, by Series of Positive Real Numbers has Positive Limit:
:$\ds \map \mu E = \sum_{n \mathop \in \N} a_n \map {\mu_n} E \ge 0$
For every $n \in \N$, also $\map {\mu_n} \O = 0$.
Therefore, it immediately follows that:
:$\ds \map \mu \O = \sum_{n \mathop \in \N} a_n \map {\mu_n} \O = \sum_{n \mathop \in \N} 0 = 0$
Finally, let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma$.
Then:
{{begin-eqn}}
{{eqn | l = \map \mu {\bigcup_{m \mathop \in \N} E_m}
      | r = \sum_{n \mathop \in \N} a_n \map {\mu_n} {\bigcup_{m \mathop \in \N} E_m}
      | c = Definition of $\mu$
}}
{{eqn | r = \sum_{n \mathop \in \N} a_n \paren {\sum_{m \mathop \in \N} \map {\mu_n} {E_m} }
      | c = The $\mu_n$ are measures
}}
{{eqn | r = \sum_{n \mathop \in \N} \sum_{m \mathop \in \N} a_n \map {\mu_n} {E_m}
}}
{{eqn | r = \sum_{m \mathop \in \N} \sum_{n \mathop \in \N} a_n \map {\mu_n} {E_m}
      | c = Tonelli's Theorem: Corollary
}}
{{eqn | r = \sum_{m \mathop \in \N} \map \mu {E_m}
      | c = Definition of $\mu$
}}
{{end-eqn}}
Therefore, having verified all three axioms, $\mu$ is a measure.
{{qed}}
Theorem: Relation Induced by Quotient Set is Equivalence

Let $S$ be a set.
Let $\RR$ be an equivalence relation on $S$.
Let $S / \RR$ be the quotient set of $S$ determined by $\RR$.
Let $\RR'$ be the equivalence relation induced by $S / \RR$ on $S$.
Then $\RR' = \RR$.

Proof:
Let $\RR$ be an equivalence relation on $S$.
Let $\tuple {x, y} \in \RR$.
By definition of equivalence class, $y \in \eqclass x \RR$ and $x \in \eqclass x \RR$.
By definition of quotient set, $x$ and $y$ both belong to the same element of $S / \RR$.
So, by definition of $\RR'$, it follows that $\tuple {x, y} \in \RR'$.
That is:
:$\tuple {x, y} \in \RR \implies \tuple {x, y} \in \RR'$
and so by definition of subset:
:$\RR \subseteq \RR'$
Now let $\tuple {x, y} \in \RR'$.
Then $y$ belongs to the same element of $S / \RR$ that $x$ does.
That is:
:$y \in \eqclass x \RR$
and so $\tuple {x, y} \in \RR$.
That is:
:$\tuple {x, y} \in \RR' \implies \tuple {x, y} \in \RR$
and so by definition of subset:
:$\RR' \subseteq \RR$
The result follows by definition of set equality.
{{qed}}
Theorem: Quotient Set Determined by Relation Induced by Partition is That Partition

Let $S$ be a set.
Let $\PP$ be a partition of $S$.
Let $\RR$ be the equivalence relation induced by $\PP$.
Then the quotient set $S / \RR$ of $S$ is $\PP$ itself.

