avewright/proofwiki-math huggingface dataset:
Theorem: Bounds for Cumulative Distribution Function

:$0 \le \map {F_X} x \le 1$ for each $x \in \R$

Proof:
From the definition of the cumulative distribution function, we have: 
:$\map {F_X} x = \map \Pr {X \le x}$
for each $x \in \R$.
We have: 
:$\O \subseteq \set {\omega \in \Omega : \map X \omega \le x} \subseteq \Omega$
So, from Measure is Monotone, we have: 
:$\map \Pr \O \le \map \Pr {X \le x} \le \map \Pr \Omega$
From the definition of a probability measure, we have: 
:$\map \Pr \O = 0$
and:
:$\map \Pr \Omega = 1$
so:
:$0 \le \map \Pr {X \le x} \le 1$
for each $x \in \R$.
{{qed}}
Category:Cumulative Distribution Functions
oiv5hpj6256y91jxk6db9l9gw4chwz2
Theorem: Cumulative Distribution Function is Increasing

:$F_X$ is an increasing function.

Proof:
Let $x, y \in \R$ have $x \le y$. 
Note that if $\omega \in \Omega$ is such that: 
:$\map X \omega \le x$ 
then: 
:$\map X \omega \le y$
so:
:$\set {\omega \in \Omega : \map X \omega \le x} \subseteq \set {\omega \in \Omega : \map X \omega \le y}$
From Measure is Monotone, we then have: 
:$\map \Pr {X \le x} \le \map \Pr {X \le y}$
That is, from the definition of cumulative distribution function, we have: 
:$\map {F_X} x \le \map {F_X} y$ 
whenever $x \le y$. 
So:
:$F_X$ is an increasing function.
{{qed}}
Theorem: Limit of Cumulative Distribution Function at Positive Infinity

:$\ds \lim_{x \mathop \to \infty} \map {F_X} x = 1$
where $\ds \lim_{x \mathop \to \infty} \map {F_X} x$ denotes the limit at $+\infty$ of $F_X$.

Proof:
From Sequential Characterization of Limit at Positive Infinity of Real Function: Corollary, we aim to show that: 
:for all increasing real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to \infty$ we have $\map {F_X} {x_n} \to 1$
at which point we will obtain: 
:$\ds \lim_{x \mathop \to \infty} \map {F_X} x = 1$
Since $\sequence {x_n}_{n \mathop \in \N}$ is increasing:
:the sequence $\sequence {\hointl {-\infty} {x_n} }_{n \mathop \in \N}$ is increasing.
We show that: 
:$\ds \bigcup_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \R$
Theorem: Limit of Cumulative Distribution Function at Negative Infinity

:$\ds \lim_{x \mathop \to -\infty} \map {F_X} x = 0$
where $\ds \lim_{x \mathop \to -\infty} \map {F_X} x$ denotes the limit at $-\infty$ of $F_X$.

Proof:
From Sequential Characterisation of Limit at Minus Infinity of Real Function: Corollary, we aim to show that: 
:for all decreasing real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to -\infty$ we have $\map {F_X} {x_n} \to 0$
at which point we will obtain: 
:$\ds \lim_{x \mathop \to -\infty} \map {F_X} x = 0$
Since $\sequence {x_n}_{n \mathop \in \N}$ is decreasing, we have: 
:the sequence $\sequence {\hointl {-\infty} {x_n} }_{n \mathop \in \N}$ is decreasing.
From Limit of Decreasing Sequence of Unbounded Below Closed Intervals with Endpoint Tending to Negative Infinity, we have: 
:$\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \O$
Let $P_X$ be the probability distribution of $X$. 
From Probability Distribution is Probability Measure, we have: 
:$P_X$ is a finite measure.
From Measure of Limit of Decreasing Sequence of Measurable Sets, we have: 
:$\ds \map {P_X} \O = \lim_{n \mathop \to \infty} \map {P_X} {\hointl {-\infty} {x_n} }$ 
So, we obtain:
{{begin-eqn}}
{{eqn | l = 0
      | r = \map {P_X} \O
      | c = Probability Distribution is Probability Measure, {{Defof|Probability Measure}}
}}
{{eqn | r = \lim_{n \mathop \to \infty} \map {P_X} {\hointl {-\infty} {x_n} }
}}
{{eqn | r = \lim_{n \mathop \to \infty} \map {F_X} {x_n}
      | c = {{Defof|Cumulative Distribution Function}}
}}
{{end-eqn}}
Since $\sequence {x_n}_{n \mathop \in \N}$ was arbitrary, we have: 
:for all decreasing real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to -\infty$ we have $\map {F_X} {x_n} \to 0$.
So, from Sequential Characterisation of Limit at Minus Infinity of Real Function: Corollary, we have: 
:$\ds \lim_{x \mathop \to \infty} \map {F_X} x = 0$
{{qed}}
Theorem: Absolute Value of Real-Valued Random Variable is Real-Valued Random Variable

