hcju/proofwikiqa huggingface dataset:
To show that $f_* \mu$ is a measure, it will suffice to check the axioms $(1)$, $(2)$ and $(3')$ for a measure. === Axiom $(1)$ === The statement of axiom $(1)$ for $f_* \mu$ is: :$\forall E' \in \Sigma': \map {f_* \mu} {E'} \ge 0$ Now observe: {{begin-eqn}} {{eqn | l = \map {f_* \mu} {E'}       | r = \map \mu {\map {f^{-1} } {E'} }       | c = {{Defof|Pushforward Measure}} }} {{eqn | o = \ge       | r = 0       | c = $\mu$ is a Measure }} {{end-eqn}} {{qed|lemma}} === Axiom $(2)$ === Let $\sequence {E'_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma$. The statement of axiom $(2)$ for $f_* \mu$ is: :$\displaystyle \map {f_* \mu} {\bigcup_{n \mathop \in \N} E'_n} = \sum_{n \mathop \in \N} \map {f_* \mu} {E'_n}$ Now compute: {{begin-eqn}} {{eqn | l = \map {f_* \mu} {\bigcup_{n \mathop \in \N} E'_n}       | r = \map \mu {\map {f^{-1} } {\bigcup_{n \mathop \in \N} E'_n} }       | c = {{Defof|Pushforward Measure}} }} {{eqn | r = \map \mu {\bigcup_{n \mathop \in \N} \map {f^{-1} } {E'_n} }       | c = Preimage of Union under Mapping: General Result }} {{eqn | r = \sum_{n \mathop \in \N} \map \mu {\map {f^{-1} } {E'_n} }       | c = $\mu$ is a Measure }} {{eqn | r = \sum_{n \mathop \in \N} \map {f_* \mu} {E'_n}       | c = {{Defof|Pushforward Measure}} }} {{end-eqn}} Note that the second equality uses Preimage of Intersection under Mapping and Preimage of Empty Set is Empty to confirm that $\sequence {\map {f^{-1} } {E'_n} }_{n \mathop \in \N}$ is pairwise disjoint: :$\map {f^{-1} } {E'_n} \cap \map {f^{-1} } {E'_m} = \map {f^{-1} } {E'_n \cap E'_m} = \map {f^{-1} } \O = \O$ {{qed|lemma}} === Axiom $(3')$ === The statement of axiom $(3')$ for $f_* \mu$ is: :$\map {f_* \mu} \O = 0$ Now compute: {{begin-eqn}} {{eqn | l = \map {f_* \mu} \O       | r = \map \mu {\map {f^{-1} } \O}       | c = {{Defof|Pushforward Measure}} }} {{eqn | r = \map \mu \O       | c = Preimage of Empty Set is Empty }} {{eqn | r = 0       | c = $\mu$ is a measure }} {{end-eqn}} {{qed|lemma}} Thus $f_* \mu$, satisfying the axioms, is seen to be a measure. {{qed}}
By Orthogonal Group is Subgroup of General Linear Group, $M \in \GL {n, \R}$. From Pushforward of Lebesgue Measure under General Linear Group, it follows that: :$M_* \lambda^n = \size {\det M^{-1} } \lambda^n$ Since $M^{-1} \in \map {\mathrm O} {n, \R}$ by Orthogonal Group is Group, Determinant of Orthogonal Matrix applies to yield: :$\size {\det M^{-1} } = 1$ Hence the result. {{qed}} {{wtd|In order to avoid circularity (through Pushforward of Lebesgue Measure under General Linear Group and Determinant as Volume of Parallelotope) the direct proof Schilling produces also needs to be covered}}
From Linear Transformation on Euclidean Space is Continuous, $M^{-1}$ is a continuous mapping. Thus from Continuous Mapping is Measurable, it is measurable, and so $M_* \lambda^n$ is defined. Now let $B \in \map {\mathcal B} {\R^n}$ be a Borel measurable set, and let $\mathbf x \in \R^n$. Then: {{begin-eqn}} {{eqn | l = \map {M_* \lambda^n} {\mathbf x + B}       | r = \map {\lambda^n} {\map {M^{-1} } {\mathbf x + B} }       | c = {{Defof|Pushforward Measure}} }} {{eqn | r = \map {\lambda^n} {\map {M^{-1} } {\mathbf x} + \map {M^{-1} } B}       | c = $M^{-1}$ is linear }} {{eqn | r = \map {\lambda^n} {\map {M^{-1} } B}       | c = Lebesgue Measure is Translation-Invariant }} {{eqn | r = \map {M_* \lambda^n} B       | c = {{Defof|Pushforward Measure}} }} {{end-eqn}} Thus $M_* \lambda^n$ is a translation-invariant measure. From Translation-Invariant Measure on Euclidean Space is Multiple of Lebesgue Measure, it follows that: :$M_* \lambda^n = \map {M_* \lambda^n} {\openint 0 1^n} \cdot \lambda^n$ Lastly, using Determinant as Volume of Parallelotope it follows that: :$\map {M_* \lambda^n} {\openint 0 1^n} = \map {\lambda^n} {\map {M^{-1} } {\openint 0 1^n} } = \size {\det M^{-1} }$ Hence the result. {{qed}}
By Characterization of Sigma-Algebra Generated by Collection of Mappings: :$\sigma \left({f}\right) = \sigma \left({f^{-1} \left({\Sigma'}\right)}\right)$ where the latter $\sigma$ denotes a $\sigma$-algebra generated by a collection of subsets. By Pre-Image Sigma-Algebra on Domain is Sigma-Algebra, $f^{-1} \left({\Sigma'}\right)$ is a $\sigma$-algebra. Hence $\sigma \left({f^{-1} \left({\Sigma'}\right)}\right) = f^{-1} \left({\Sigma'}\right)$. {{qed}}
=== Necessary Condition === Suppose that $f$ is $\mathcal{B}^n \, / \, \mathcal{B}^m$-measurable. It is to be shown that for $1 \le i \le m$, $f_i: \R^n \to \R$ is $\mathcal{B}^n \, / \, \mathcal B$-measurable. By Mapping Measurable iff Measurable on Generator and Characterization of Euclidean Borel Sigma-Algebra, it suffices to show that: :$f_i^{-1} \left({\left({a \,.\,.\, b}\right)}\right) \in \mathcal{B}^n$ for every open real interval $\left({a \,.\,.\, b}\right)$. Since $f \left({\mathbf x}\right) = \mathbf y$ iff, for all $i$, $f_i \left({\mathbf x}\right) = y_i$, it follows that: :$\mathbf x \in f^{-1} \left({\mathbf y}\right) \iff \forall i: \mathbf x \in f_i^{-1} \left({y_i}\right)$ i.e., iff $\mathbf x \in \displaystyle \bigcap_{i \mathop = 1}^m f_i^{-1} \left({y_i}\right)$. By Preimage of Union under Mapping, it follows that for any $B \in \mathcal{B}^m$: :$\mathbf x \in f^{-1} \left({B}\right) \iff \exists \mathbf y \in B: \forall i: \mathbf x \in f_i^{-1} \left({y_i}\right)$ In particular, consider the open set $B = \left\{{\mathbf y \in \R^n: y_i \in \left({a \,.\,.\, b}\right)}\right\}$. Then by construction of $B$, $\mathbf x \in f_i^{-1} \left({\left({a \,.\,.\, b}\right)}\right)$ iff $f \left({x}\right) \in B$. Therefore, $f_i^{-1} \left({\left({a \,.\,.\, b}\right)}\right) = f^{-1} \left({B}\right)$. Since $B$ is open, it is also measurable by Characterization of Euclidean Borel Sigma-Algebra. Thus $f_i^{-1} \left({\left({a \,.\,.\, b}\right)}\right) \in \mathcal{B}^n$ as $f$ was assumed $\mathcal{B}^n \, / \, \mathcal{B}^m$-measurable. {{MissingLinks|To appropriate versions of Definition:Preimage}} {{qed|lemma}} === Sufficient Condition === Suppose that all $f_i$ are $\mathcal{B}^n \, / \, \mathcal B$-measurable. It is to be shown that $f$ is $\mathcal{B}^n \, / \, \mathcal{B}^m$-measurable. By Mapping Measurable iff Measurable on Generator and Characterization of Euclidean Borel Sigma-Algebra, it suffices to show that: :$f^{-1} \left({\left[[{\mathbf a \,.\,.\, \mathbf b}\right))}\right) \in \mathcal{B}^n$ for every half-open $m$-rectangle $\left[[{\mathbf a \,.\,.\, \mathbf b}\right))$. Now observe, for all $\mathbf x \in \R^n$: {{begin-eqn}} {{eqn|l = f \left({\mathbf x}\right)      |o = \in      |r = \left[ [{\mathbf a \,.\,.\, \mathbf b}\right)) }} {{eqn|ll= \iff      |l = \forall i: f_i \left({\mathbf x}\right)      |o = \in      |r = \left[{a_i \,.\,.