Proof:
Let $P \subseteq S$ such that $P \in \PP$.
Let $x \in P$.
Then:
{{begin-eqn}}
{{eqn | l = y
      | o = \in
      | r = \eqclass x \RR
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \tuple {x, y}
      | o = \in
      | r = \RR
      | c = {{Defof|Equivalence Class}}
}}
{{eqn | ll= \leadstoandfrom
      | l = y
      | o = \in
      | r = P
      | c = {{Defof|Equivalence Relation Induced by Partition}}
}}
{{end-eqn}}
Therefore:
:$P = \eqclass x \RR$
and so:
:$P \in S / \RR$
and so:
:$\PP \subseteq S / \RR$
Now let $x \in S$.
As $\PP$ is a partition:
:$\exists P \in \PP: x \in P$
Then by definition of $\RR$:
:$\tuple {x, y} \in \RR \iff y \in \eqclass x \RR$
{{begin-eqn}}
{{eqn | l = y
      | o = \in
      | r = P
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \tuple {x, y}
      | o = \in
      | r = \RR
      | c = {{Defof|Equivalence Relation Induced by Partition}}
}}
{{eqn | ll= \leadstoandfrom
      | l = y
      | o = \in
      | r = \eqclass x \RR
      | c = {{Defof|Equivalence Class}}
}}
{{end-eqn}}
Therefore:
:$\eqclass x \RR = P$
and so:
:$\eqclass x \RR \in \PP$
That is:
:$\SS / \RR \subseteq \PP$
It follows by definition of set equality that:
:$\SS / \RR = P$
Hence the result.
{{qed}}
Theorem: Null Sets Closed under Subset

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $N \in \Sigma$ be a $\mu$-null set, and let $M \in \Sigma$ be a subset of $N$.
Then $M$ is also a $\mu$-null set.

Proof:
As $\mu$ is a measure, $\map \mu M \ge 0$.
Also, by Measure is Monotone, $\map \mu M \le \map \mu N = 0$.
Hence $\map \mu M = 0$, and $M$ is a $\mu$-null set.
{{qed}}
Theorem: Null Sets Closed under Subset

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $N \in \Sigma$ be a $\mu$-null set, and let $M \in \Sigma$ be a subset of $N$.
Then $M$ is also a $\mu$-null set.

Proof:
For each $x, y \in \HH$, define $\EE_{x, y} : \map \BB X \to \C$ by:
:$\map {\EE_{x, y} } A = \innerprod {\map \EE A x} y$
for each $A \in \map \BB X$.
We then have:
:$\map {\EE_{x, y} } {E'} = 0$
for each $x, y \in \HH$.
From Null Sets Closed under Subset, we have $\map {\EE_{x, y} } E = 0$ for each $x, y \in \HH$.
That is, $\innerprod {\map \EE E x} y = 0$ for all $x, y \in \HH$.
From Linear Subspace Dense iff Zero Orthocomplement, we have $\HH^\bot = \set { {\mathbf 0}_\HH}$ and hence $\map \EE E x = {\mathbf 0}_\HH$ for each $x \in \HH$.
Hence $\map \EE E = {\mathbf 0}_{\map B \HH}$.
{{qed}}
Theorem: Null Sets Closed under Subset

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $N \in \Sigma$ be a $\mu$-null set, and let $M \in \Sigma$ be a subset of $N$.
Then $M$ is also a $\mu$-null set.

Proof:
Let $\le_{\map B \HH}$ be the canonical preordering on $\map B \HH$.
From Measure is Monotone: Resolution of the Identity, we have:
:$\map \EE E \le_{\map B \HH} \map \EE {E'} = {\mathbf 0}_{\map B \HH}$
On the other hand from Bounds on Projection in Unital C*-Algebra, we have:
:${\mathbf 0}_{\map B \HH} \le_{\map B \HH} \map \EE E$
From Canonical Preordering of C*-Algebra is Antisymmetric, we then have:
:$\map \EE E = {\mathbf 0}_{\map B \HH}$
{{qed}}
Theorem: Null Sets Closed under Countable Union

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\sequence {N_n}_{n \mathop \in \N}$ be a sequence of $\mu$-null sets.
Then $N := \ds \bigcup_{n \mathop \in \N} N_n$ is also a $\mu$-null set.

