avewright/proofwiki-math huggingface dataset:
Theorem: Central Subgroup is Normal

Let $G$ be a group.
Let $H$ be a central subgroup of $G$.
Then $H$ is a normal subgroup of $G$.

Proof:
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.
{{questionable|Being a normal subgroup of a subgroup does not mean being a normal subgroup of the whole group}}
It follows from Subgroup of Abelian Group is Normal that $Z$ is a normal subgroup of $G$.
{{Qed}}
Theorem: Central Subgroup is Normal

Let $G$ be a group.
Let $H$ be a central subgroup of $G$.
Then $H$ is a normal subgroup of $G$.

Proof:
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 | q = \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 | ll= \leadsto
      | l = x h x^{-1}
      | o = \in
      | r = H
      | c = as $h \in H$
}}
{{eqn | ll= \leadsto
      | l = H
      | o = \lhd
      | r = G
      | c = {{Defof|Normal Subgroup}}
}}
{{end-eqn}}
{{Qed}}
Theorem: Angle Between Vectors in Terms of Dot Product

The angle between two non-zero vector quantities can be calculated by:
:$\theta = \arccos \dfrac {\mathbf v \cdot \mathbf w} {\norm {\mathbf v} \norm {\mathbf w} }$
where:
:$\mathbf v \cdot \mathbf w$ represents the dot product of $\mathbf v$ and $\mathbf w$ 
:$\norm {\, \cdot \,}$ represents vector length
:$\arccos$ represents arccosine.

Proof:
{{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 = \map \arccos {\cos \theta}
      | r = \map \arccos {\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 = \map \arccos {\frac {\mathbf v \cdot \mathbf w} {\norm {\mathbf v} \norm {\mathbf w} } }
      | c = Composite of Bijection with Inverse is Identity Mapping
}}
{{end-eqn}}
{{qed}}
Theorem: Angle Between Non-Zero Vectors Always Defined

Let $\mathbf v$ and $\mathbf w$ be non-zero vectors in real Euclidean space $\R^n$.
Then the angle between $\mathbf v$ and $\mathbf w$ is always defined.

Proof:
=== 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}}
Theorem: Dynkin System Contains Empty Set

Let $X$ be a set, and let $\DD$ be a Dynkin system on $X$.
Then the empty set $\O$ is an element of $\DD$.

Proof:
As $\DD$ is a Dynkin system, $X \in \DD$.
By Set Difference with Self is Empty Set, $X \setminus X = \O$.
Hence, by property $(2)$ of a Dynkin system, $\O = X \setminus X \in \DD$.
{{qed}}
Theorem: Dynkin System Closed under Disjoint Union

Let $X$ be a set, and let $\DD$ be a Dynkin system on $X$.
Let $D, E \in \DD$ be disjoint.
Then the union $D \cup E$ is also an element of $\DD$.

Proof:
Define $D_1 = D, D_2 = E$, and for $n \ge 2$, $D_n = \O$.
Then by Dynkin System Contains Empty Set:
:$\forall n \in \N: D_n \in \DD$
Also, by Intersection with Empty Set, it follows that $\sequence {D_n}_{n \mathop \in \N}$ is a pairwise disjoint sequence.
Hence, by property $(3)$ of a Dynkin system:
:$\ds D \cup E = \bigcup_{n \mathop \in \N} D_n \in \DD$
{{qed}}
Theorem: Sigma-Algebra is Dynkin System

Let $X$ be a set, and let $\Sigma$ be a $\sigma$-algebra on $X$.
Then $\Sigma$ is a Dynkin system on $X$.

Proof:
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}}
Theorem: Existence and Uniqueness of Dynkin System Generated by Collection of Subsets

Let $X$ be a set.
Let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.
Then $\map \delta \GG$, the Dynkin system generated by $\GG$, exists and is unique.

Proof:
=== Existence ===
By Power Set is Dynkin System, there exists at least one Dynkin system containing $\GG$.
Next, let $\Bbb D$ be the collection of Dynkin systems containing $\GG$:
:$\Bbb D := \set {\DD': \GG \subseteq \DD', \text{$\DD'$ is a Dynkin system} }$
By Intersection of Dynkin Systems is Dynkin System, $\DD := \bigcap \Bbb D$ is a Dynkin system.
Also, by Set Intersection Preserves Subsets:
:$\GG \subseteq \DD$
Now let $\DD'$ be a Dynkin system containing $\GG$.
By construction of $\DD$, and Intersection is Subset: General Result:
:$\DD \subseteq \DD'$
{{qed|lemma}}
Theorem: Generated Sigma-Algebra Contains Generated Dynkin System

Let $X$ be a set.
Let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.
Then $\map \delta \GG \subseteq \map \sigma \GG$.
Here $\delta$ denotes generated Dynkin system, and $\sigma$ denotes generated $\sigma$-algebra.