\, b_i}\right) }} {{eqn|ll= \iff      |l = \mathbf x      |o = \in      |r = \bigcap_{i \mathop = 1}^m f_i^{-1} \left[{a_i \,.\,.\, b_i}\right) }} {{end-eqn}} Thus: :$f^{-1} \left({\left[[{\mathbf a \,.\,.\, \mathbf b}\right))}\right) = \displaystyle \bigcap_{i \mathop = 1}^m f_i^{-1} \left[{a_i \,.\,.\, b_i}\right)$ Now the $f_i^{-1} \left[{a_i \,.\,.\, b_i}\right)$ are measurable sets since the $f_i$ are $\mathcal{B}^n \, / \, \mathcal B$-measurable. Since $\mathcal{B}^n$ is a $\sigma$-algebra, Sigma-Algebra Closed under Countable Intersection applies, and it follows that: :$f^{-1} \left({\left[[{\mathbf a \,.\,.\, \mathbf b}\right))}\right) \in \mathcal{B}^n$ {{qed}}
Since $\mathcal G \subseteq \sigma \left({\mathcal G}\right)$, it follows immediately that: :$f^{-1} \left({\mathcal G}\right) \subseteq f^{-1} \left({\sigma \left({\mathcal G}\right)}\right)$ By Pre-Image Sigma-Algebra on Domain is Sigma-Algebra, the latter is a $\sigma$-algebra, and so by Generated Sigma-Algebra Preserves Subset, it follows that: :$\sigma \left({f^{-1} \left({\mathcal G}\right)}\right) \subseteq f^{-1} \left({\sigma \left({\mathcal G}\right)}\right)$ Conversely, by definition of generated $\sigma$-algebra, we have: :$f^{-1} \left({\mathcal G}\right) \subseteq \sigma \left({f^{-1} \left({\mathcal G}\right)}\right)$ Hence from Mapping Measurable iff Measurable on Generator, it follows that $f$ is $\sigma \left({f^{-1} \left({\mathcal G}\right)}\right) \, / \, \sigma \left({\mathcal G}\right)$-measurable. But this by definition of measurable mapping means that: :$f^{-1} \left({\sigma \left({\mathcal G}\right)}\right) \subseteq \sigma \left({f^{-1} \left({\mathcal G}\right)}\right)$ Hence the result, by definition of set equality. {{qed}}
By definition, $F_\mu$ is a Stieltjes function {{iff}} it is increasing and left-continuous. === $F_\mu$ is Increasing === Let $x, y \in \R$ such that $x \le y$. It is apparent that for all $z \in \R$: :$z \le 0 \implies F_\mu \left({z}\right) \le 0$ :$z \ge 0 \implies F_\mu \left({z}\right) \ge 0$ Therefore, only the cases $x \le y \le 0$ and $0 \le x \le y$ remain. In the first of these cases, note that: :$\left[{y \,.\,.\, 0}\right) \subseteq \left[{x \,.\,. 0}\right)$ Hence from Measure is Monotone, it follows that: :$F_\mu \left({x}\right) = -\mu \left({\left[{x \,.\,.\, 0}\right)}\right) \le -\mu \left({\left[{y \,.\,.\, 0}\right)}\right) = F_\mu \left({y}\right)$ The remaining case is analogous, using the observation that: :$\left[{0 \,.\,.\, x}\right) \subseteq \left[{0 \,.\,.\, y}\right)$ Hence $F_\mu$ is increasing. {{qed|lemma}} === $F_\mu$ is Left-Continuous === Suppose that $x > 0$. Suppose further that $x_k \uparrow x$ is an increasing sequence with limit $x$, and that $0 < x_1 < x$. Then as an increasing sequence of sets, we have: :$\left[{0 \,.\,.\, x_k}\right) \uparrow \left[{0 \,.\,.\, x}\right)$ Now we compute: {{begin-eqn}} {{eqn | l = \lim_{k \to \infty} F_\mu \left({x_k}\right)       | r = \lim_{k \to \infty} \mu \left({\left[{0 \,.\,.\, x_k}\right)}\right)       | c = {{Defof|Stieltjes Function of Measure}} }} {{eqn | r = \mu \left({\left[{0 \,.\,.\, x}\right)}\right)       | c = Characterization of Measures: $(3)$ }} {{eqn | r = F_\mu \left({x}\right) }} {{end-eqn}} Next, suppose $x = 0$. Suppose further that $x_k \uparrow 0$ is an increasing sequence with limit $0$. Now, we have the decreasing sequence of sets $\left[{x_k \,.\,.