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. 
Let $X$ be a real-valued random variable. 
Then $\size X$ is a real-valued random variable.

Proof:
Since $X$ is a real-valued random variable, $X$ is $\Sigma$-measurable.
From Absolute Value of Measurable Function is Measurable, $\size X$ is $\Sigma$-measurable.
So $\size X$ is a real-valued random variable.
{{qed}}
Category:Random Variables
e13aoey7fx5ems1r5rfwtvrr4sbpnp6
Theorem: Positive Part of Real-Valued Random Variable is Real-Valued Random Variable

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. 
Let $X$ be a real-valued random variable. 
Then the positive part $X^+$ of $X$ is a real-valued random variable.

Proof:
Since $X$ is a real-valued random variable, $X$ is $\Sigma$-measurable.
From Function Measurable iff Positive and Negative Parts Measurable, $X^+$ is $\Sigma$-measurable.
So $X^+$ is a real-valued random variable.
{{qed}}
Category:Random Variables
Category:Positive Parts
pm3g79v0jvdzyrqtauuibf8k4a9yiai
Theorem: Negative Part of Real-Valued Random Variable is Real-Valued Random Variable

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. 
Let $X$ be a real-valued random variable. 
Then the negative part $X^-$ of $X$ is a real-valued random variable.

Proof:
Since $X$ is a real-valued random variable, $X$ is $\Sigma$-measurable.
From Function Measurable iff Positive and Negative Parts Measurable, $X^-$ is $\Sigma$-measurable.
So $X^-$ is a real-valued random variable.
{{qed}}
Category:Random Variables
Category:Negative Parts
suuv132d4rwhvo0gzxracwbbjw9ppte
Theorem: Probability Distribution is Probability Measure

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. 
Let $\struct {S, \Sigma'}$ be a measurable space.
Let $X$ be a random variable on $\struct {\Omega, \Sigma, \Pr}$ taking values in $\struct {S, \Sigma'}$. 
Let $P_X$ be the probability distribution of $X$.
Then: 
:$P_X$ is a probability measure on $\struct {S, \Sigma'}$.

Proof:
From the definition of probability distribution, we have: 
:$P_X = X_* \Pr$
where $X_* \Pr$ is the pushforward $X_* \Pr$ of $\Pr$, under $X$, on $\Sigma'$.
From Pushforward Measure is Measure, we have: 
:$P_X$ is a measure.
We then have: 
{{begin-eqn}}
{{eqn	| l = \map {P_X} S
		| r = \map \Pr {X^{-1} \sqbrk S}
		| c = {{Defof|Probability Distribution}}
}}
{{eqn	| r = \map \Pr \Omega
}}
{{eqn	| r = 1
		| c = {{Defof|Probability Measure}}
}}
{{end-eqn}}
so:
:$P_X$ is a probability measure. 
{{qed}}
Category:Probability Distributions
j66xxlhvl6z2izj1jrcfi1zy0xltxd5
Theorem: Limit of Decreasing Sequence of Unbounded Below Closed Intervals with Endpoint Tending to Negative Infinity

Let $\sequence {x_n}_{n \mathop \in \N}$ be a decreasing sequence with $x_n \to -\infty$.
Then: 
:$\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \O$
That is: 
:$\hointl {-\infty} {x_n} \downarrow \O$
where $\downarrow$ denotes the limit of decreasing sequence of sets.

