nyuuzyou/wikis huggingface dataset:
Frobenius's Theorem/Lemma 1

== Lemma ==
Let $\struct {A, \oplus}$ be a quadratic real algebra.
Then:
:$\quad U = \set {u \in A \setminus \R: u^2 \in \R} \cup \set 0$
is a linear subspace of $A$.
== Proof ==
$U$ is closed under scalar multiplication.
We have to show that $u, v \in U$ implies $u + v \in U$.
Let $\set {u, v, 1}$ be a linearly dependent set.
Then there exist constants $a, b \in \R$ such that $u = av + b$.
Then $u^2 = a^2 v^2 + 2 a b v + b^2$.
Since $u^2$, $a^2 v^2$, and $b^2$ are all real, it follows that $2 a b v \in \R$, that is, $v \in \R$.
Since $0$ is the only real element of $U$, it follows that $v = 0$.
Reversing $u$ and $v$ in the preceding argument shows that also $u = 0$, so that $u + v = 0 \in U$.
Now let $\set {u, v, 1}$ be a linearly independent set.
We have:
:$\paren {u + v}^2 + \paren {u - v}^2 = 2 u^2 + 2 v^2 \in \R$
On the other hand, as $A$ is quadratic there exist $\lambda, \mu \in \R$ such that:
:$\paren {u + v}^2 - \lambda \paren {u + v} \in \R$
:$\paren {u - v}^2 - \mu \paren {u - v} \in R$
Hence:
:$\lambda \paren {u + v} + \mu \paren {u - v} \in \R$
However, we have that $\set {u, v, 1}$ is linearly independent.
Therefore $\lambda + \mu = \lambda - \mu = 0$, and so $\lambda = \mu = 0$.
This proves that $u \pm v \in U$.
Thus $U$ is indeed a linear subspace of $A$.
Category:Frobenius's Theorem
Frobenius's Theorem/Lemma 2

== Lemma ==
Let $\struct {A, \oplus}$ be a quadratic real algebra.
Then:
:$\forall u, v \in U: u v + v u \in \R$
== Proof ==
From Lemma 1:
:$\forall u, v \in U: u v + v u = \paren {u + v}^2 - u^2 - v^2 \in \R$.
Category:Frobenius's Theorem
Frobenius's Theorem/Lemma 3

== Lemma ==
Let $\struct {A, \oplus}$ be a quadratic real algebra.
Then:
:$A = \R \oplus U$
== Proof ==
Let $a \in A \setminus \R$.
Then:
:$\exists \nu \in \R: a^2 - \nu a \in \R$
Therefore, if we set
:$u = a - \dfrac \nu 2 \in U$
then $u^2 = a^2 - \nu a + \nu^2/4 \in \R$, so
:$a = \dfrac \nu 2 + u \in \R \oplus U$
which proves the assertion.
Category:Frobenius's Theorem
Frobenius's Theorem/Lemma 4

== Lemma ==
Let $\struct {A, \oplus}$ be a quadratic real algebra.
If $A$ is also a division algebra, then every non-zero $u \in U$ can be written as $u = \alpha v$ with $\alpha \in \R$ and $v^2 = -1$.
== Proof ==
This follows directly from the definition of division algebra.
Also, we have that $u^2 \in \R$.
 $u^2 = 0$.
Since $u$ is non-zero and since $A$ is a division algebra, there exists an inverse $u^{-1}$ such that $u u^{-1} = 1$.
But then $u = u 1 = u u u^{-1} = 0 \cdot u^{-1} = 0$, which cannot be since $u$ was assumed to be non-zero.
Now suppose $u^2 > 0$.
Then there exists an $\alpha \in \R$ such that $\alpha^2 = u^2$.
But then $\paren {u - \alpha} \paren {u + \alpha} = u^2 - \alpha^2$ would be $0$.
Since $A$ is assumed to be a division algebra, this would imply that either $u = \alpha$ or $u = -\alpha$.
This is impossible, since $u \in U$.
So:
:$u^2 < 0$
and so:
:$u^2 = -\alpha^2$
with $0 \ne \alpha \in \R$.
Thus, $v = \alpha^{-1} u$ is a desired element.
Category:Frobenius's Theorem
Frobenius's Theorem/Lemma 5