\, 0}\right) \downarrow \varnothing$. Reasoning as above, but applying Characterization of Measures: $(3'')$ instead yields the result in this case. Finally, suppose $x < 0$ and $x_k \uparrow x$ is an increasing sequence with limit $x$. Observe the decreasing sequence of sets $\left[{x_k \,.\,.\, 0}\right) \downarrow \left[{x \,.\,.\, 0}\right)$. This time, applying Characterization of Measures: $(3')$ yields the result. Note the additional assumption on $\mu$ was in fact used, in combination with Measure is Monotone. This is necessary to establish the conditions for the last two applications of Characterization of Measures. From the Trichotomy Law for Real Numbers and Proof by Cases, the result now follows. {{qed}}
Suppose $f: \R^1 \to \R$. Define: :$\map f x = y$ :$\Delta y = \map f {x + \Delta x} - \Delta x$ From the definition of the derivative of a real function, we can say that $f$ is differentiable {{iff}}: :$\dfrac {\Delta y} {\Delta x} \to \dfrac {\d y} {\d x}$ as $\Delta x \to 0$. Clearly, this is equivalent to saying that $f$ is differentiable {{iff}}: :$\dfrac {\Delta y} {\Delta x} - \dfrac {\d y} {\d x} = \varepsilon$ as $\varepsilon \to 0$, where $\varepsilon \in \R$ is some real number. Solving this equation for $\Delta y$: :$\Delta y = \dfrac {\d y} {\d x} \Delta x + \varepsilon \Delta x$ That is, {{iff}} the real function is differentiable $\varepsilon \to 0$ as $\Delta x \to 0$. {{qed|lemma}} Now consider $f: \R^n \to \R$, $n > 1$. From the definition of differentiability of a real-valued function, $f$ is differentiable {{iff}}: :$\map {\Delta f} {\mathbf x} = \map {\nabla f} {\mathbf x} \cdot \Delta \mathbf x + \begin{bmatrix} \\ \varepsilon_1 \\ \varepsilon_2 \\ \vdots \\ \varepsilon_n \end {bmatrix} \cdot \Delta \mathbf x$ such that $\begin{bmatrix} \\ \varepsilon_1 \\ \varepsilon_2 \\ \vdots \\ \varepsilon_n \end{bmatrix} \to \mathbf 0$ as $\Delta \mathbf x \to \mathbf 0$. {{wtd|needs to change to accommodate the new (more general) definition of differentiability}}  Observe that: {{begin-eqn}} {{eqn | l = \map {\nabla f} {\mathbf x} \cdot \Delta \mathbf x + \begin {bmatrix} \\ \varepsilon_1 \\ \varepsilon_2 \\ \vdots \\ \varepsilon_n \end {bmatrix} \cdot \Delta \mathbf x       | r = \begin {bmatrix} \frac {\map {\partial f} {\mathbf x} } {\partial x_1} \\ \frac {\map {\partial f} {\mathbf x} } {\partial x_2} \\ \vdots \\ \frac {\map {\partial f} {\mathbf x} } {\partial x_n} \end {bmatrix} \cdot \begin {bmatrix} \Delta x_1 \\ \Delta x_2 \\ \vdots \\ \Delta x_n \end {bmatrix} + \begin {bmatrix} \\ \varepsilon_1 \\ \varepsilon_2 \\ \vdots \\ \varepsilon_n \end {bmatrix} \cdot \begin {bmatrix} \Delta x_1 \\ \Delta x_2 \\ \vdots \\ \Delta x_n \end {bmatrix} }} {{eqn | r = \sum_{i \mathop = 1}^n \frac {\map {\partial f} {\mathbf x} } {\partial x_i} \Delta x_i + \sum_{i \mathop = 1}^n \varepsilon_i \Delta x_i       | c = {{Defof|Dot Product}} }} {{end-eqn}} where $\begin {bmatrix} \\ \varepsilon_1 \\ \varepsilon_2 \\ \vdots \\ \varepsilon_n \end {bmatrix} \to \begin {bmatrix} \\ 0 \\ 0 \\ \vdots \\ 0 \end {bmatrix}$ as $\begin {bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end {bmatrix} \to  \begin {bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}$. {{qed}}