Proof:
{{AimForCont}} suppose that:
:$\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} \ne \O$
Let: 
:$\ds x \in \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n}$
Then: 
:$x \in \hointl {-\infty} {x_n}$ for each $n$.
From the definition of a sequence diverging to $-\infty$:
:there exists $N \in \N$ such that $x_N < x$.
But then: 
:$x \not \in \hointl {-\infty} {x_N}$
contradicting that: 
:$x \in \hointl {-\infty} {x_n}$ for each $n$.
So we have reached a contradiction, and we have: 
:$\ds \bigcap_{n \mathop = 1}^\infty \hointl {-\infty} {x_n} = \O$
{{qed}}
Category:Decreasing Sequences of Sets
9bbjvjbyifry3vx132vg7vzk78nxr0b
Theorem: Subsequence of Real Sequence Diverging to Positive Infinity Diverges to Positive Infinity

Let $\sequence {x_n}_{n \mathop \in \N}$ be a real sequence with $x_n \to +\infty$.
Let $\sequence {x_{n_j} }_{j \mathop \in \N}$ be a subsequence of $\sequence {x_n}_{n \mathop \in \N}$.
Then: 
:$x_{n_j} \to +\infty$

Proof:
Let $M > 0$ be a real number.
From the definition of divergence to $+\infty$, there exists $N \in \N$ such that: 
:$x_n > M$
for each $n \ge N$.
From Strictly Increasing Sequence of Natural Numbers, we have: 
:$n_j \ge j$ for each $j$.
So, we have: 
:$x_{n_j} > M$
for each $j \ge N$.
Since $M$ was arbitrary, we have: 
:$x_{n_j} \to \infty$
by the definition of divergence to $+\infty$.
{{qed}}
Category:Divergent Sequences
Category:Limits of Sequences
2xyj26679fhzq5zc1zwom9l06e9q7n7
Theorem: Subsequence of Real Sequence Diverging to Negative Infinity Diverges to Negative Infinity

Let $\sequence {x_n}_{n \mathop \in \N}$ be a real sequence with $x_n \to -\infty$.
Let $\sequence {x_{n_j} }_{j \mathop \in \N}$ be a subsequence of $\sequence {x_n}_{n \mathop \in \N}$.
Then: 
:$x_{n_j} \to -\infty$

Proof:
Let $M > 0$ be a real number.
From the definition of divergence to $-\infty$, there exists $N \in \N$ such that: 
:$x_n < -M$
for each $n \ge N$.
From Strictly Increasing Sequence of Natural Numbers, we have: 
:$n_j \ge j$ for each $j$.
So, we have: 
:$x_{n_j} < -M$
for each $j \ge N$.
Since $M$ was arbitrary, we have: 
:$x_{n_j} \to -\infty$
by the definition of divergence to $-\infty$.
{{qed}}
Category:Divergent Sequences
Category:Limits of Sequences
bll8qwk726ks77peo22meqkni0q6gei
Theorem: Area of Equilateral Triangle

Let $T$ be an equilateral triangle.
Let the length of one side of $T$ be $s$.
Let $\AA$ be the area of $T$.
Then:
:$\AA = \dfrac {s^2 \sqrt 3} 4$

Proof:
:300px
From Area of Triangle in Terms of Two Sides and Angle:
:$\AA = \dfrac {s^2} 2 \sin 60 \degrees$
From Sine of $60 \degrees$:
:$\sin 60 \degrees = \dfrac {\sqrt 3} 2$
The result follows.
{{qed}}
Category:Areas of Triangles
Category:Equilateral Triangles
gb7x34wc74pnf5o6tyja9n96y41qp6d
Theorem: Sequential Characterization of Limit at Minus Infinity of Real Function

Let $f : \R \to \R$ be a real function. 
Let $L$ be a real number. 
Then: 
:$\ds \lim_{x \to -\infty} \map f x = L$ 
{{iff}}: 
:for all real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to -\infty$ we have $\map f {x_n} \to L$
where:
:$\ds \lim_{x \mathop \to -\infty} \map f x$ denotes the limit at $-\infty$ of $f$.