== Lemma ==
Let $A$ be a quadratic real division algebra.
Let:
:$U = \left\{{u \in A \setminus \R: u^2 \in \R}\right\} \cup \left\{{0}\right\}$
where $\setminus$ denotes set difference.
Suppose $e_1, \ldots, e_k \in U$ are such that:
: $\forall i \le k: e_i^2 = -1$
: $\forall i, j \le k, i \ne j: e_i e_j = -e_j e_i$
If $U$ is not equal to the linear span of $e_1, \ldots, e_k$, then there exists $e_{k+1} \in U$ such that:
: $e_{k+1}^2 = -1$
: $\forall i \le k: e_i e_{k+1} = -e_{k+1} e_i$
== Proof ==
From Lemma 1, $U$ is a linear subspace of $A$ such that:
:$\forall u, v \in U: u v + v u \in \R$
:$A = \R \oplus U$
Pick $u \in U$ that is not contained in the linear span of $e_1, \ldots, e_k$.
By Lemma 1, we can set $\alpha_i = \dfrac 1 2 \left({u e_i + e_i u}\right) \in \R$.
Note that $v = u + \alpha_1 e_1 + \cdots + \alpha_k e_k$ satisfies $e_i v = v e_i$ for all $i \le k$.
Let $e_{k+1}$ be a scalar multiple of $v$ such that $e_{k+1}^2 = -1$ (whose existence follows from Lemma 1: Assertion 4).
Then $e_{k+1}$ has all desired properties.
Category:Frobenius's Theorem
Frobenius Endomorphism on Field is Injective

== Theorem ==
Let $p$ be a prime number.
Let $F$ be a field of characteristic $p$.
Then the Frobenius endomorphism $\Frob: F \to F$ is injective.
== Proof ==
We have:
:$\map \Frob 1 = 1$
By Ring Homomorphism from Field is Monomorphism or Zero Homomorphism, $\Frob$ is injective.
== Also see ==
* Definition:Perfect Field
Category:Frobenius Endomorphisms
From Medial Straight Line arises Infinite Number of Irrational Straight Lines

== Theorem ==
== Proof ==
:
Let $A$ be a medial straight line.
It is to be shown that there is an infinite number of irrational straight lines, and none of them are:
:Binomial
:First bimedial
:Second bimedial
:Major
:The side of a rational plus a medial area
:The side of the sum of two medial areas
:Apotome
:First apotome of a medial
:Second apotome of a medial
:Minor
:That which produces a medial whole with a rational area
:That which produces a medial whole with a medial area
Let the above set be called $\KK$.
Let $B$ be a rational straight line.
Let $C^2$ be equal to $A \cdot B$.
Therefore by:
:
and from:
:
it follows that:
:$C$ is irrational.
Consider a square on any of $\KK$.
None of them, when applied to a rational straight line, produces as breadth a medial straight line.
Let $D^2$ be equal to $B \cdot C$.
Therefore by:
:
and from:
:
it follows that:
:$D$ is irrational.
Consider a square on any of $\KK$.
None of them, when applied to a rational straight line, produces $C$ as breadth.
This process can be continued .
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Frullani's Integral

== Theorem ==
Let $a, b > 0$. 
Let $f$ be a function continuously differentiable on the non-negative real numbers. 
Suppose that $\ds \map f \infty = \lim_{x \mathop\to \infty} \map f x$ exists, and is finite.
Then: 
:$\ds \int_0^\infty \frac {\map f {a x} - \map f {b x} } x \rd x = \paren {\map f \infty - \map f 0} \ln \frac a b$
== Proof == 
 
== Sources ==
* : $\S 15$: Miscellaneous Definite Integrals: $15.118$
* 
Category:Definite Integrals
Fréchet-Darmois-Cramér-Rao Inequality

== Theorem ==
Let $S$ be a sample of $n$ observations from a probability distribution with frequency function $\map f {x, \theta}$.
Let it be assumed that certain regularity conditions apply.
Let it also be assumed that the extremes do not depend on $\theta$.
The lower bound of the variance of an estimator $T$ of a parameter $\theta$ is given by $\dfrac 1 I$ where:
:$I = -n \map E {\dfrac {\partial^2 \ln f} {\partial \theta^2} }$
== Proof ==
== Also known as ==
== Historical Note ==
== Sources ==
* 
* 
* 
* 
Category:Fréchet-Darmois-Cramér-Rao Inequality
Category:Variance
Category:Estimators
Category:Information (Estimation Theory)
Fréchet-Darmois-Cramér-Rao Inequality/Also known as

== Fréchet-Darmois-Cramér-Rao Inequality: Also known as ==
The Fréchet-Darmois-Cramér-Rao Inequality is usually known as:
:the Cramér-Rao Inequality
:the Cramér-Rao Bound
:the Cramér-Rao Lower Bound.
== Sources ==
* 
* 
* 
* 
Category:Fréchet-Darmois-Cramér-Rao Inequality
Fréchet Filter is Filter