By hypothesis, $f$ is differentiable. From Characterization of Differentiability: {{begin-eqn}} {{eqn | l = \Delta  z       | r = \sum_{i \mathop = 1}^n \frac {\partial z} {\partial x_i} \Delta x_i + \sum_{i \mathop = 1}^n \epsilon_i \Delta x_i       | c = $\forall i: 1 \le i \le n: \epsilon_i \to 0$ as $\Delta x_i \to 0$ }} {{end-eqn}} Let $\Delta t \ne 0$ and divide both sides of the equation by $\Delta t$: {{begin-eqn}} {{eqn | l = \frac {\Delta z} {\Delta t}       | r = \sum_{i \mathop = 1}^n \frac {\partial z} {\partial x_i} \frac {\Delta x_i} {\Delta t} + \sum_{i \mathop = 1}^n \epsilon_i \frac {\Delta x_i} {\Delta t}       | c = $\forall i: 1 \le i \le n: \epsilon_i \to 0$ as $\Delta x_i \to 0$ }} {{end-eqn}} Recall that each $x_i$ was defined to be differentiable {{WRT|Differentiation}} $t$, that is, that each $\dfrac {\d x_i} {\d t}$ exists. Then $\Delta x_i \to 0$ as $\Delta t \to 0$. Therefore: {{begin-eqn}} {{eqn | l = \frac {\d z} {\d t}       | r = \sum_{i \mathop = 1}^n \frac {\partial z} {\partial x_i} \frac {\d x_i} {\d t} + \sum_{i \mathop = 1}^n 0 \frac {\d x_i} {\d t} }} {{eqn | r = \sum_{i \mathop = 1}^n \frac {\partial z} {\partial x_i} \frac {\d x_i} {\d t}       | c = as $\Delta t \to 0$ }} {{end-eqn}} {{qed}} {{proofread}}
=== Sufficient Condition === Let $T \subseteq S$ such that $m := \inf \left({T}\right) \in T$. By definition, $m$ is a lower bound of $T$ and so: :$\forall x \in T: m \preccurlyeq x$ That is: :$\neg \exists x \in T: x \prec m$ Thus by definition $x$ is a minimal element of $T$. Thus by definition $S$ is well-founded and so is a well-ordered set. {{qed|lemma}} === Necessary Condition === Now suppose $T$ is a well-ordered set. Then from Well-Ordering is Total Ordering it follows immediately that $\left({S, \preccurlyeq}\right)$ is a totally ordered set. {{qed}}
It is immediate from the definition of $\mu_f$ that $\mu_f \left({\varnothing}\right) = 0$. Now suppose that for some half-open interval $\left[{a \,.\,.\, b}\right)$ one has: :$\left[{a \,.\,.\, b}\right) = \displaystyle \bigcup_{n \mathop \in \N} \left[{b_n \,.\,.\, b_{n+1}}\right)$ where $b_0 = a$ and $\displaystyle \lim_{n \mathop \to \infty} b_n = b$. Then we compute: {{begin-eqn}} {{eqn | l = \sum_{n \mathop \in \N} \mu_f \left({\left[{b_n \,.\,.\, b_{n + 1} }\right)}\right)       | r = \sum_{n \mathop \in \N} f \left({b_{n + 1} }\right) - f \left({b_n}\right)       | c = Definition of $\mu_f$ }} {{eqn | r = \lim_{n \to \infty} f \left({b_{n + 1} }\right) - f \left({b_0}\right)       | c = Telescoping Series }} {{eqn | r = f \left({b}\right) - f \left({a}\right)       | c = Definition of $\left({b_n}\right)_n$ }} {{eqn | r = \mu_f \left({\left[{a \,.\,.\, b}\right)}\right)       | c = Definition of $\mu_f$ }} {{end-eqn}} which verifies the second condition for a pre-measure. Hence $\mu_f$ is indeed a pre-measure. {{qed}} Category:Measure Theory bxb62y1d8clttga1ln1strkxwpv20zc
We intend to use Carathéodory's Theorem (Measure Theory). To this end, observe that by Characterization of Euclidean Borel Sigma-Algebra, we have: :$\mathcal B \left({\R}\right) = \sigma \left({\mathcal{J}_{ho}}\right)$ From Pre-Measure of Finite Stieltjes Function is Pre-Measure, $\mu_f$ is a pre-measure. Note that for all $n \in \N$, we have: :$\mu_f \left({\left[{-n \,.\,.\, n}\right)}\right) = f \left({n}\right) - f \left({-n}\right) < +\infty$ and $\left[{-n \,.\,.\, n}\right) \uparrow \R$ as an increasing sequence of sets. All these facts combine with Carathéodory's Theorem (Measure Theory) to establish existence and uniqueness of $\mu$. {{qed}}