Proof:
=== Necessary Condition ===
Suppose that: 
:$\ds \lim_{x \to -\infty} \map f x = L$
Let $\sequence {x_n}_{n \mathop \in \N}$ be a real sequence with $x_n \to -\infty$. 
Let $\epsilon > 0$. 
From the definition of limit at $-\infty$, we have:
:there exists $M > 0$ such that for all $x < -M$ we have $\size {\map f x - L} < \epsilon$.
Since $\sequence {x_n}_{n \mathop \in \N}$ diverges to $-\infty$, we have: 
:there exists $N \in \N$ such that $x_n < -M$ for all $n \ge N$. 
That is, for all $n \ge N$, we have: 
:$\size {\map f {x_n} - L} < \epsilon$ 
Since $\epsilon$ was arbitrary, we have: 
:$\map f {x_n} \to L$ 
Since $\sequence {x_n}_{n \mathop \in \N}$ was arbitrary: 
:for all real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to -\infty$ we have $\map f {x_n} \to L$. 
So, if: 
:$\ds \lim_{x \to -\infty} \map f x = L$
then:
:for all real sequences $\sequence {x_n}_{n \mathop \in \N}$ with $x_n \to -\infty$ we have $\map f {x_n} \to L$. 
{{qed|lemma}}
Theorem: Lebesgue Infinity-Space is Subset of Tempered Distribution Space

Let $\map {L^\infty} \R$ be the Lebesgue infinity-space.
Let $\map {\SS'} \R$ be the tempered distribution space.
Then in the distributional sense it holds that:
:$\map {L^\infty} \R \subseteq \map {\SS'} \R$
That is, every Schwartz distribution defined by an element of $\map {L^\infty} \R$ is a tempered distribution.

Proof:
Let $f \in \map {L^\infty} \R$.
Let $\phi \in \map \SS \R$ be a Schwartz test function.
Let $T_f : \map \SS \R \to \R$ be a mapping such that:
:$\ds \map {T_f} \phi = \int_\R \map f x \map \phi x \rd x$
Then:
{{begin-eqn}}
{{eqn | l = \size {\map {T_f} \phi}
      | r = \size {\int_\R \map f x \map \phi x \rd x}
}}
{{eqn | o = \le
      | r = \norm f_\infty \int_{-\infty}^\infty \size {\map \phi x} \rd x
      | c = {{Defof|Supremum Norm}}
}}
{{eqn | r = \norm f_\infty \int_{-\infty}^\infty \frac 1 {1 + x^2} \paren {\size {\map \phi x} + x^2 \size {\map \phi x} } \rd x
}}
{{eqn | o = \le
      | r = \norm f_\infty \paren {\int_{-\infty}^\infty \frac 1 {1 + x^2} \rd x} \paren {\norm \phi_\infty + \norm {x^2 \phi}_\infty}
}}
{{end-eqn}}
Let $\sequence {\phi_n}_{n \mathop \in \N}$ be a sequence in $\map \SS \R$.
Suppose, $\sequence {\phi_n}_{n \mathop \in \N}$ converges in $\map \SS \R$ to $\mathbf 0$:
:$\phi_n \stackrel \SS \longrightarrow \mathbf 0$
That is:
:$\ds \forall l, m \in \N : \lim_{n \mathop \to \infty} \sup_{x \mathop \in \R} \size {x^l \map {\phi_n^{\paren m} } x} = 0$
Specifically:
:$\ds \lim_{n \mathop \to \infty} \sup_{x \mathop \in \R} \size {\map {\phi_n} x} = 0$
:$\ds \lim_{n \mathop \to \infty} \sup_{x \mathop \in \R} \size {x^2 \map {\phi_n} x} = 0$
Then:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \map {T_f} {\phi_n}
      | o = \le
      | r = \norm f_\infty \paren {\int_{-\infty}^\infty \frac 1 {1 + x^2} \rd x} \lim_{n \mathop \to \infty} \paren {\norm {\phi_n}_\infty + \norm {x^2 \phi_n}_\infty}
}}
{{eqn | r = 0
}}
{{end-eqn}}
On the other hand:
:$\map {T_f} {\mathbf 0} = 0$
where $\mathbf 0$ denotes the zero Schwartz test function.
Hence:
:$T_f \in \map {\SS'} \R$
{{qed}}
Theorem: Linear Combination of Real-Valued Random Variables is Real-Valued Random Variable

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ and $Y$ be real-valued random variables on $\struct {\Omega, \Sigma, \Pr}$.
Let $\alpha$ and $\beta$ be real numbers.
Then: 
:$\alpha X + \beta Y$ is a real-valued random variable.