== Theorem ==
Let $S$ be an infinite set.
Let $\FF$ be the Fréchet filter on $S$.
Then $\FF$ is a filter on $S$.
== Proof ==
By definition, $\FF$ must satisfy the filter on set axioms:
=== $\paren {\text F 1}$ ===
By Set Difference with Self is Empty Set:
:$S \setminus S = \O$
By Cardinality of Empty Set, $\O$ is finite.
Therefore, $S$ is cofinite in $S$ by definition.
Thus, by definition of Fréchet filter:L
:$S \in \FF$
=== $\paren {\text F 2}$ ===
By Set Difference with Empty Set is Self:
:$S \setminus \O = S$
But , $S$ is infinite.
Therefore, $\O$ is not cofinite in $S$ by definition.
Thus, by definition of Fréchet filter:
:$\O \notin \FF$
=== $\paren {\text F 3}$ ===
Let $U, V \in \FF$.
Then, by definition of Fréchet filter, $U$ and $V$ are cofinite in $S$.
That is, $S \setminus U$ and $S \setminus V$ are finite.
Thus, by Union of Finite Sets is Finite, $\paren {S \setminus U} \cup \paren {S \setminus V}$ is finite.
But by Difference with Intersection:
:$S \setminus \paren {U \cap V} = \paren {S \setminus U} \cup \paren {S \setminus V}$
Therefore, $S \setminus \paren {U \cap V}$ is finite.
Thus, by definition, $U \cap V$ is cofinite in $S$.
Then, by definition of Fréchet filter, $U \cap V \in \FF$.
=== $\paren {\text F 4}$ ===
Let $U \in \FF$.
Let $V$ be a set such that $U \subseteq V \subseteq S$.
Then, by Set Difference with Subset is Superset of Set Difference:
:$S \setminus V \subseteq S \setminus U$
By definition of Fréchet filter, $U$ is cofinite in $S$.
That is, $S \setminus U$ is finite.
But then, by Subset of Finite Set is Finite, $S \setminus V$ is finite.
Thus, $V$ is cofinite in $S$ by definition.
Therefore, $V \in \FF$, by definition of Fréchet filter.
As $\FF$ satisfies the filter on set axioms, it is a filter on $S$ by definition.
Category:Fréchet Filters
Fréchet Space (Functional Analysis) is Complete Metric Space

== Theorem ==
Let $\struct {\R^\omega, d}$ be the Fréchet space on $\R^\omega$.
Then $\struct {\R^\omega, d}$ is a complete metric space.
== Proof ==
From Fréchet Space (Functional Analysis) is Metric Space, $\struct {\R^\omega, d}$ is a metric space.
It remains to be demonstrated that $\struct {\R^\omega, d}$ is complete.
Let $\sequence {x^{\paren n}}_{n \mathop \in \N}$ be an arbitrary Cauchy sequence in $\R^\omega$, where as an exception we denote the index of the Cauchy sequence as a superscript.
Recall that each $\forall n \in \N: x^{\paren n} \in \R^\omega$ itself is a sequence in $\R$ by the definition of $\R^\omega$.
More precisely, each $x^{\paren n} \in \R^\omega$ can be written as
:$x^{\paren n} := \sequence { {x_i}^{\paren n} }_{i \mathop \in \N}$
where ${x_i}^{\paren n} \in \R$ for all $i \in \N$.
=== Lemma ===
For each $j \in \N$, the real sequence $ \sequence { {x_j}^{\paren n}}_{n \mathop \in \N} $ is convergent.
=== Proof ===
Let $j \in \N$.
In view of Cauchy's convergence criterion, it suffices to show that $\sequence { {x_j}^{\paren n}} _{n \mathop \in \N} $ is a Cauchy sequence in $\R$.
Let $\epsilon > 0$ be arbitrary.
Since $ \sequence {x^{\paren n}} _{n\in\N} $ is a Cauchy sequence in $\R^\omega$, there is a $N \in \N$ such that:
:$\forall m, n \in \N: m, n \ge N: \map d {x^{\paren m}, x^{\paren n}} < \dfrac {2^{-j} \epsilon} {1 + \epsilon}$
Thus:
:$\dfrac {2^{-j} \size {{x_j}^{\paren m} - {x_j}^{\paren n}} } {1 + \size {{x_j}^{\paren m} - {x_j}^{\paren n}} } \le \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {{x_i}^{\paren m} - {x_i}^{\paren n}} } {1 + \size {{x_i}^{\paren m} - {x_i}^{\paren n}} } = \map d {x_m, x_n} < \dfrac {2^{-j} \epsilon} {1 + \epsilon}$
which implies:
:$\size { {x_j}^{\paren m} - {x_j}^{\paren n}} < \epsilon$
Altogether:
:$\forall m, n \in \N: m, n \ge N: \size {{x_j}^{\paren m} - {x_j}^{\paren n}} < \epsilon$
Let $\ds x^{\paren \infty} := \sequence {{x_i}^{\paren \infty}}_{i \mathop \in \N} \in \R^\omega$, where $\ds {x_i}^{\paren \infty} := \lim_{n \mathop \to \infty} {x_i}^{\paren n}$ exists for each $i \in \N$ by the preceding lemma.
It remains to show that $x^{\paren \infty}$ is the limit of $\sequence {x^{\paren n}}_{n \in \N}$ in $\R^\omega$.
To this end, let $\epsilon > 0$ be arbitrary.
For each $i \in \N$, there is a $N_i \in \N$ such that:
:$\forall n \ge N_i: \size { {x_i}^{\paren \infty} - {x^{\paren n}} _i} < \dfrac \epsilon 4$
Furthermore, $M \in \N$ be so large that $\dfrac 1 {2^M} < \dfrac \epsilon 2$.
Finally, let $N := \max \set {N_0, N_1, \ldots, N_M}$.
Then, for all $n \ge N$:
== Sources ==
* : Part $\text {II}$: Counterexamples: $37$. Fréchet Space: $7$
Category:Fréchet Product Metric
Category:Complete Metric Spaces
Fréchet Space (Functional Analysis) is Metric Space