From Pre-Measure of Finite Stieltjes Function Extends to Unique Measure, it suffices to verify that: :$\mu_{f_\mu} \left({\left[{a \,.\,.\, b}\right)}\right) = \mu \left({\left[{a \,.\,.\, b}\right)}\right)$ for all half-open intervals $\left[{a \,.\,.\, b}\right)$. Now we have: {{begin-eqn}} {{eqn|l = \mu_{f_\mu} \left({\left[{a \,.\,.\, b}\right)}\right)      |r = f_\mu \left({b}\right) - f_\mu \left({a}\right)      |c = Definition of $\mu_{f_\mu}$ }} {{end-eqn}} If either $a = 0$ or $b = 0$, the result follows immediately from the definition of $f_\mu$. Now suppose that $a < b < 0$. Then: {{begin-eqn}} {{eqn|l = f_\mu \left({b}\right) - f_\mu \left({a}\right)      |r = \mu \left({\left[{a \,.\,.\, 0}\right)}\right) - \mu \left({\left[{b \,.\,.\, 0}\right)}\right)      |c = Definition of $f_\mu$ }} {{eqn|r = \mu \left({\left[{a \,.\,.\, b}\right)}\right)      |c = Measure of Set Difference with Subset }} {{end-eqn}} Finally, let $0 < a < b$. Then: {{begin-eqn}} {{eqn|l = f_\mu \left({b}\right) - f_\mu \left({a}\right)      |r = \mu \left({\left[{0 \,.\,.\, b}\right)}\right) - \mu \left({\left[{0 \,.\,.\, a}\right)}\right)      |c = Definition of $f_\mu$ }} {{eqn|r = \mu \left({\left[{a \,.\,.\, b}\right)}\right)      |c = Measure of Set Difference with Subset }} {{end-eqn}} The final case $a < 0 < b$ is a trivial consequence of Measure is Finitely Additive Function. Hence it must be that $\mu_{f_\mu} = \mu$. {{qed}}
Consider the definition of $\mathcal C$ as a limit of a decreasing sequence. In the notation as introduced there, we see that each $S_n$ is a collection of disjoint closed intervals. From Closed Set Measurable in Borel Sigma-Algebra, these are measurable sets. Furthermore, each $S_n$ is finite. Hence by Sigma-Algebra Closed under Union, it follows that $C_n := \displaystyle \bigcup S_n$ is measurable as well. Then, as we have: :$\mathcal C = \displaystyle \bigcap_{n \mathop \in \N} C_n$ it follows from Sigma-Algebra Closed under Countable Intersection that $\mathcal C$ is measurable. The $C_n$ also form a decreasing sequence of sets with limit $\mathcal C$. Thus, from Characterization of Measures: $(3')$, it follows that: :$\lambda \left({\mathcal C}\right) = \displaystyle \lim_{n \to \infty} \lambda \left({C_n}\right)$ It is not too hard to show that, for all $n \in \N$: :$\lambda \left({C_n}\right) = \left({\dfrac 2 3}\right)^n$ {{finish|yes, I know}} Now we have by Sequence of Powers of Number less than One that: :$\displaystyle \lim_{n \mathop \to \infty} \left({\frac 2 3}\right)^n = 0$ and the result follows. {{qed}}
=== Necessary Condition === Let $g$ be a $\map \sigma f \, / \, \map {\mathcal B} \R$-measurable function. We need to construct a measurable $\tilde g$ such that $g = \tilde g \circ f$. Let us proceed in the following fashion: :Establish the result for $g$ a characteristic function; :Establish the result for $g$ a simple function; :Establish the result for all $g$ So let $g = \chi_E$ be a characteristic function. By Characteristic Function Measurable iff Set Measurable, it follows that $E$ is $\sigma \left({f}\right)$-measurable. Thus there exists some $A \in \Sigma$ such that $E = \map {f^{-1} } A$. Again by Characteristic Function Measurable iff Set Measurable, we have $\chi_A: Y \to \R$ is measurable. It follows that $\chi_E = \chi_A \circ f$, and $\tilde g := \chi_A$ works. Now let $g = \displaystyle \sum_{i \mathop = 1}^n a_i \chi_{E_i}$ be a simple function. Let $A_i$ be associated to $E_i$ as above. Then we have: {{begin-eqn}} {{eqn | l = \sum_{i \mathop = 1}^n a_i \chi_{E_i}       | r = \sum_{i \mathop = 1}^n a_i \paren {\chi_{A_i} \circ f}       | c = by the result for characteristic functions }} {{eqn | r = \paren {\sum_{i \mathop = 1}^n a_i \chi_{A_i} } \circ f       | c = Composition of Mappings is Linear }} {{end-eqn}} Now $\displaystyle \sum_{i \mathop = 1}^n a_i \chi_{A_i}$ is a simple function, hence measurable by Simple Function is Measurable. Therefore, it suffices as a choice for $\tilde g$. Next, let $g \ge 0$ be a measurable function. By Measurable Function Pointwise Limit of Simple Functions, we find simple functions $g_j$ such that: :$\displaystyle \lim_{j \mathop \to \infty} g_j = g$ Applying the previous step to each $g_j$, we find a sequence of $\tilde g_j$ satisfying: :$\displaystyle \lim_{j \mathop \to \infty} \tilde g_j \circ f = g$ From Composition with Pointwise Limit it follows that we have, putting $\tilde g := \displaystyle \lim_{j \mathop \to \infty} \tilde g_j$: :$\displaystyle \lim_{j \mathop \to \infty} \tilde g_j \circ f = \tilde g \circ f$ {{explain|Address why (if?!) $\tilde g$ is well defined}} An application of Pointwise Limit of Measurable Functions is Measurable yields $\tilde g$ measurable. Thus we have provided a suitable $\tilde g$ for every $g$, such that: :$g = \tilde g \circ f$ as desired. {{qed}} === Sufficient Condition === Suppose that such a $\tilde g$ exists. Note that $f$ is $\map \sigma f \, / \, \Sigma$-measurable by definition of $\map \sigma f$. The result follows immediately from Composition of Measurable Mappings is Measurable. {{qed}}
Each of $(2)$ up to $(5')$ is equivalent to $(1)$ by combining Mapping Measurable iff Measurable on Generator and Generators for Extended Real Sigma-Algebra. {{qed}}
Let $\overline B \in \overline{\mathcal B}$. Then by Extended Real Sigma-Algebra Induces Borel Sigma-Algebra on Reals, we have: :$\overline B \cap \R \in \mathcal B \left({\R}\right)$ We also have, by definition of the extended real numbers $\overline \R$, that: :$\overline \R \setminus \R = \left\{{+\infty, -\infty}\right\}$ and therefore, $\overline B \setminus \R \subseteq \left\{{+\infty, -\infty}\right\}$. Here, $\setminus$ signifies set difference. By Set Difference Union Intersection: :$\overline B = \left({\overline B \setminus \R}\right) \cup \left({\overline B \cap \R}\right)$ Therefore, any $\overline B \in \overline{\mathcal B}$ is of the purported form $B \cup S$ with $B \in \mathcal B \left({\R}\right)$ and $S \in \mathcal S$. It remains to show that any such set is in fact an element of $\overline{\mathcal B}$. Since any $B \in \mathcal B \left({\R}\right)$ is naturally also in $\overline{\mathcal B}$, it suffices to show that: :$\mathcal S \subseteq \overline{\mathcal B}$ by applying Sigma-Algebra Closed under Union. From Closed Set Measurable in Borel Sigma-Algebra, it will now suffice to show that: :$\varnothing, \left\{{+\infty}\right\}, \left\{{-\infty}\right\}, \left\{{+\infty, -\infty}\right\}$ are all closed sets in $\overline \R$. That they are follows from Extended Real Number Space is Hausdorff and Finite Subspace of Hausdorff Space is Closed. {{qed}}
We have Euclidean Space is Subspace of Extended Real Number Space. The result follows from Borel Sigma-Algebra of Subset is Trace Sigma-Algebra. {{qed}}