Proof:
Since $X$ and $Y$ are real-valued random variables, we have: 
:$X$ and $Y$ are $\Sigma$-measurable functions.
From Pointwise Scalar Multiple of Measurable Function is Measurable, we have: 
:$\alpha X$ and $\beta Y$ are $\Sigma$-measurable.
From Pointwise Sum of Measurable Functions is Measurable, we have: 
:$\alpha X + \beta Y$ is $\Sigma$-measurable.
So:
:$\alpha X + \beta Y$ is a real-valued random variable.
{{qed}}
Category:Random Variables
Category:Linear Combination of Real-Valued Random Variables is Real-Valued Random Variable
3bl08zr34sybx81al1k1iy7zg3eonbc
Theorem: Linear Combination of Real-Valued Random Variables is Real-Valued Random Variable/General Result

Let $n \in \N$. 
Let $\sequence {X_i}_{i \mathop \in \N}$ be a sequence of real-valued random variables on $\struct {\Omega, \Sigma, \Pr}$.
Let $\sequence {\alpha_i}_{i \mathop \in \N}$ be a sequence of real numbers. 
Then: 
:$\ds \sum_{i \mathop = 1}^n \alpha_i X_i$ is a real-valued random variable.

Proof:
We proceed by induction.
For all $n \in \N$ let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 1}^n \alpha_i X_i$ is $\Sigma$-measurable.
Theorem: Absolutely Continuous Random Variable is Continuous

Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be an absolutely continuous random variable on $\struct {\Omega, \Sigma, \Pr}$. 
Then $X$ is a continuous random variable.

Proof:
Let $F_X$ be the cumulative distribution function for $X$. 
From the definition of an absolutely continuous random variable, we have: 
:$F_X$ is absolutely continuous.
From Absolutely Continuous Real Function is Continuous, we then have: 
:$F_X$ is continuous.
Then, from the definition of a continuous random variable, we have: 
:$X$ is a continuous random variable.
{{qed}}
Category:Absolutely Continuous Random Variables
Category:Continuous Random Variables
smjado41v7fdgapvfukl0yj18yn4u0o
Theorem: Fourier Transform of Tempered Distribution on 1-Lebesgue Space equals Tempered Distribution of Fourier Transform of defining Function

Let $\map {L^1} \R$ be the Lebesgue $1$-space.
Let $f \in \map {L^1} \R$.
Let $T_f \in \map {\SS'} \R$ be a tempered distribution associated with $f$.
Then:
:$\hat T_f = T_{\hat f}$
where:
:$\hat T_f$ denotes the Fourier transform of the tempered distribution $T_f$
:$\hat f$  denotes the Fourier transform of the real function $f$.