== Theorem ==
Let $\struct {\R^\omega, d}$ be the Fréchet space on $\R^\omega$.
Then $\struct {\R^\omega, d}$ is a metric space.
== Proof ==
It is to be demonstrated that $d$ satisfies all the metric space axioms.
Recall from the definition of the Fréchet space that the distance function $d: \R^\omega \times \R^\omega \to \R$ is defined on $\R^\omega$ as:
:$\forall x, y \in \R^\omega: \map d {x, y} = \ds \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {x_i - y_i} } {1 + \size {x_i - y_i} }$
where:
:$x := \sequence {x_i}_{i \mathop \in \N} = \tuple {x_0, x_1, x_2, \ldots}$
:$y := \sequence {y_i}_{i \mathop \in \N} = \tuple {y_0, y_1, y_2, \ldots}$
denote arbitrary elements of $\R^\omega$.
First it is confirmed that Fréchet Product Metric is Absolutely Convergent on arbitrary $x$ and $y$, as follows.
First note that:
:$\dfrac {\size {x_i - y_i} } {1 + \size {x_i - y_i} } = 1 - \dfrac 1 {1 + \size {x_i - y_i} }$
implies that we have:
:$0 \le \dfrac {\size {x_i - y_i} } {1 + \size {x_i - y_i} } < 1$
By the Ratio Test we have that the series is absolutely convergent.
An important property of absolutely convergent series is that any rearrangement of the terms of the series is also convergent and equal to the original series.
This will be used in the proof of .
=== Proof of ===
So holds for $d$.
=== Proof of ===
So holds for $d$.
=== Proof of ===
Thus for such a $k \in \N$:
It has been established during the course of demonstrating compliance with (and is in any case trivially obvious) that:
:$x_j = y_j \implies \dfrac {2^{-j} \size {x_j - x_j} } {1 + \size {x_j - x_j} } = 0$
Thus $\ds \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {x_i - x_i} } {1 + \size {x_i - x_i} }$ contains at least one term:
:$\dfrac {2^{-k} \size {x_k - x_k} } {1 + \size {x_k - x_k} } > 0$
and any number of other terms:
:$\dfrac {2^{-j} \size {x_j - x_j} } {1 + \size {x_j - x_j} } = 0$
Hence:
:$\ds \sum_{i \mathop \in \N} \dfrac {2^{-i} \size {x_i - x_i} } {1 + \size {x_i - x_i} } > 0$
for $x \ne y$.
So holds for $d$.
=== Proof of ===
Consider the real function:
:$\forall a \in \R_{\ge 0}: \map f a = \dfrac a {1 + a}$
We want to show $f$ satisfies:
:$\map f a \le \map f {a'}$
for $a \ge 0$ and $a' \ge a$.
The first inequality follows since:
:$\map f a = 1 - \dfrac 1 {1 + a}$
with $\dfrac 1 {1 + a}$ decreasing as $a$ increases.
We also want to show it satisfies :
:$\map f a \le \map f {a + b} \le \map f a + \map f b$
for $a, b > 0$.
The first inequality here follows from the above inequality.
For the second inequality here we note $\dfrac {\map f a} a = \dfrac 1 {1 + a}$ gives:
:$\dfrac {\map f a} a \ge \dfrac {\map f {a + b} } {a + b}$
:$\dfrac {\map f b} b \ge \dfrac {\map f {a + b} } {a + b}$
for $a, b > 0$
from which we obtain $\map f a + \map f b \ge \dfrac {\map f {a + b} } {a + b} = \map f {a + b}$ (the inequality is immediately verified in cases where $a$ or $b$ is $0$).
We now use these inequalities to prove :
So holds for $d$.
Thus $d$ satisfies all the metric space axioms and so is a metric.
Category:Fréchet Product Metric
Full Rook Matrix is Nonsingular