Proof:
By Sigma-Algebra is Dynkin System, $\map \sigma \GG$ is a Dynkin system.
The definition of $\map \delta \GG$ now ensures that $\map \delta \GG \subseteq \map \sigma \GG$.
{{qed}}
Theorem: Dynkin System Closed under Intersections is Sigma-Algebra

Let $X$ be a set, and let $\DD$ be a Dynkin system on $X$.
Suppose that $\DD$ satisfies the following condition:
:$(1):\quad \forall D, E \in \DD: D \cap E \in \DD$
That is, $\DD$ is closed under intersection.
Then $\DD$ is a $\sigma$-algebra.

Proof:
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:
:$\ds E_n := D_n \cap \paren {X \setminus \bigcup_{m \mathop < n} D_m}$
By Dynkin System Closed under Union and $(1)$, it follows that $E_n \in \DD$ for all $n \in \N$.
Theorem: Carathéodory's Theorem (Measure Theory)

Let $X$ be a set.
Let $\SS \subseteq \powerset X$ be a semi-ring of subsets of $X$.
Let $\mu: \SS \to \overline \R$ be a pre-measure on $\SS$.
Let $\map \sigma \SS$ be the $\sigma$-algebra generated by $\SS$.
Then $\mu$ extends to a measure $\mu^*$ on $\map \sigma \SS$.

Proof:
{{ProofWanted}}
{{Namedfor|Constantin Carathéodory|cat = Carathéodory}}
Theorem: Congruence Modulo Normal Subgroup is Congruence Relation

Let $\struct {G, \circ}$ be a group.
Let $N$ be a normal subgroup of $G$.
Then congruence modulo $N$ is a congruence relation for the group operation $\circ$.

Proof:
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 = {{Group-axiom|1}}
}}
{{eqn | r = \paren {x \circ \paren {y \circ y'^{-1} } } \circ x'^{-1}
      | c = {{Group-axiom|1}}
}}
{{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}}
Theorem: Intersection Measure is Measure

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $F \in \Sigma$ be a measurable set.
Then the intersection measure $\mu_F$ is a measure on the measurable space $\struct {X, \Sigma}$.

Proof:
Verify the axioms for a measure in turn for $\mu_F$:
Theorem: Lebesgue Measure is Invariant under Translations

Let $\lambda^n$ be the $n$-dimensional Lebesgue measure on $\R^n$ equipped with the Borel $\sigma$-algebra $\map \BB {\R^n}$.
Let $\mathbf x \in \R^n$.
Then $\lambda^n$ is translation invariant; that is, for all $B \in \map \BB {\R^n}$, have:
:$\map {\lambda^n} {\mathbf x + B} = \map {\lambda^n} B$
where $\mathbf x + B$ is the set $\set {\mathbf x + \mathbf b: \mathbf b \in B}$.

