hcju/proofwikiqa huggingface dataset:
=== Case 1 === Suppose that $\mathbf v$ and $\mathbf w$ are not scalar multiples of each other. :400px From Construction of Triangle from Given Lengths, it is sufficient to show that sum of the lengths of any two sides is greater than the third side. Consider the side with length $\norm {\mathbf v}$. From the triangle inequality for vectors: {{begin-eqn}} {{eqn | l = \norm {\mathbf v}       | r = \norm {\mathbf {w + v - w} } }} {{eqn | o = <       | r = \norm {\mathbf w} + \norm {\mathbf {v - w} } }} {{end-eqn}} Note that the equality is a strict inequality because the vectors are not scalar multiples of each other. {{explain|Pick a better way to explain this than linking to the above}} Consider the side with length $\norm {\mathbf w}$. {{begin-eqn}} {{eqn | l = \norm {\mathbf w}       | r = \norm {\mathbf {v + w - v} } }} {{eqn | o = <       | r = \norm {\mathbf v} + \norm {\mathbf {w - v} } }} {{eqn | r = \norm {\mathbf v} + \norm {\mathbf {v - w} } }} {{end-eqn}} Lastly, Consider the side with length $\norm {\mathbf v - \mathbf w}$. {{begin-eqn}} {{eqn | l = \norm {\mathbf {v - w} }       | r = \norm {\mathbf {v + \paren {-w} } } }} {{eqn | o = <       | r = \norm {\mathbf v} + \norm {\mathbf {-w} } }} {{eqn | r = \norm {\mathbf v} + \norm {\mathbf w} }} {{end-eqn}} {{qed|lemma}} === Case 2 === Suppose that $\mathbf v$ and $\mathbf w$ ''are'' scalar multiples of each other. Then the existence of $\theta$ follows directly from the definition of the angle between vectors that are scalar multiples of each other. {{explain|Link to the above definition, or word it here.}} {{qed}}
As $\mathcal D$ is a Dynkin system, $X \in \mathcal D$. By Set Difference with Self is Empty Set, $X \setminus X = \varnothing$. Hence, by property $(2)$ of a Dynkin system, $\varnothing = X \setminus X \in \mathcal D$. {{qed}}
Define $D_1 = D, D_2 = E$, and for $n \ge 2$, $D_n = \varnothing$. Then by Dynkin System Contains Empty Set, $D_n \in \mathcal D$ for all $n \in \N$. Also, by Intersection with Empty Set, it follows that $\left({D_n}\right)_{n \in \N}$ is a pairwise disjoint sequence. Hence, by property $(3)$ of a Dynkin system, have: :$D \cup E = \displaystyle \bigcup_{n \mathop \in \N} D_n \in \mathcal D$ {{qed}}
The axioms $(1)$ and $(2)$ for both $\sigma$-algebras and Dynkin systems are identical. Dynkin system axiom $(3)$ is seen to be a specification of $\sigma$-algebra axiom $(3)$ to pairwise disjoint sequences. Hence $\Sigma$ is trivially a Dynkin system on $X$. {{qed}}
=== Existence === By Power Set is Dynkin System, there exists at least one Dynkin system containing $\mathcal G$. Next, let $\Bbb D$ be the collection of Dynkin systems containing $\mathcal G$: :$\Bbb D := \left\{{\mathcal{D}': \mathcal G \subseteq \mathcal{D}', \text{$\mathcal{D}'$ is a Dynkin system}}\right\}$ By Intersection of Dynkin Systems is Dynkin System, $\mathcal D := \bigcap \Bbb D$ is a Dynkin system. Also, by Set Intersection Preserves Subsets: : $\mathcal G \subseteq \mathcal D$ Now let $\mathcal{D}'$ be a Dynkin system containing $\mathcal G$. By construction of $\mathcal D$, and Intersection is Subset: General Result: : $\mathcal D \subseteq \mathcal{D}'$ {{qed|lemma}} === Uniqueness === Suppose both $\mathcal{D}_1$ and $\mathcal{D}_2$ are Dynkin systems generated by $\mathcal G$. Then property $(2)$ for these Dynkin systems implies both $\mathcal{D}_1 \subseteq \mathcal{D}_2$ and $\mathcal{D}_2 \subseteq \mathcal{D}_1$. By definition of set equality: : $\mathcal{D}_1 = \mathcal{D}_2$ {{qed}}
