hcju/proofwikiqa huggingface dataset:
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}}
$\operatorname{N} \left({\mathbf A}\right) \subseteq \R^n$, by construction. We have: * $\mathbf 0 \in \operatorname{N} \left({\mathbf A}\right)$, from Null Space Contains Zero Vector * $\forall \mathbf v, \mathbf w \in \operatorname{N} \left({\mathbf A}\right): \mathbf v + \mathbf w \in \operatorname{N} \left({\mathbf A}\right)$, from Null Space Closed under Vector Addition * $\forall\mathbf v \in \operatorname{N} \left({\mathbf A}\right), \lambda \in \R: \lambda \mathbf v \in \operatorname{N} \left({\mathbf A}\right)$, from Null Space Closed under Scalar Multiplication The result follows from Vector Subspace of Real Vector Space. {{qed}}
Since $\norm x = \map d {x, \mathbf 0}$, the result follows directly from Distance Function of Metric Space is Continuous. {{qed}} Category:Norm Theory Category:Continuous Mappings nwsfsr3ucq08u3e6eq3x8ugflydugc2
Let $\dfrac a b = \dfrac c d \in \Q$. Thus: :$a d = b c \in \Z$ Then: {{begin-eqn}} {{eqn | l = \nu_p^\Z \left({a}\right) + \nu_p^\Z \left({d}\right)       | r = \nu_p^\Z \left({a d}\right)       | c = P-adic Valuation on Integers is Valuation: Axiom $V1$ }} {{eqn | r = \nu_p^\Z \left({b c}\right) }} {{eqn | r = \nu_p^\Z \left({c}\right) + \nu_p^\Z \left({b}\right)       | c = P-adic Valuation on Integers is Valuation: Axiom $V1$ }} {{eqn | n = 1       | ll= \leadsto       | l = \nu_p^\Z \left({a}\right) - \nu_p^\Z \left({b}\right)       | r = \nu_p^\Z \left({c}\right) - \nu_p^\Z \left({d}\right) }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = \nu_p^\Q \left({\frac a b}\right)       | r = \nu_p^\Z \left({a}\right) - \nu_p^\Z \left({b}\right)       | c = {{Defof|P-adic Valuation|$p$-adic Valuation}} }} {{eqn | r = \nu_p^\Z \left({c}\right) - \nu_p^\Z \left({d}\right)       | c = from $(1)$ }} {{eqn | r = \nu_p^\Q \left({\dfrac c d}\right)       | c = {{Defof|P-adic Valuation|$p$-adic Valuation}} }} {{end-eqn}} Thus, by definition, $\nu_p: \Q \to \Z \cup \left\{{+\infty}\right\}$ is well defined. {{qed}} Category:P-adic Valuations jl2hw7civs0ob7rw2jf4ru71dlgotmh
=== 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}}
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}} 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}} 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}} 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}} By 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}} By corollary of Characterisation of Cauchy Sequence in Non-Archimedean 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 $c \in \Q$: :$\displaystyle \lim_{n \mathop \to \infty} x_n = c$ in $\struct {\Q, \norm {\,\cdot\,}_p}$ By product rule for convergent sequences then: :$\displaystyle \lim_{n \mathop \to \infty} x_n^k = c^k$ Hence: :$c^k = a \in \Z$. By 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}}
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 to Surjection from Natural Numbers iff Countable 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}}