Proof:
As $\mu$ is a measure:
:$\map \mu N \ge 0$
Also:
{{begin-eqn}}
{{eqn | l = \map \mu N
      | o = \le
      | r = \sum_{n \mathop \in \N} \map \mu {N_n}
      | c = Measure is Countably Subadditive
}}
{{eqn | r = \sum_{n \mathop \in \N} 0
      | c = as the $N_n$ are $\mu$-null sets
}}
{{eqn | r = 0
}}
{{end-eqn}}
Hence necessarily:
:$\map \mu N = 0$
and $N$ is a $\mu$-null set.
{{qed}}
Theorem: Absolute Value induces Equivalence Compatible with Integer Multiplication

Let $\Z$ be the set of integers.
Let $\RR$ be the relation on $\Z$ defined as:
:$\forall x, y \in \Z: \struct {x, y} \in \RR \iff \size x = \size y$
where $\size x$ denotes the absolute value of $x$.
Then $\RR$ is a congruence relation for integer multiplication.

Proof:
From Absolute Value Function on Integers induces Equivalence Relation, $\RR$ is an equivalence relation.
Let:
:$\size {x_1} = \size {x_2}$
:$\size {y_1} = \size {y_2}$
Then by Absolute Value Function is Completely Multiplicative:
{{begin-eqn}}
{{eqn | l = \size {x_1 y_1}
      | r = \size {x_1} \size {y_1}
      | c = 
}}
{{eqn | r = \size {x_2} \size {y_2}
      | c = 
}}
{{eqn | r = \size {x_2 y_2}
      | c = 
}}
{{end-eqn}}
That is:
:$\paren {x_1 y_1, x_2 y_2} \in \RR$
That is, $\RR$ is a congruence relation for integer multiplication.
{{qed}}
Theorem: Completion Theorem (Measure Space)

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Then there exists a completion $\struct {X, \Sigma^*, \bar \mu}$ of $\struct {X, \Sigma, \mu}$.

Proof:
We give an explicit construction of $\struct {X, \Sigma^*, \bar \mu}$.
To this end, define $\NN$ to be the collection of subsets of $\mu$-null sets:
:$\NN := \set {N \subseteq X: \exists M \in \Sigma: \map \mu M = 0, N \subseteq M}$
Now, we define:
:$\Sigma^* := \set {E \cup N: E \in \Sigma, N \in \NN}$
and assert $\Sigma^*$ is a $\sigma$-algebra.
By Empty Set is Null Set, $\O \in \NN$, and thus by Union with Empty Set:
:$\forall E \in \Sigma: E \cup \O = E \in \Sigma^*$
that is to say, $\Sigma \subseteq \Sigma^*$.
As a consequence, $X \in \Sigma^*$.
Now, suppose that $E \cup N \in \Sigma^*$, and $N \subseteq M, M \in \Sigma$. Then:
{{begin-eqn}}
{{eqn | l = X \setminus \paren {E \cup N}
      | r = \paren {X \setminus E} \cap \paren {X \setminus N}
      | c = De Morgan's Laws: Difference with Union
}}
{{eqn | r = \paren {X \setminus E} \cap \paren {\paren {X \setminus M} \cup \paren {M \setminus N} }
      | c = Union of Relative Complements of Nested Subsets
}}
{{eqn | r = \paren {\paren {X \setminus E} \cap \paren {X \setminus M} } \cup \paren {\paren {X \setminus E} \cap \paren {M \setminus N} }
      | c = Intersection Distributes over Union
}}
{{eqn|o =}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \paren {\paren {X \setminus E} \cap \paren {X \setminus M} }
      | o = \in
      | r = \Sigma
      | c = $E, M \in \Sigma$, Sigma-Algebra Closed under Intersection
}}
{{eqn | l = \paren {X \setminus E} \cap \paren {M \setminus N}
      | o = \subseteq
      | r = M
      | c = Set Difference is Subset, Set Intersection Preserves Subsets
}}
{{eqn | ll= \leadsto
      | l = X \setminus \paren {E \cup N}
      | o = \in
      | r = \Sigma^*
}}
{{end-eqn}}
Finally, let $\sequence {E_n}_{n \mathop \in \N}$ and $\sequence {N_n}_{n \mathop \in \N}$ be sequences in $\Sigma$ and $\NN$, respectively.
Let $\sequence {M_n}_{n \mathop \in \N}$ be a sequence of $\mu$-null sets such that:
:$\forall n \in \N: N_n \subseteq M_n$
Then, compute:
{{begin-eqn}}
{{eqn | l = \bigcup_{n \mathop \in \N} \paren {E_n \cup N_n}
      | r = \paren {\bigcup_{n \mathop \in \N} E_n} \cup \paren {\bigcup_{n \mathop \in \N} N_n}
      | c = Set Union is Self-Distributive: Families of Sets
}}
{{eqn | l = \bigcup_{n \mathop \in \N} N_n
      | o = \subseteq
      | r = \bigcup_{n \mathop \in \N} M_n
      | c = Set Union Preserves Subsets
}}
{{end-eqn}}
From Null Sets Closed under Countable Union, also:
:$\ds \map \mu {\bigcup_{n \mathop \in \N} M_n} = 0$
hence it follows that:
:$\ds \bigcup_{n \mathop \in \N} N_n \in \NN$
Next, as $\Sigma$ is a $\sigma$-algebra, it follows that:
:$\ds \bigcup_{n \mathop \in \N} E_n \in \Sigma$
and finally, we conclude:
:$\ds \bigcup_{n \mathop \in \N} \paren {E_n \cup N_n} \in \Sigma^*$
Therefore, we have shown that $\Sigma^*$ is a $\sigma$-algebra.
Next, define $\bar \mu: \Sigma^* \to \overline{\R}_{\ge 0}$ by:
:$\map {\bar \mu} {E \cup N} := \map \mu E$
It needs verification that this well-defines $\bar \mu$.
Theorem: Restricted Measure is Measure