By Sigma-Algebra is Dynkin System, $\sigma \left({\mathcal G}\right)$ is a Dynkin system. The definition of $\delta \left({\mathcal G}\right)$ now ensures that $\delta \left({\mathcal G}\right) \subseteq \sigma \left({\mathcal G}\right)$. {{qed}}
The first two conditions for a Dynkin system are identical to those for a $\sigma$-algebra. Hence it is only required to verify that $(1)$ implies that $\DD$ is closed under arbitrary countable unions. So let $\sequence {D_n}_{n \mathop \in \N}$ be a sequence in $\DD$. Now define the sequence $\sequence {E_n}_{n \mathop \in \N}$ by: :$\displaystyle E_n := D_n \cap \paren {X \setminus \bigcup_{m \mathop < n} D_m}$ By Dynkin System Closed under Union and $(1)$, it follows that $E_n \in \DD$ for all $n \in \N$. === Lemma === For all $n \in \N$, it holds that: :$\displaystyle \bigcup_{k \mathop = 0}^n E_k = \bigcup_{k \mathop = 0}^n D_k$ === Proof of Lemma === {{finish|tedium, best postponed until after the implementation of the new extension}} The lemma, combined with the definition of the $E_n$, gives immediately that for all $n \in \N$: :$\displaystyle E_n \in X \setminus \bigcup_{m \mathop < n} D_m = X \setminus \bigcup_{m \mathop < n} E_m$ whence the $E_n$ are pairwise disjoint. Another consequence is that: :$\displaystyle \bigcup_{n \mathop \in \N} D_n = \bigcup_{n \mathop \in \N} E_n$ Now since the $E_n$ are pairwise disjoint, it follows that: :$\displaystyle \bigcup_{n \mathop \in \N} E_n \in \DD$ which, combined with above equality, concludes in: :$\displaystyle \bigcup_{n \mathop \in \N} D_n \in \DD$ Therefore, $\DD$ is closed under countable unions, making it a $\sigma$-algebra. {{qed}}
From Sigma-Algebra is Dynkin System and the definition of generated Dynkin system, it follows that: :$\map \delta \GG \subseteq \map \sigma \GG$ Let $D \in \map \delta \GG$, and define: :$\delta_D := set {E \subseteq X: E \cap D \in \map \delta \GG}$ Let us verify that these $\delta_D$ form Dynkin systems. First of all, note $X \cap D = D$, hence $X \in \delta_D$. Next, compute, for any $E \in \delta_D$: {{begin-eqn}} {{eqn | l = \paren {X \setminus E} \cap D       | r = \paren {\paren {X \setminus E} \cap D} \cup \paren {\paren {X \setminus D} \cap D}       | c = Set Difference Intersection with Second Set is Empty Set, Union with Empty Set }} {{eqn | r = \paren {\paren {X \setminus E} \cup \paren {X \setminus D} } \cap D       | c = Intersection Distributes over Union }} {{eqn | r = \paren {X \setminus \paren {E \cap D} } \cap D       | c = De Morgan's Laws: Difference with Intersection }} {{eqn | r = \paren {X \setminus \paren {E \cap D} } \cap \paren {X \setminus \paren {X \setminus D} }       | c = Set Difference with Set Difference }} {{eqn | r = X \setminus \paren {\paren {E \cap D} \cup \paren {X \setminus D} }       | c = De Morgan's Laws: Difference with Union }} {{end-eqn}} Now from Intersection is Associative, Set Difference Intersection with Second Set is Empty Set and Intersection with Empty Set: :$\paren {E \cap D} \cap \paren {X \setminus D} = E \cap \paren {D \cap \paren {X \setminus D} } = E \cap \O = \O$ Thus, since $E \cap D, X \setminus D \in \map \delta \GG$, it follows that their disjoint union is as well. Finally, combining the above, it follows that: :$\paren {X \setminus E} \cap D \in \map \delta \GG$ Thus: :$E \in \delta_D \implies X \setminus E \in \delta_D$ Finally, let $\sequence {E_n}_{n \mathop \in \N}$ be a pairwise disjoint sequence of sets in $\delta_D$. Then it is immediate that $\sequence {E_n \cap D}_{n \mathop \in \N}$ is also pairwise disjoint. Hence: {{begin-eqn}} {{eqn | l = \paren {\bigcup_{n \mathop \in \N} E_n} \cap D       | r = \bigcup_{n \mathop \in \N} \paren {E_n \cap D}       | c = Intersection Distributes over Union }} {{end-eqn}} and since the $E_n \cap D$ are in $\map \delta \GG$ by assumption, this disjoint union is also in $\map \delta \GG$. Therefore, $\delta_D$ is a Dynkin system. Now by definition of generated Dynkin system, $\GG \subseteq \map \delta \GG$. By this observation and $(1)$, we immediately obtain, for any $G \in \GG$: :$\GG \subseteq \delta_G$ Therefore, by definition of generated Dynkin system, for all $G \in \GG$: :$\map \delta \GG \subseteq \delta_G$ Hence, for any $D \in \map \delta \GG$ and $G \in \GG$: :$D \cap G \in \map \delta \GG$ Thus we have established that, for all $D \in \map \delta \GG$: :$\GG \subseteq \delta_D$ whence, by the definition of $\delta_D$ and generated Dynkin system: :$\map \delta \GG \subseteq \delta_D$ That is to say: :$\forall D, E \in \map \delta \GG: D \cap E \in \map \delta \GG$ Hence, by Dynkin System Closed under Intersections is Sigma-Algebra, $\map \delta \GG$ is a $\sigma$-algebra. Thus, it follows that $\map \sigma \GG \subseteq \map \delta \GG$. Hence the result, by definition of set equality. {{qed}}
Let $x \mathrel {\RR_N} y$ denote that $x$ and $y$ are in the same coset, that is: :$x \mathrel {\RR_N} y \iff x \circ N = y \circ N$ as specified in the definition of congruence modulo $N$. Let $x \mathrel {\RR_N} x'$ and $y \mathrel {\RR_N} y'$. To demonstrate that $\RR_N$ is a congruence relation for $\circ$, we need to show that: :$\paren {x \circ y} \mathrel {\RR_N} \paren {x' \circ y'}$ So: {{begin-eqn}} {{eqn | l = \paren {x \circ y} \circ \paren {x' \circ y'}^{-1}       | r = \paren {x \circ y} \circ \paren {y'^{-1} \circ x'^{-1} }       | c = Inverse of Group Product }} {{eqn | r = \paren {\paren {x \circ y} \circ y'^{-1} } \circ x'^{-1}       | c =  }} {{eqn | r = \paren {x \circ \paren {y \circ y'^{-1} } } \circ x'^{-1}       | c =  }} {{end-eqn}} By Cosets are Equal iff Product with Inverse in Subgroup: :$x \circ x'^{-1} \in N$ and $y \circ y'^{-1} \in N$ Thus: :$\paren {x \circ y} \circ \paren {x' \circ y'}^{-1} \in x \circ H \circ x'^{-1}$ But we also have that: {{begin-eqn}} {{eqn | l = x \circ H \circ x'^{-1}       | r = H \circ x \circ x'^{-1}       | c = {{Defof|Normal Subgroup}} }} {{eqn | o = \subseteq       | r = H \circ H       | c = {{Defof|Subset Product}}: $x \circ x'^{-1} \in H$ }} {{eqn | r = H       | c = Product of Subgroup with Itself }} {{end-eqn}} That is: :$\paren {x \circ y} \circ \paren {x' \circ y'}^{-1} \in H$ and so: :$\paren {x \circ y} \mathrel {\RR_N} \paren {x' \circ y'}$ Hence the result, by definition of congruence relation. {{qed}}