Proof:
Let $\phi \in \map \SS \R$ be a Schwartz test function.
{{begin-eqn}}
{{eqn | l = \map { {\hat T}_f} \phi
      | r = \map {T_f} {\hat \phi}
      | c = {{Defof|Fourier Transform of Tempered Distribution}}
}}
{{eqn | r = \int_\R \map f x \map {\hat \phi} x \rd x
      | c = {{Defof|Tempered Distribution}}
}}
{{eqn | r = \int_{-\infty}^\infty \map f x \paren {\int_{-\infty}^\infty \map \phi \xi e^{-2\pi i x \xi} \rd \xi } \rd x
      | c = {{Defof|Fourier Transform of Real Function}}
}}
{{eqn | r = \int_{-\infty}^\infty \int_{-\infty}^\infty \map f x \map \phi \xi e^{-2\pi i x \xi} \rd \xi \rd x
}}
{{eqn | r = \int_{-\infty}^\infty \map \phi \xi \paren {\int_{-\infty}^\infty \map f x e^{-2\pi i x \xi} \rd x} \rd \xi
      | c = Fubini's Theorem
}}
{{eqn | r = \int_{-\infty}^\infty \map \phi \xi \map {\hat f} \xi \rd \xi
      | c = {{Defof|Fourier Transform of Real Function}}
}}
{{end-eqn}}
We have that the Fourier Transform of 1-Lebesgue Space Function is Bounded.
Then, from Lebesgue Infinity-Space is Subset of Tempered Distribution Space it follows that $\ds \int_{-\infty}^\infty \map \phi \xi \map {\hat f} \xi \rd \xi$ is a tempered distribution.
That is, by definition of tempered distribution:
:$\ds \int_{-\infty}^\infty \map \phi \xi \map {\hat f} \xi \rd \xi = T_{\hat f}$
{{qed}}
Theorem: Random Vector is Random Variable

Let $n \in \N$. 
Let $\struct {X, \Sigma, \Pr}$ be a probability space.
Let $\struct {S_1, \Sigma_1}, \struct {S_2, \Sigma_2}, \ldots, \struct {S_n, \Sigma_n}$ be measurable spaces.
Let: 
:$\ds S = \prod_{i \mathop = 1}^n S_i$
Let: 
:$\ds \Sigma' = \bigotimes_{i \mathop = 1}^n \Sigma_i$ 
where $\ds \bigotimes_{i \mathop = 1}^n \Sigma_i$ denotes the product $\sigma$-algebra of $\Sigma_1, \Sigma_2, \ldots, \Sigma_n$. 
For each $i \in \set {1, 2, \ldots, n}$:
:let $X_i$ be a random variable on $\struct {\Omega, \Sigma, \Pr}$ taking values in $\struct {S_i, \Sigma_i}$. 
Define a function $\mathbf X : \Omega \to S$ by: 
:$\map {\mathbf X} \omega = \tuple {\map {X_1} \omega, \map {X_2} \omega, \ldots, \map {X_n} \omega}$
for each $\omega \in \Omega$.
Then $\mathbf X$ is a random variable on $\struct {\Omega, \Sigma, \Pr}$ taking values in $\struct {S, \Sigma'}$.

Proof:
Note that from the definition of product $\sigma$-algebra: finite case, we have:
:$\ds \Sigma' = \map \sigma {\set {\prod_{i \mathop = 1}^n S_i : S_i \in \Sigma_i \text { for each } i \in \set {1, 2, \ldots, n} } }$
where $\sigma$ denotes the $\sigma$-algebra generated by a collection of subsets. 
We aim to apply Mapping Measurable iff Measurable on Generator.
We therefore aim to show that: 
:${\mathbf X}^{-1} \sqbrk B \in \Sigma$ 
for each: 
:$\ds B \in \set {\prod_{i \mathop = 1}^n E_i : E_i \in \Sigma_i \text { for each } i \in \set {1, 2, \ldots, n} }$
Let: 
:$\ds B \in \set {\prod_{i \mathop = 1}^n E_i : E_i \in \Sigma_i \text { for each } i \in \set {1, 2, \ldots, n} }$
Then, for each $i \in \set {1, 2, \ldots, n}$ there exists $E_i \in \Sigma_i$ such that: 
:$\ds B = \prod_{i \mathop = 1}^n E_i$ 
Then, for $\omega \in \Omega$, we have $\map {\mathbf X} \omega \in B$ {{iff}}:
:$\ds \map {X_i} \omega \in E_i$ for each $i \in \set {1, 2, \ldots, n}$.
That is, $\omega \in {\mathbf X}^{-1} \sqbrk B$ {{iff}}: 
:$\omega \in {X_i}^{-1} \sqbrk {E_i}$ for each $i \in \set {1, 2, \ldots, n}$.
So:
:$\ds {\mathbf X}^{-1} \sqbrk B = \bigcap_{i \mathop = 1}^n {X_i}^{-1} \sqbrk {E_i}$
Since $X_i$ is $\Sigma/\Sigma_i$-measurable, we have: 
:${X_i}^{-1} \sqbrk {E_i} \in \Sigma$ 
for each $i \in \set {1, 2, \ldots, n}$.  
From Sigma-Algebra Closed under Finite Intersection, we therefore have: 
:$\ds \bigcap_{i \mathop = 1}^n {X_i}^{-1} \sqbrk {E_i} \in \Sigma$
so:
:$\ds {\mathbf X}^{-1} \sqbrk B \in \Sigma$ 
which was the demand. 
From Mapping Measurable iff Measurable on Generator, we then have: 
:$\mathbf X$ is $\Sigma/\Sigma'$-measurable.
So:
:$\mathbf X$ is a random variable on $\struct {\Omega, \Sigma, \Pr}$ taking values in $\struct {S, \Sigma'}$
as required.
{{qed}}
Theorem: Fourier Transform of Dirac Delta Distribution