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}} === Axiom $(\text V 1)$ === Let $m, n \in \Z$. Let $m = 0$ or $n = 0$. Then: :$\map {\nu_p^\Z} m = +\infty$ or: :$\map {\nu_p^\Z} n = +\infty$ Also: :$n m = 0$ and hence: {{begin-eqn}} {{eqn | l = \map {\nu_p^\Z} {n m}       | r = +\infty       | c =  }} {{eqn | r = \map {\nu_p^\Z} n + \map {\nu_p^\Z} m       | c =  }} {{end-eqn}} Let $n m \ne 0$. Then by definition of the restricted $p$-adic valuation: :$p^{\map {\nu_p^\Z} n} \divides n$ :$p^{\map {\nu_p^\Z} n + 1} \nmid n$ Also: :$p^{\map {\nu_p^\Z} m} \divides m$ :$p^{\map {\nu_p^\Z} m + 1} \nmid m$ Hence: :$p^{\map {\nu_p^\Z} n + \map {\nu_p^\Z} m} \divides n m$ :$p^{\map {\nu_p^\Z} n + \map {\nu_p^\Z} m + 1} \nmid n m$ So: :$\map {\nu_p^\Z} {n m} = \map {\nu_p^\Z} n + \map {\nu_p^\Z} m$ {{qed|lemma}} === Axiom $(\text V 2)$ === By definition of the restricted $p$-adic valuation: :$\forall n \in \Z: \map {\nu_p^\Z} n = +\infty \iff n = 0$ {{qed|lemma}} === Axiom $(\text V 3)$ === Let $m, n \in \Z$. {{WLOG}} let: :$p^\alpha \divides n$ :$p^\beta \divides m$ where $\alpha \ge \beta$. Then $\exists t \in \Z, k \in \Z$ such that: {{begin-eqn}} {{eqn | l = n + m       | r = p^\alpha k + p^\beta t }} {{eqn | r = p^\beta \paren {p^{\alpha - \beta} k + t} }} {{end-eqn}} Thus: :$p^\beta \divides \paren {m + n}$ {{qed|lemma}} Hence by the definition of $\nu_p^\Z$: :$\map {\nu_p^\Z} {m + n} \ge \min \set {\map {\nu_p^\Z} m, \map {\nu_p^\Z} n}$ {{qed}} Category:P-adic Valuations j80iwlsbxd05fx5n9kq4c749emu0pk9
Let us verify the measure axioms $(1)$, $(2)$ and $(3')$ for $\mu$. === Proof of $(1)$ === Let $S \in \Sigma$. Then $\map \mu S = 0 \ge 0$. {{qed|lemma}} === Proof of $(2)$ === It is to be shown that (for a sequence $\sequence {S_n}_{n \in \N}$ of pairwise disjoint sets): :$\displaystyle \sum_{n \mathop = 1}^\infty \map \mu {S_n} = \map \mu {\bigcup_{n \mathop = 1}^\infty S_n}$ Now by definition of $\mu$: :$\displaystyle \map \mu {S_n} = \map \mu {\bigcup_{n \mathop = 1}^\infty S_n} = 0$ Thus, the desired equation becomes: :$\displaystyle \sum_{n \mathop = 1}^\infty 0 = 0$ which trivially holds. {{qed|lemma}} === Proof of $(3')$ === Note that $\O \in \Sigma$ as $\Sigma$ is a $\sigma$-algebra. Hence $\map \mu \O = 0$. {{qed|lemma}} The axioms are fulfilled, and it follows that $\mu$ is a measure. {{qed}} Category:Measure Theory itkkj40p9yi6ytnllp3v6fvmhkx5o4f
Let us verify the measure axioms $(1)$, $(2)$ and $(3')$ for $\mu$. === Proof of $(3')$ === We have by definition $\map \mu \O = 0$. {{qed|lemma}} === Proof of $(1)$ === Let $S \in \Sigma$. Then: :$\map \mu S = + \infty > 0$ for $S \ne \O$ :$\map \mu S = 0 \ge 0$ for $S = \O$ So $\map \mu S \ge 0$ for all $S \in \Sigma$. {{qed|lemma}} === Proof of $(2)$ === It is to be shown that (for a sequence $\sequence {S_n}_{n \in \N}$ of pairwise disjoint sets): :$\displaystyle \sum_{n \mathop = 1}^\infty \map \mu {S_n} = \map \mu {\bigcup_{n \mathop = 1}^\infty S_n}$ Suppose $S_n = \O$ for all $n \in \N$. Then by definition of $\mu$: :$\displaystyle \map \mu {S_n} = \map \mu {\bigcup_{n \mathop = 1}^\infty S_n} = \map \mu \O = 0$ Thus, the desired equation becomes: :$\displaystyle \sum_{n \mathop = 1}^\infty 0 = 0$ which trivially holds. Suppose $S_n \ne \O$ for some $n \in \N$. Then $S_n \subseteq \displaystyle \bigcup_{n \mathop = 1}^\infty