Verify the axioms for a measure in turn for $\mu_F$: === Axiom $(1)$ === The statement of axiom $(1)$ for $\mu_F$ is: :$\forall E \in \Sigma: \mu_F \left({E}\right) \ge 0$ For every $E \in \Sigma$ have: {{begin-eqn}} {{eqn|l = \mu_F \left({E}\right)      |r = \mu \left({E \cap F}\right)      |c = Definition of $\mu_F$ }} {{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 $\Sigma$. The statement of axiom $(2)$ for $\mu_F$ is: :$\displaystyle \mu_F \left({\bigcup_{n \mathop \in \N} E_n}\right) = \sum_{n \mathop \in \N} \mu_F \left({E_n}\right)$ This is verified by the following computation: {{begin-eqn}} {{eqn|l = \mu_F \left({\bigcup_{n \mathop \in \N} E_n}\right)      |r = \mu \left({ \left({\bigcup_{n \mathop \in \N} E_n}\right) \cap F}\right)      |c = Definition of $\mu_F$ }} {{eqn|r = \mu \left({ \bigcup_{n \mathop \in \N} \left({E_n \cap F}\right) }\right)      |c = Intersection Distributes over Union: General Result }} {{eqn|r = \sum_{n \mathop \in \N} \mu \left({E_n \cap F}\right)      |c = $\mu$ is a measure }} {{eqn|r = \sum_{n \mathop \in \N} \mu_F \left({E_n}\right)      |c = Definition of $\mu_F$ }} {{end-eqn}} {{qed|lemma}} === Axiom $(3')$ === The statement of axiom $(3')$ for $\mu_F$ is: :$\mu_F \left({\varnothing}\right) = 0$ By Intersection with Empty Set, $\varnothing \cap F = \varnothing$. Hence: :$\mu_F \left({\varnothing}\right) = \mu \left({\varnothing \cap F}\right) = 0$ because $\mu$ is a measure. {{qed|lemma}} Having verified a suitable set of axioms, it follows that $\mu_F$ is a measure. {{qed}}
Denote with $\tau_{\mathbf x}: \R^n \to \R^n$ the translation by $\mathbf x$. From Translation in Euclidean Space is Measurable Mapping, $\tau_{\mathbf x}$ is $\mathcal B \left({\R^n}\right) \, / \, \mathcal B \left({\R^n}\right)$-measurable. Consider the pushforward measure $\lambda^n_{\mathbf x} := \left({\tau_{\mathbf x}}\right)_* \lambda^n$ on $\mathcal B \left({\R^n}\right)$. By Characterization of Euclidean Borel Sigma-Algebra, it follows that: :$\mathcal B \left({\R^n}\right) = \sigma \left({\mathcal{J}^n_{ho}}\right)$ where $\sigma$ denotes generated $\sigma$-algebra, and $\mathcal{J}^n_{ho}$ is the set of half-open $n$-rectangles. Let us verify the four conditions for Uniqueness of Measures, applied to $\lambda^n$ and $\lambda^n_{\mathbf x}$. Condition $(1)$ follows from Half-Open Rectangles Closed under Intersection. Condition $(2)$ is achieved by the sequence of half-open $n$-rectangles given by: :$J_k := \left[{-k \,.\,.\, k}\right)^n$ For condition $(3)$, let $\lefthalf-open $n$-rectangle. Since: :$\tau_{\mathbf x}^{-1} \left({\left[[{\mathbf a \,.\,.\, \mathbf b}\right))}\right) = \mathbf x + \left[[{\mathbf a \,.\,.\, \mathbf b}\right)) = \left[[{\mathbf {a + x} \,.\,.\, \mathbf {b + x}}\right))$ we have: {{begin-eqn}} {{eqn|l = \lambda^n_{\mathbf x} \left({\left[\left[{\mathbf a \,.\,.\, \mathbf b}\right)\right)}\right)      |r = \lambda^n \left({\left[\left[{\mathbf {a + x} \,.\,.\, \mathbf {b + x} }\right)\right)}\right)      |c = Definition of pushforward measure }} {{eqn|r = \prod_{i \mathop = 1}^n \left({\left({b_i + x_i}\right) - \left({a_i + x_i}\right)}\right)      |c = Definition of Lebesgue measure }} {{eqn|r = \prod_{i \mathop = 1}^n \left({b_i - a_i}\right) }} {{eqn|r = \lambda^n \left({\left[\left[{\mathbf a \,.\,.\, \mathbf b}\right)\right)}\right)      |c = Definition of Lebesgue measure }} {{end-eqn}} Finally, since: :$\displaystyle \lambda^n \left({J_k}\right) = \prod_{i \mathop = 1}^n \left({k - \left({-k}\right)}\right) = \left({2 k}\right)^n$ the last condition, $(4)$, is also satisfied. Whence Uniqueness of Measures implies that: :$\lambda^n_{\mathbf x} = \lambda^n$ and since for all $B \in \mathcal B \left({\R^n}\right)$ we have: :$\mathbf x + B = \tau_{\mathbf x}^{-1} \left({B}\right)$ this precisely boils down to: :$\lambda^n \left({\mathbf x + B}\right) = \lambda^n \left({B}\right)$ {{qed}}