Proof:
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 $\map \BB {\R^n} \, / \, \map \BB {\R^n}$-measurable.
Consider the pushforward measure $\lambda^n_{\mathbf x} := \paren {\tau_{\mathbf x} }_* \lambda^n$ on $\map \BB {\R^n}$.
By Characterization of Euclidean Borel Sigma-Algebra, it follows that:
:$\map \BB {\R^n} = \map \sigma {\JJ^n_{ho} }$
where $\sigma$ denotes generated $\sigma$-algebra, and $\JJ^n_{ho}$ 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:
:$J_k := \hointr {-k} k^n$
For condition $(3)$, let $\horectr {\mathbf a} {\mathbf b} \in \JJ^n_{ho}$ be a half-open $n$-rectangle.
Since:
:$\map {\tau_{\mathbf x}^{-1} } {\horectr {\mathbf a} {\mathbf b} } = \mathbf x + \horectr {\mathbf a} {\mathbf b} = \horectr {\mathbf {a + x} } {\mathbf {b + x} }$
we have:
{{begin-eqn}}
{{eqn | l = \map {\lambda^n_{\mathbf x} } {\horectr {\mathbf a} {\mathbf b} }
      | r = \map {\lambda^n} {\horectr {\mathbf {a + x} } {\mathbf {b + x} } }
      | c = {{Defof|Pushforward Measure}}
}}
{{eqn | r = \prod_{i \mathop = 1}^n \paren {\paren {b_i + x_i} - \paren {a_i + x_i} }
      | c = {{Defof|Lebesgue Measure}}
}}
{{eqn | r = \prod_{i \mathop = 1}^n \paren {b_i - a_i}
}}
{{eqn | r = \map {\lambda^n} {\horectr {\mathbf a} {\mathbf b} }
      | c = {{Defof|Lebesgue Measure}}
}}
{{end-eqn}}
Finally, since:
:$\ds \map {\lambda^n} {J_k} = \prod_{i \mathop = 1}^n \paren {k - \paren {-k} } = \paren {2 k}^n$
the last condition, $(4)$, is also satisfied.
Whence Uniqueness of Measures implies that:
:$\lambda^n_{\mathbf x} = \lambda^n$
and since for all $B \in \map \BB {\R^n}$ we have:
:$\mathbf x + B = \map {\tau_{\mathbf x}^{-1} } B$
this precisely boils down to:
:$\map {\lambda^n} {\mathbf x + B} = \map {\lambda^n} B$
{{qed}}
Theorem: Translation Invariant Measure on Euclidean Space is Multiple of Lebesgue Measure

Let $\mu$ be a measure on $\R^n$ equipped with the Borel $\sigma$-algebra $\map \BB {\R^n}$.
Suppose that $\mu$ is translation invariant.
Also, suppose that $\kappa := \map \mu {\hointr 0 1^n} < +\infty$.
Then $\mu = \kappa \lambda^n$, where $\lambda^n$ is the $n$-dimensional Lebesgue measure.

Proof:
From Characterization of Euclidean Borel Sigma-Algebra, we have:
:$\map \BB {\R^n} = \map \sigma {\JJ^n_{ho, \text {rat} } }$
where $\JJ^n_{ho, \text {rat} }$ denotes the collection of half-open $n$-rectangles with rational endpoints.
So let $J = \horectr {\mathbf a} {\mathbf b} \in \JJ^n_{ho, \text {rat} }$.
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 $\hointr 0 {\dfrac 1 M}^n$, in that:
:$\ds J = \bigcup_{i \mathop = 1}^{\map k J} \mathbf x_i + \hointr 0 {\dfrac 1 M}^n$
for some $\map k J \in \N$ and suitable $\mathbf x_i$, where:
:$\mathbf x_i + \hointr 0 {\dfrac 1 M}^n := \horectr {\mathbf x_i} {\mathbf x_i + \dfrac 1 M}$
Using that $\mu$ is translation invariant, this means:
:$\map \mu J = \map k J \, \map \mu {\hointr 0 {\dfrac 1 M}^n}$
Also, by Lebesgue Measure is Translation Invariant:
:$\map {\lambda^n} J = \map k J \, \map {\lambda^n} {\hointr 0 {\dfrac 1 M}^n}$
A moment's thought shows us that the half-open $n$-rectangle $I = \hointr 0 1^n$ may be covered by $M^n$ copies of $\hointr 0 {\dfrac 1 M}^n$, so that:
:$\map k I = M^n$
For brevity, write $I / M$ for $\hointr 0 {\dfrac 1 M}^n$.
Now, using that:
:$\ds \map {\lambda^n} {I / M} = \prod_{i \mathop = 1}^n \frac 1 M = \frac 1 {M^n}$
and $\map \mu I = \kappa$, compute:
{{begin-eqn}}
{{eqn | l = \map \mu J
      | r = \map k J \map \mu {I / M}
      | c = Definition of $\map k J$
}}
{{eqn | r = \frac {\map k J} {M^n} \paren {M^n \map \mu {I / M} }
      | c = $M^n / M^n = 1$
}}
{{eqn | r = \frac {\map k J} {M^n} \map \mu I
      | c = $M^n = \map k I$
}}
{{eqn | r = \frac {\kappa \, \map k J} {M^n}
      | c = Definition of $\kappa$
}}
{{eqn | r = \kappa \, \map k J \, \map {\lambda^n} {I / M}
}}
{{eqn | r = \kappa \, \map {\lambda^n} J
      | c = Definition of $\map k J$
}}
{{end-eqn}}
Therefore, $\map \mu J = \kappa \, \map {\lambda^n} J$ for all $J \in \JJ^n_{ho, \text {rat} }$.
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 $\hointr {-k} k^n \mathop \uparrow \R^n$
Finally, for $(4)$, we recall that $\kappa < +\infty$.
Thus, by Uniqueness of Measures, $\mu = \kappa \lambda^n$.
{{qed}}
Theorem: Dynkin System Closed under Set Difference with Subset

Let $X$ be a set.
Let $\DD$ be a Dynkin system on $X$.
Let $D, E \in \DD$ and suppose that $E \subseteq D$.
Then the set difference $D \setminus E$ is also an element of $\DD$.