S_n \ne \O$. :$\displaystyle \map \mu {S_n} = \map \mu {\bigcup_{n \mathop = 1}^\infty S_n} = + \infty$ :$\displaystyle \sum_{n \mathop = 1}^\infty \map \mu {S_n} = \map \mu {S_n} = + \infty = \map \mu {\bigcup_{n \mathop = 1}^\infty S_n}$ Thus $\displaystyle \sum_{n \mathop = 1}^\infty \map \mu {S_n} = \map \mu {\bigcup_{n \mathop = 1}^\infty S_n}$. {{qed|lemma}} The axioms are fulfilled, and it follows that $\mu$ is a measure. {{qed}} Category:Measure Theory tb9f5zr4v95oh5kjlohqra29tgmz5ny
Verifying the axioms $(1)$, $(2)$ and $(3')$ for a measure in turn: === Axiom $(1)$ === The statement of axiom $(1)$ for $a \mu + b \nu$ is: :$\forall E \in \Sigma: \left({a \mu + b \nu}\right) \left({E}\right) \ge 0$ Let $E \in \Sigma$. Then $\mu \left({E}\right), \nu \left({E}\right) \ge 0$ as $\mu$ and $\nu$ are measures. Hence, $a \mu \left({E}\right) \ge 0$ as $a \ge 0$. Also, $b \nu \left({E}\right) \ge 0$ since $b \ge 0$. Therefore it follows that: :$a \mu \left({E}\right) + b \nu \left({E}\right) \ge 0$ as desired. {{qed|lemma}} === Axiom $(2)$ === Let $\left({E_n}\right)_{n \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma$. The statement of axiom $(2)$ for $a \mu + b \nu$ is: :$\displaystyle \left({a \mu + b \nu}\right) \left({\bigcup_{n \mathop \in \N} E_n}\right) = \sum_{n \mathop \in \N} \left({a \mu + b \nu}\right) \left({E_n}\right)$ So let us do a direct computation: {{begin-eqn}} {{eqn | l = \left({a \mu + b \nu}\right) \left({\bigcup_{n \mathop \in \N} E_n}\right)       | r = a \mu \left({\bigcup_{n \mathop \in \N} E_n}\right) + b \nu \left({\bigcup_{n \mathop \in \N} E_n}\right)       | c = {{Defof|Pointwise Addition}} }} {{eqn | r = a \sum_{n \mathop \in \N} \mu \left({E_n}\right) + b \sum_{n \mathop \in \N} \nu \left({E_n}\right)       | c = $\mu$ and $\nu$ are measures and satisfy $(2)$ }} {{eqn | r = \sum_{n \mathop \in \N} a \mu \left({E_n}\right) + b \nu \left({E_n}\right)       | c = Combined Sum Rule for Real Sequences }} {{eqn | r = \sum_{n \mathop \in \N} \left({a \mu + b \nu}\right) \left({E_n}\right) }} {{end-eqn}} which establishes $a \mu + b \nu$ satisfies $(2)$. {{qed|lemma}} {{handwaving|glosses over $+\infty$ intricacies, but that can be dealt with more generally for monotone sequences in $\overline{\R}$}} === Axiom $(3')$ === The statement of axiom $(3')$ for $a \mu + b \nu$ is: :$\left({a \mu + b \nu}\right) \left({\varnothing}\right) = 0$ This is verified by the following: {{begin-eqn}} {{eqn | l = \left({a \mu + b \nu}\right) \left({\varnothing}\right)       | r = a \mu \left({\varnothing}\right) + b \nu \left({\varnothing}\right)       | c = {{Defof|Pointwise Addition}} }} {{eqn | r = a \cdot 0 + b \cdot 0       | c = $\mu$ and $\nu$ are measures and satisfy $(3')$ }} {{eqn | r = 0 }} {{end-eqn}} Thus, $a \mu + b \nu$ satisfies $(3')$. {{qed|lemma}} Having verified an appropriate set of axioms, it follows that $a \mu + b \nu$ is a measure. {{qed}}