From Characterization of Euclidean Borel Sigma-Algebra, we have: :$\mathcal B \left({\R^n}\right) = \sigma \left({\mathcal{J}^n_{ho,\text{rat}}}\right)$ where $\mathcal{J}^n_{ho,\text{rat}}$ denotes the collection of half-open $n$-rectangles with rational endpoints. So let $J = \left[[{\mathbf a \,.\,.\, \mathbf b}\right)) \in \mathcal{J}^n_{ho,\text{rat}}$. Let $M \in \N$ be a common denominator of the $a_i, b_i$ (which are rational by assumption). We may then cover $J$ by finitely many pairwise disjoint half-open $n$-rectangles $\left[{0 \,.\,.\, \dfrac 1 M}\right)^n$, in that: :$\displaystyle J = \bigcup_{i \mathop = 1}^{k \left({J}\right)} \mathbf x_i + \left[{0 \,.\,.\, \frac 1 M}\right)^n$ for some $k \left({J}\right) \in \N$ and suitable $\mathbf x_i$, where: :$\mathbf x_i + \left[{0 \,.\,.\, \dfrac 1 M}\right)^n := \left[\left[{\mathbf x_i \,.\,.\, \mathbf x_i + \dfrac 1 M}\right)\right)$ Using that $\mu$ is translation-invariant, this means: :$\mu \left({J}\right) = k \left({J}\right) \, \mu \left({\left[{0 \,.\,.\, \dfrac 1 M}\right)^n}\right)$ Also, by Lebesgue Measure is Translation-Invariant: :$\lambda^n \left({J}\right) = k \left({J}\right) \, \lambda^n \left({\left[{0 \,.\,.\, \dfrac 1 M}\right)^n}\right)$ A moment's thought shows us that the half-open $n$-rectangle $I = \left[{0 \,.\,.\, 1}\right)^n$ may be covered by $M^n$ copies of $\left[{0 \,.\,.\, \dfrac 1 M}\right)^n$, so that: :$k \left({I}\right) = M^n$ For brevity, write $I/M$ for $\left[{0 \,.\,.\, \dfrac 1 M}\right)^n$. Now, using that: :$\displaystyle \lambda^n \left({I/M}\right) = \prod_{i \mathop = 1}^n \frac 1 M = \frac 1 {M^n}$ and $\mu \left({I}\right) = \kappa$, compute: {{begin-eqn}} {{eqn|l = \mu \left({J}\right)      |r = k \left({J}\right) \mu \left({I/M}\right)      |c = Definition of $k \left({J}\right)$ }} {{eqn|r = \frac {k \left({J}\right)} {M^n} \left({M^n \mu \left({I/M}\right)}\right)      |c = $M^n / M^n = 1$ }} {{eqn|r = \frac {k \left({J}\right)} {M^n} \mu \left({I}\right)      |c = $M^n = k \left({I}\right)$ }} {{eqn|r = \frac {\kappa \, k \left({J}\right)} {M^n}      |c = Definition of $\kappa$ }} {{eqn|r = \kappa \, k \left({J}\right) \, \lambda^n \left({I/M}\right) }} {{eqn|r = \kappa \, \lambda^n \left({J}\right)      |c = Definition of $k \left({J}\right)$ }} {{end-eqn}} Therefore, $\mu \left({J}\right) = \kappa \, \lambda^n \left({J}\right)$ for all $J \in \mathcal{J}^n_{ho,\text{rat}}$. Let us quickly verify the other conditions for Uniqueness of Measures. For $(1)$, we have Half-Open Rectangles Closed under Intersection. For $(2)$, observe the exhausting sequence $\left[{-k \,.\,.\, k}\right)^n \mathop \uparrow \R^n$ Finally, for $(4)$, we recall that $\kappa < +\infty$. Thus, by Uniqueness of Measures, $\mu = \kappa \lambda^n$. {{qed}}