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\Sigma'$ be a sub-$\sigma$-algebra of $\Sigma$.
Then the restricted measure $\mu \restriction_{\Sigma'}$ is a measure on the measurable space $\struct {X, \Sigma'}$.

Proof:
Verify the axioms for a measure in turn for $\mu \restriction_{\Sigma'}$:
Theorem: Restricting Measure Preserves Finiteness

Let $\left({X, \Sigma, \mu}\right)$ be a measure space.
Let $\mu$ be a finite measure.
Let $\Sigma'$ be a sub-$\sigma$-algebra of $\Sigma$.
Then the restricted measure $\mu \restriction_{\Sigma'}$ is also a finite measure.

Proof:
By Restricted Measure is Measure, $\mu \restriction_{\Sigma'}$ is a measure.
Now by definition of $\mu \restriction_{\Sigma'}$, have:
:$\mu \restriction_{\Sigma'} \left({X}\right) = \mu \left({X}\right) < \infty$
as $\mu$ is a finite measure.
Hence $\mu \restriction_{\Sigma'}$ is also a finite measure.
{{qed}}
Theorem: Measure Space has Exhausting Sequence of Finite Measure iff Cover by Sets of Finite Measure

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Then there exists an exhausting sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ such that:
:$\forall n \in \N: \map \mu {E_n} < \infty$
{{iff}} there exists a sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ such that:
:$(1): \quad \ds \bigcup_{n \mathop \in \N} E_n = X$
:$(2): \quad \forall n \in \N: \map \mu {E_n} < +\infty$

Proof:
=== Necessary Condition ===
Suppose that there exists an exhausting sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ such that:
::$\forall n \in \N: \map \mu {E_n} < \infty$
Let $\sequence {F_n}_{n \mathop \in \N}$ be an exhausting sequence in $\Sigma$ such that:
:$\forall n \in \N: \map \mu {F_n} < +\infty$
Then as $\sequence {F_n}_{n \mathop \in \N}$ is exhausting, have:
:$\ds \bigcup_{n \mathop \in \N} F_n = X$
It follows that the sequence $\sequence {F_n}_{n \mathop \in \N}$ satisfies $(1)$ and $(2)$.
{{qed|lemma}}