Let us verify the conditions for a measure in turn. Let $E \in \Sigma$. Then for all $n \in \N$, $\mu_n \left({E}\right) \ge 0$ and so $a_n \mu_n \left({E}\right) \ge 0$. Therefore, by Series of Positive Real Numbers has Positive Limit: :$\displaystyle \mu \left({E}\right) = \sum_{n \mathop \in \N} a_n \mu_n \left({E}\right) \ge 0$ For every $n \in \N$, also $\mu_n \left({\varnothing}\right) = 0$. Therefore, it immediately follows that: :$\displaystyle \mu \left({\varnothing}\right) = \sum_{n \mathop \in \N} a_n \mu_n \left({\varnothing}\right) = \sum_{n \mathop \in \N} 0 = 0$ Finally, let $\left({E_n}\right)_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma$. Then: {{begin-eqn}} {{eqn | l = \mu \left({\bigcup_{m \mathop \in \N} E_m}\right)       | r = \sum_{n \mathop \in \N} a_n \mu_n \left({\bigcup_{m \mathop \in \N} E_m}\right)       | c = Definition of $\mu$ }} {{eqn | r = \sum_{n \mathop \in \N} a_n \left({\sum_{m \mathop \in \N} \mu_n \left({E_m}\right)}\right)       | c = The $\mu_n$ are measures }} {{eqn | r = \sum_{n \mathop \in \N} \sum_{m \mathop \in \N} a_n \mu_n \left({E_m}\right) }} {{eqn | r = \sum_{m \mathop \in \N} \sum_{n \mathop \in \N} a_n \mu_n \left({E_m}\right)       | c = Double Series of Positive Real Numbers }} {{eqn | r = \sum_{m \mathop \in \N} \mu \left({E_m}\right)       | c = Definition of $\mu$ }} {{end-eqn}} Therefore, having verified all three axioms, $\mu$ is a measure. {{qed}}
Let $\mathcal R$ be an equivalence relation on $S$. Let $\left({x, y}\right) \in \mathcal R$. By definition of equivalence class, $y \in \left[\!\left[{x}\right]\!\right]_\mathcal R$ and $x \in \left[\!\left[{x}\right]\!\right]_\mathcal R$. By definition of quotient set, $x$ and $y$ both belong to the same element of $S / \mathcal R$. So, by definition of $\mathcal R'$, it follows that $\left({x, y}\right) \in \mathcal R'$. That is: :$\left({x, y}\right) \in \mathcal R \implies \left({x, y}\right) \in \mathcal R'$ and so by definition of subset: :$\mathcal R \subseteq \mathcal R'$ Now let $\left({x, y}\right) \in \mathcal R'$. Then $y$ belongs to the same element of $S / \mathcal R$ that $x$ does. That is: :$y \in \left[\!\left[{x}\right]\!\right]_\mathcal R$ and so $\left({x, y}\right) \in \mathcal R$. That is: :$\left({x, y}\right) \in \mathcal R' \implies \left({x, y}\right) \in \mathcal R$ and so by definition of subset: :$\mathcal R' \subseteq \mathcal R$ The result follows by definition of set equality. {{qed}}
Let $P \subseteq S$ such that $P \in \mathcal P$. Let $x \in P$. Then: {{begin-eqn}} {{eqn | l=y       | o=\in       | r=\left[\!\left[{x}\right]\!\right]_\mathcal R       | c= }} {{eqn | ll=\iff       | l=\left({x, y}\right)       | o=\in       | r=\mathcal R       | c=by definition of equivalence class }} {{eqn | ll=\iff       | l=y       | o=\in       | r=P       | c=by definition of relation induced by $\mathcal P$ }} {{end-eqn}} Therefore: : $P = \left[\!\left[{x}\right]\!\right]_\mathcal R$ and so: : $P \in S / \mathcal R$ and so: : $\mathcal P \subseteq S / \mathcal R$ Now let $x \in S$. As $\mathcal P$ is a partition: :$\exists P \in \mathcal P: x \in P$ Then by definition of $\mathcal R$: :$\left({x, y}\right) \in \mathcal R \iff y \in \left[\!\left[{x}\right]\!\right]_\mathcal R$ {{begin-eqn}} {{eqn | l=y       | o=\in       | r=P       | c= }} {{eqn | ll=\iff       | l=\left({x, y}\right)       | o=\in       | r=\mathcal R       | c=by definition of relation induced by $\mathcal P$ }} {{eqn | ll=\iff       | l=y       | o=\in       | r=\left[\!\left[{x}\right]\!\right]_\mathcal R       | c=by definition of equivalence class }} {{end-eqn}} Therefore: : $\left[\!\left[{x}\right]\!\right]_\mathcal R = P$ and so: : $\left[\!\left[{x}\right]\!\right]_\mathcal R \in \mathcal P$ That is: : $\mathcal S / \mathcal R \subseteq P$ It follows by definition of set equality that: :$\mathcal S / \mathcal R = P$ Hence the result. {{qed}}