For brevity, write for example $E^c$ for $\complement_X \left({E}\right) = X \setminus E$. We reason as follows: {{begin-eqn}} {{eqn | l = D \setminus E       | r = D \cap E^c       | c = Set Difference as Intersection with Relative Complement }} {{eqn | r = \left({D^c \cup E}\right)^c       | c = De Morgan's Laws: Complement of Union, Relative Complement of Relative Complement }} {{end-eqn}} Now this implies that $D \setminus E \in \mathcal D$ {{iff}} $D^c \cup E \in \mathcal D$. It is already known that $D^c$ and $E$ are in $\mathcal D$ by axiom $(2)$ for a Dynkin system. Since $E \subseteq D$, it follows that $D^c \cap E = \varnothing$, and thus Dynkin System Closed under Disjoint Union applies to give: :$D^c \cup E \in \mathcal D$ which, combined with above reasoning, yields $D \setminus E \in \mathcal D$. {{qed}}
From Cartesian Product of Intersections: General Case, we have: :$\displaystyle \left(({\mathbf a \,.\,.\, \mathbf b}\right)) \cap \left(({\mathbf c \,.\,.\, \mathbf d}\right)) = \prod_{i \mathop = 1}^n \left({a_i \,.\,.\, b_i}\right) \cap \left({c_i \,.\,.\, d_i}\right)$ Therefore, it suffices to show that the intersection of two open intervals is again an open interval. Now let $x \in \left({a_i \,.\,.\, b_i}\right) \cap \left({c_i \,.\,.\, d_i}\right)$. Then $x$ is subject to: * $x > a_i$ and $x > c_i$, i.e., $x > \max \left\{{a_i, c_i}\right\}$ * $x < b_i$ and $x < d_i$, i.e., $x < \min \left\{{b_i, d_i}\right\}$ and we see that these conditions are satisfied precisely when: :$x \in \left({\max \left\{{a_i, c_i}\right\} \,.\,.\, \min \left\{{b_i, d_i}\right\}}\right)$ Thus, we conclude: :$\left({a_i \,.\,.\, b_i}\right) \cap \left({c_i \,.\,.\, d_i}\right) = \left({\max \left\{{a_i, c_i}\right\} \,.\,.\, \min \left\{{b_i, d_i}\right\}}\right)$ showing that indeed the intersection is an open interval. Combining this with the above reasoning, it follows that indeed: :$\left(({\mathbf a \,.\,.\, \mathbf b}\right)) \cap \left(({\mathbf c \,.\,.\, \mathbf d}\right))$ is an open $n$-rectangle. {{qed}} Category:Analysis ol5b6dzhrmypwf027n6us41e07ogiqq
Define $f: \R^n \to \R^n$ by $f \left({\mathbf x}\right) := \dfrac 1 t \mathbf x$. Then for all $\mathbf x \in \R^n$, $f^{-1} \left({\mathbf x}\right) = \left\{{t \mathbf x}\right\}$, where $f^{-1}$ denotes the preimage of $f$. Thus $t \cdot B = \displaystyle \bigcup_{\mathbf b \mathop \in B} f^{-1} \left({\mathbf b}\right) = f^{-1} \left({B}\right)$, where the last equality holds by definition of preimage. It follows that the statement of the theorem comes down to showing that $f$ is $\mathcal B \, / \, \mathcal B$-measurable. By Characterization of Euclidean Borel Sigma-Algebra the half-open $n$-rectangles $\mathcal{J}_{ho}^n$ generate $\mathcal B$. Applying Mapping Measurable iff Measurable on Generator, it suffices to demonstrate: :$\forall J \in \mathcal{J}_{ho}^n: f^{-1} \left({J}\right) \in \mathcal B$ Now for a half-open $n$-rectangle $J = \left[[{\mathbf a \,.\,.\, \mathbf b}\right))$, it follows from the consideration on $f^{-1}$ above, that: :$f^{-1} \left({J}\right) = \left[[{t \mathbf a \,.\,.\, t \mathbf b}\right))$ which is again in $\mathcal{J}_{ho}^n$ and so, in particular, in $\mathcal B$. Thus $f$ is $\mathcal B \, / \, \mathcal B$-measurable, i.e.: :$\forall B \in \mathcal B: t \cdot B = f^{-1} \left({B}\right) \in \mathcal B$ which was to be demonstrated. {{qed}}