Proof:
For brevity, write for example $E^\complement$ for $\relcomp X E = X \setminus E$.
We reason as follows:
{{begin-eqn}}
{{eqn | l = D \setminus E
      | r = D \cap E^\complement
      | c = Set Difference as Intersection with Relative Complement
}}
{{eqn | r = \paren {D^\complement \cup E}^\complement
      | c = De Morgan's Laws: Complement of Union, Relative Complement of Relative Complement
}}
{{end-eqn}}
Now this implies that $D \setminus E \in \DD$ {{iff}} $D^\complement \cup E \in \DD$.
It is already known that $D^\complement$ and $E$ are in $\DD$ by axiom $(2)$ for a Dynkin system.
Since $E \subseteq D$, it follows that $D^\complement \cap E = \O$, and thus Dynkin System Closed under Disjoint Union applies to give:
:$D^\complement \cup E \in \DD$
which, combined with above reasoning, yields $D \setminus E \in \DD$.
{{qed}}
Theorem: Open Rectangles Closed under Intersection

Let $\paren {\openint {\mathbf a} {\mathbf b} }$ and $\paren {\openint {\mathbf c} {\mathbf d} }$ be open $n$-rectangles.
Then $\paren {\openint {\mathbf a} {\mathbf b} } \cap \paren {\openint {\mathbf c} {\mathbf d} }$ is also an open $n$-rectangle.

Proof:
From Cartesian Product of Intersections: General Case, we have:
:$\ds \paren {\openint {\mathbf a} {\mathbf b} } \cap \paren {\openint {\mathbf c} {\mathbf d} } = \prod_{i \mathop = 1}^n \openint {a_i} {b_i} \cap \openint {c_i} {d_i}$
Therefore, it suffices to show that the intersection of two open intervals is again an open interval.
Now let $x \in \openint {a_i} {b_i} \cap \openint {c_i} {d_i}$.
Then $x$ is subject to:
:$x > a_i$ and $x > c_i$, that is: $x > \max \set {a_i, c_i}$
:$x < b_i$ and $x < d_i$, that is: $x < \min \set {b_i, d_i}$
and we see that these conditions are satisfied {{iff}}:
:$x \in \openint {\max \set {a_i, c_i} } {\min \set {b_i, d_i} }$
Thus, we conclude:
:$\openint {a_i} {b_i} \cap \openint {c_i} {d_i} = \openint {\max \set {a_i, c_i} } {\min \set {b_i, d_i} }$
showing that indeed the intersection is an open interval.
Combining this with the above reasoning, it follows that indeed:
:$\paren {\openint {\mathbf a} {\mathbf b} } \cap \paren {\openint {\mathbf c} {\mathbf d} }$
is an open $n$-rectangle.
{{qed}}
Category:Analysis
1cepjj14i0c00q59k2bf81jjdw8b3de
Theorem: Euclidean Borel Sigma-Algebra Closed under Scalar Multiplication

Let $\map \BB {\R^n}$ be the Borel $\sigma$-algebra on $\R^n$.
Let $B \in \BB$, and let $t \in \R_{>0}$.
Then also $t \cdot B := \set {t \mathbf b: \mathbf b \in B} \in \BB$.