As $\mu$ is a measure, $\mu \left({M}\right) \ge 0$. Also, by Measure is Monotone, $\mu \left({M}\right) \le \mu \left({N}\right) = 0$. Hence $\mu \left({M}\right) = 0$, and $M$ is a $\mu$-null set. {{qed}}
As $\mu$ is a measure, $\mu \left({N}\right) \ge 0$. Also: {{begin-eqn}} {{eqn|l = \mu \left({N}\right)      |o = \le      |r = \sum_{n \mathop \in \N} \mu \left({N_n}\right)      |c = Measure is Countably Subadditive }} {{eqn|r = \sum_{n \mathop \in \N} 0      |c = The $N_n$ are $\mu$-null sets }} {{eqn|r = 0 }} {{end-eqn}} Hence necessarily $\mu \left({N}\right) = 0$, and $N$ is a $\mu$-null set. {{qed}}
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 definition of absolute value: {{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}}
Verify the axioms for a measure in turn for $\mu \restriction_{\Sigma'}$: === Axiom $(1)$ === The statement of axiom $(1)$ for $\mu \restriction_{\Sigma'}$ is: :$\forall E' \in \Sigma': \mu \restriction_{\Sigma'} \left({E'}\right) \ge 0$ Now, for every $E' \in \Sigma'$, compute: {{begin-eqn}} {{eqn|l = \mu \restriction_{\Sigma'} \left({E'}\right)      |r = \mu \left({E'}\right)      |c = Definition of $\mu \restriction_{\Sigma'}$ }} {{eqn|o = \ge      |r = 0      |c = $\mu$ is a measure }} {{end-eqn}} {{qed|lemma}} === Axiom $(2)$ === Let $\left({E'_n}\right)_{n \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma'$. Then the statement of axiom $(2)$ for $\mu \restriction_{\Sigma'}$ is: :$\displaystyle \mu \restriction_{\Sigma'} \left({\bigcup_{n \mathop \in \N} E'_n}\right) = \sum_{n \mathop \in \N} \mu \restriction_{\Sigma'} \left({E'_n}\right)$ One can show this by means of the following computation: {{begin-eqn}} {{eqn|l = \mu \restriction_{\Sigma'} \left({\bigcup_{n \mathop \in \N} E'_n}\right)      |r = \mu \left({\bigcup_{n \mathop \in \N} E'_n}\right)      |c = Definition of $\mu \restriction_{\Sigma'}$ }} {{eqn|r = \sum_{n \mathop \in \N} \mu \left({E'_n}\right)      |c = $\mu$ is a measure }} {{eqn|r = \sum_{n \mathop \in \N} \mu \restriction_{\Sigma'} \left({E'_n}\right)      |c = Definition of $\mu \restriction_{\Sigma'}$ }} {{end-eqn}} {{qed|lemma}} === Axiom $(3')$ === The statement of axiom $(3')$ for $\mu \restriction_{\Sigma'}$ is: :$\mu \restriction_{\Sigma'} \left({\varnothing}\right) = 0$ By Sigma-Algebra Contains Empty Set, $\varnothing \in \Sigma'$. Hence: :$\mu \restriction_{\Sigma'} \left({\varnothing}\right) = \mu \left({\varnothing}\right) = 0$ because $\mu$ is a measure. {{qed|lemma}} Having verified a suitable set of axioms, it follows that $\mu \restriction_{\Sigma'}$ is a measure. {{qed}}
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}}