It follows from Rescaling is Linear Transformation that the mapping $\mathbf x \mapsto t \mathbf x$ is a linear transformation. Denote $t \, \mathbf I_n$ for the matrix associated to this linear transformation by Linear Transformation as Matrix Product. From Determinant of Rescaling Matrix, it follows that: :$\det \left({t \, \mathbf I_n}\right) = t^n$ From Inverse of Rescaling Matrix, $t \, \mathbf I_n$ is the inverse of $t^{-1} \mathbf I_n$. Thus, it follows that: {{begin-eqn}} {{eqn|l = \lambda^n \left({t \cdot B}\right)      |r = \lambda^n \left({\left({t \, \mathbf I_n }\right) \left({B}\right)}\right) }} {{eqn|r = \lambda^n \left({\left({t^{-1} \, \mathbf I_n }\right)^{-1} \left({B}\right)}\right)      |c = Inverse of Group Inverse }} {{eqn|r = \left({t^{-1} \, \mathbf I_n }\right)_* \lambda^n \left({B}\right)      |c = Definition of pushforward measure }} {{eqn|r = \left\vert{\det \, \left({ \left({t^{-1} \, \mathbf I_n }\right)^{-1} }\right) }\right\vert \cdot \lambda^n \left({B}\right)      |c = Pushforward of Lebesgue Measure under General Linear Group }} {{end-eqn}} {{explain|For Inverse of Group Inverse to be used, it has to be established that it is indeed a group.}} Now recall $\det \, \left({\left({t^{-1} \, \mathbf I_n }\right)^{-1}}\right) = \det \, \left({t \, \mathbf I_n}\right) = t^n$. Since $t > 0$, $\left\vert{t^n}\right\vert = t^n$, and the result follows. {{qed}}
Consider the pushforward measure $\theta_* \mu$ on $\left({X, \Sigma}\right)$. By definition, this makes equation $(3)$ come down to: :$\theta_* \mu \restriction_{\mathcal G} = \mu \restriction_{\mathcal G}$ where $\restriction$ denotes restriction. The suppositions $(1)$, $(2)$ and $(3)$ together constitute precisely the prerequisites to Uniqueness of Measures. Hence $\theta_* \mu = \mu$, i.e., $\mu$ is $\theta$-invariant. {{qed}}
Fix $H \in \HH$. Define, for $E \in \Sigma$: :$\map \mu E := \map \Pr {E \cap H}$ :$\map \nu E := \map \Pr E \map \Pr H$ Then by Intersection Measure is Measure and Restricted Measure is Measure, $\mu$ is a measure on $\Sigma$. Namely, it is the intersection measure $\Pr_H$ restricted to $\Sigma$, that is $\Pr_H \restriction_\Sigma$. Next, by Linear Combination of Measures and Restricted Measure is Measure, $\nu$ is also a measure on $\Sigma$. Namely, it is the restricted measure $\map \Pr H \Pr \restriction_\Sigma$. Let $\GG' := \GG \cup \set X$. It is immediate that $\GG'$ is also a $\cap$-stable generator for $\Sigma$. By assumption $(1)$, $\mu$ and $\nu$ coincide on $\GG$ (since $\map \Pr X = 1$). From Restricting Measure Preserves Finiteness, $\mu$ and $\nu$ are also finite measures. Hence, $\GG$ contains the exhausting sequence of which every term equals $X$. Having verified all conditions, Uniqueness of Measures applies to yield $\mu = \nu$. Now fix $E \in \Sigma$ and define, for $E' \in \Sigma'$: :$\map {\mu'_E} {E'} := \map \Pr {E \cap E'}$ :$\map {\nu'_E} {E'} := \map \Pr E \map \Pr {E'}$ Mutatis mutandis, above consideration applies again, and we conclude by Uniqueness of Measures: :$\mu'_E = \nu'_E$ for all $E \in \Sigma$.  That is, expanding the definition of the measures $\mu'_E$ and $\nu'_E$: :$\forall E \in \Sigma: \forall E' \in \Sigma': \map \Pr {E \cap E'} = \map \Pr E \map \Pr {E'}$ This is precisely the statement that $\Sigma$ and $\Sigma'$ are $\Pr$-independent $\sigma$-algebras. {{qed}}