Let $\delta \in \map {\SS'} \R$ be the Dirac delta distribution.
Let $\mathbf 1 : \map \SS \R \to \R$ be the constant tempered distribution such that for all $\phi \in \map \SS \R$ we have:
:$\ds \map {\mathbf 1} \phi = \int_{-\infty}^\infty 1 \cdot \map \phi x \rd x$
Then in the distributional sense it holds that:
:$\hat \delta = \mathbf 1$
where the hat denotes the Fourier transform of a tempered distribution.

Proof:
Let $\phi \in \map \SS \R$ be a Schwartz test function.
Then:
{{begin-eqn}}
{{eqn | l = \map {\hat \delta} \phi
      | r = \map \delta {\hat \phi}
      | c = {{Defof|Fourier Transform of Tempered Distribution}}
}}
{{eqn | r = \map {\hat \phi} 0
      | c = {{Defof|Tempered Dirac Delta Distribution}}
}}
{{eqn | r = \int_{- \infty}^\infty \map \phi x e^{-2 \pi i 0 x} \rd x
      | c = {{Defof|Fourier Transform of Real Function}}
}}
{{eqn | r = \int_{- \infty}^\infty \map \phi x \cdot 1 \rd x
}}
{{eqn | r = \map {\mathbf 1} \phi
}}
{{end-eqn}}
{{qed}}
Theorem: Fourier Transform of Derivative of Tempered Distribution

Let $T \in \map {\SS'} \R$ be a tempered distribution.
Let $\xi \in \R$ be a real number.
Let the hat denote the Fourier transform.
Then in the distributional sense it holds that:
:$\hat {\paren{T'} }  = 2 \pi i \xi \hat T$

Proof:
Let $\phi \in \map \SS \R$ be a Schwartz test function.
Then:
{{begin-eqn}}
{{eqn | l = \map {\hat {\paren {T'} } } {\map \phi x}
      | r = \map {T'} {\map {\hat \phi} x}
      | c = {{Defof|Fourier Transform of Tempered Distribution}}
}}
{{eqn | r = \map {T'} {\int_{-\infty}^\infty \map \phi \xi e^{-2\pi i \xi x} }
      | c = {{Defof|Fourier Transform of Real Function}}
}}
{{eqn | r = -\map T {\dfrac \d {\d x} \int_{-\infty}^\infty \map \phi \xi e^{-2\pi i \xi x} }
      | c = {{Defof|Derivative of Tempered Distribution}}
}}
{{eqn | r = -\map T {-2\pi i \xi \int_{-\infty}^\infty \map \phi \xi e^{-2\pi i \xi x} }
}}
{{eqn | r = 2\pi i \xi \map T {\int_{-\infty}^\infty \map \phi \xi e^{-2\pi i \xi x} }
      | c = {{Defof|Tempered Distribution}}
}}
{{eqn | r = 2\pi i \xi \map T {\map {\hat \phi} x}
      | c = {{Defof|Fourier Transform of Real Function}}
}}
{{eqn | r = 2\pi i \xi \map {\hat T} {\map \phi x}
      | c = {{Defof|Fourier Transform of Tempered Distribution}}
}}
{{end-eqn}}
{{qed}}