=== Necessary Condition === Let $\mu$ be a $\sigma$-finite measure. Let $\left({F_n}\right)_{n \in \N}$ be an exhausting sequence in $\Sigma$ such that: :$\forall n \in \N: \mu \left({F_n}\right) < \infty$ Then as $\left({F_n}\right)_{n \in \N}$ is exhausting, have: :$\displaystyle \bigcup_{n \mathop \in \N} F_n = X$ It follows that the sequence $\left({F_n}\right)_{n \in \N}$ satisfies $(1)$ and $(2)$. {{qed|lemma}} === Sufficient Condition === Let $\mu$ be any measure. Let $\left({E_n}\right)_{n \in \N}$ be a sequence satisfying $(1)$ and $(2)$. Define $F_n := \displaystyle \bigcup_{k \mathop = 1}^n E_k$. Then by Sigma-Algebra Closed under Union: : $F_n \in \Sigma$ for all $n \in \N$ Also, by Set is Subset of Union: : $F_{n+1} = F_n \cup E_{n+1}$, hence $F_n \subseteq F_{n+1}$ The definition of the $F_n$ assures that: : $X = \displaystyle \bigcup_{n \mathop \in \N} E_n = \bigcup_{n \mathop \in \N} F_n$ Hence $\left({F_n}\right)_{n \in \N}$ is an exhausting sequence in $\Sigma$. Furthermore, compute, for any $n \in \N$: {{begin-eqn}} {{eqn|l = \mu \left({F_n}\right)      |r = \mu \left({\bigcup_{k \mathop = 1}^n E_k}\right)      |c = Definition of $F_n$ }} {{eqn|o = \le      |r = \sum_{k \mathop = 1}^n \mu \left({E_k}\right)      |c = Measure is Subadditive }} {{eqn|o = <      |r = +\infty      |c = By $(2)$ }} {{end-eqn}} Hence, by definition, $\mu$ is $\sigma$-finite. Thus, $\left({X, \Sigma, \mu}\right)$ is also $\sigma$-finite. {{qed}}
Let $H$ be a central subgroup of $G$. By definition of central subgroup: :$H \subseteq \map Z G$ where $\map Z G$ is the center of $G$. Thus we have that $H$ is a group which is a subset of $\map Z G$. Therefore by definition $H$ is a subgroup of $\map Z G$. We also have from Center of Group is Abelian Subgroup that $\map Z G$ is an abelian group. It follows from Subgroup of Abelian Group is Normal that $Z$ is a normal subgroup of $G$. {{Qed}}
Let $H$ be a central subgroup of $G$. By definition of central subgroup: :$H \subseteq \map Z G$ where $\map Z G$ is the center of $G$. Then: {{begin-eqn}} {{eqn | ll= \forall x \in G: \forall h \in H:       | l = x h x^{-1}       | r = x x^{-1} h       | c = as $h \in H \implies h \in \map Z G$ }} {{eqn | r = h       | c =  }} {{eqn | lll=\leadsto       | l = x h x^{-1}       | o = \in       | r = H       | c = as $h \in H$ }} {{eqn | lll=\leadsto       | l = H       | o = \lhd       | r = G       | c = {{Defof|Normal Subgroup}} }} {{end-eqn}} {{Qed}}
{{begin-eqn}} {{eqn | l = \norm {\mathbf v} \norm {\mathbf w} \cos \theta       | r = \mathbf v \cdot \mathbf w       | c = Cosine Formula for Dot Product }} {{eqn | l = \cos \theta       | r = \frac {\mathbf v \cdot \mathbf w} {\norm {\mathbf v} \norm {\mathbf w} }       | c = because $\mathbf v, \mathbf w \ne \mathbf 0 \implies \norm {\mathbf v}, \norm {\mathbf w} \ne 0$ }} {{eqn | l = \arccos \paren {\cos \theta}       | r = \arccos \paren {\frac {\mathbf v \cdot \mathbf w} {\norm {\mathbf v} \norm {\mathbf w} } }       | c = because $0 \le \theta \le \pi$ $\implies -1 \le \cos \theta \le 1$ }} {{eqn | l = \theta       | r = \arccos \paren {\frac {\mathbf v \cdot \mathbf w} {\norm {\mathbf v} \norm {\mathbf w} } }       | c = Composite of Bijection with Inverse is Identity Mapping }} {{end-eqn}} {{qed}}