Proof:
Define $f: \R^n \to \R^n$ by $\map f {\mathbf x} := \dfrac 1 t \mathbf x$.
Then for all $\mathbf x \in \R^n$, $\map {f^{-1} } {\mathbf x} = \set {t \mathbf x}$, where $f^{-1}$ denotes the preimage of $f$.
Thus $\ds t \cdot B = \bigcup_{\mathbf b \mathop \in B} \map {f^{-1} } {\mathbf b} = \map {f^{-1} } B$, where the last equality holds by definition of preimage.
It follows that the statement of the theorem comes down to showing that $f$ is $\BB \, / \, \BB$-measurable.
By Characterization of Euclidean Borel Sigma-Algebra the half-open $n$-rectangles $\JJ_{ho}^n$ generate $\BB$.
Applying Mapping Measurable iff Measurable on Generator, it suffices to demonstrate:
:$\forall J \in \JJ_{ho}^n: \map {f^{-1} } J \in \BB$
Now for a half-open $n$-rectangle $J = \horectr {\mathbf a} {\mathbf b}$, it follows from the consideration on $f^{-1}$ above, that:
:$\map {f^{-1} } J = \horectr {t \mathbf a} {t \mathbf b}$
which is again in $\JJ_{ho}^n$ and so, in particular, in $\BB$.
Thus $f$ is $\BB \, / \, \BB$-measurable, that is:
:$\forall B \in \BB: t \cdot B = \map {f^{-1} } B \in \BB$
which was to be demonstrated.
{{qed}}
Theorem: Lebesgue Measure of Scalar Multiple

Let $\lambda^n$ be the $n$-dimensional Lebesgue measure on $\R^n$ equipped with the Borel $\sigma$-algebra $\map \BB {\R^n}$.
Let $B \in \BB$.
Let $t \in \R_{>0}$.
Then:
:$\map {\lambda^n} {t \cdot B} = t^n \map {\lambda^n} B$
where $t \cdot B$ is the set $\set {t \mathbf b: \mathbf b \in B}$.

Proof:
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:
:$\map \det {t \, \mathbf I_n} = t^n$
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 = \map {\lambda^n} {t \cdot B}
     |r = \map {\lambda^n} {\map {\paren {t \, \mathbf I_n} } B}
}}
{{eqn|r = \map {\lambda^n} {\map {\paren {t^{-1} \, \mathbf I_n}^{-1} } B}
     |c = Inverse of Group Inverse
}}
{{eqn|r = \map {\paren {\paren {t^{-1} \, \mathbf I_n}_* \lambda^n} } B
     |c = {{Defof|Pushforward Measure}}
}}
{{eqn|r = \size {\map \det {\paren {t^{-1} \, \mathbf I_n}^{-1} } } \cdot \map {\lambda^n} B
     |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 $\map \det {\paren {t^{-1} \, \mathbf I_n}^{-1} } = \map \det {t \, \mathbf I_n} = t^n$.
Since $t > 0$, $\size {t^n} = t^n$, and the result follows.
{{qed}}
Theorem: Measure Invariant on Generator is Invariant

Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\theta: X \to X$ be an $\Sigma / \Sigma$-measurable mapping.
Suppose that $\Sigma$ is generated by $\GG \subseteq \powerset X$.
Also, let $\GG$ satisfy the following:
:$(1): \quad \forall G, H \in \GG: G \cap H \in \GG$
:$(2): \quad$ There exists an exhausting sequence $\sequence {G_n}_{n \mathop \in \N} \uparrow X$ in $\GG$ such that:
:::$\quad \forall n \in \N: \map \mu {G_n} < +\infty$
Suppose furthermore that, for all $G \in \GG$, $\mu$ satisfies:
:$(3): \quad \map \mu {\theta^{-1} \sqbrk G} = \map \mu G$
Then $\mu$ is a $\theta$-invariant measure.

Proof:
Consider the pushforward measure $\theta_* \mu$ on $\struct {X, \Sigma}$.
By definition, this makes equation $(3)$ come down to:
:$\theta_* \mu \restriction_\GG = \mu \restriction_\GG$
where $\restriction$ denotes restriction.
The suppositions $(1)$, $(2)$ and $(3)$ together constitute precisely the prerequisites to Uniqueness of Measures.
Hence $\theta_* \mu = \mu$, that is, $\mu$ is $\theta$-invariant.
{{qed}}
Theorem: Sigma-Algebras with Independent Generators are Independent

Let $\struct {\Omega, \EE, \Pr}$ be a probability space.
Let $\Sigma, \Sigma'$ be sub-$\sigma$-algebras of $\EE$.
Suppose that $\GG, \HH$ are $\cap$-stable generators for $\Sigma, \Sigma'$, respectively.
Suppose that, for all $G \in \GG, H \in \HH$:
:$(1): \quad \map \Pr {G \cap H} = \map \Pr G \map \Pr H$
Then $\Sigma$ and $\Sigma'$ are $\Pr$-independent.

Proof:
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 $E' \in \Sigma'$:
:$\map {\mu'_E} {E'} := \map \Pr {E \cap E'}$
:$\map {\nu'_E} {E'} := \map \Pr E \map \Pr {E'}$
And {{mutatis}}, 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}}