== Theorem ==
A full rook matrix is nonsingular.
== Proof ==
Let $\mathbf A$ be a full rook matrix.
By definition, $\mathbf A$ is an instance of a permutation matrix.
By Determinant of Permutation Matrix, it follows that $\det \mathbf A = \pm 1$.
By Matrix is Nonsingular iff Determinant has Multiplicative Inverse:
:$\mathbf A$ is nonsingular.
== Sources ==
* : Chapter $2$: 'And you do addition?': $\S 2.2$: Functions on vectors: $\S 2.2.4$: Multiplication and inverses: Ponderable $2.2.2 \ \text{(a)}$
Category:Rook Matrices
Category:Nonsingular Matrices
Fully Characteristic Subgroup is Characteristic/Proof 1

== Theorem ==
== Proof ==
By definition of a fully characteristic subgroup,
$H$ is invariant under every group endomorphism of $G$.
By Group Automorphism is Endomorphism, every group automorphism is a group endomorphism.
Hence, invariance over every group endomorphism implies invariance over every group automorphism.
And by definition of a characteristic subgroup,
$H$ is characteristic.
Category:Fully Characteristic Subgroup is Characteristic
Fully Normal Space is Normal Space

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a fully normal space.
Then $T$ is a normal space.
== Proof ==
From the definition, $T$ is fully normal :
:$T$ is fully $T_4$
:$T$ is a $T_1$ (Fréchet) space.
We have that a fully $T_4$ space is also a $T_4$ space.
So:
:$T$ is a $T_4$ Space
:$T$ is a $T_1$ (Fréchet) space
which is precisely the definition of a normal space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Paracompactness
Category:Fully Normal Spaces
Category:Normal Spaces
Fully Normal Space is Paracompact

== Theorem ==
Let $T = \struct {S, \tau}$ be a fully normal space.
Then $T$ is paracompact.
== Proof ==
From the definition, $T$ is fully normal :
:$T$ is fully $T_4$
:$T$ is a $T_1$ (Fréchet) space.
Then $T$ is fully $T_4$ every open cover of $S$ has a star refinement.
Let $\UU$ be an open cover for $T$.
Then from the definition, there exists a cover $\VV$ for $T$ such that:
:$\ds \forall x \in S: \exists U \in \UU: \paren {\bigcup \set {V \in \VV: x \in V} } \subseteq U$
Recall from the definition of paracompact:
:$T$ is paracompact every open cover of $S$ has an open refinement which is locally finite.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Paracompactness
Category:Fully Normal Spaces
Category:Paracompact Spaces
Fully T4 Space is T4 Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a fully $T_4$ space.
Then $T$ is a $T_4$ space.
== Proof ==
Recall that $\struct {S, \tau}$ is a $T_4$ space :
:$\forall A, B \in \map \complement \tau, A \cap B = \O: \exists U, V \in \tau: A \subseteq U, B \subseteq V, U \cap V = \O$ 
That is, for any two disjoint closed sets $A, B \subseteq X$ there exist disjoint open sets $U, V \in \tau$ containing $A$ and $B$ respectively.
Consider the open cover $\UU = \set {\map \complement A, \map \complement B}$.
Since $T$ is fully $T_4$, there exists a star refinement $\VV$ of $\UU$.
Define $\VV_A$ and $\VV_B$ by:
:$\VV_A := \set {V \in \VV: A \cap V \ne \O}$
:$\VV_B := \set {V \in \VV: B \cap V \ne \O}$
and subsequently $U_A$ and $U_B$ by:
:$U_A := \bigcup \VV_A$
:$U_B := \bigcup \VV_B$
Now given $a \in A$, since $\VV$ is a cover, there exists $V \in \VV$ such that $a \in V$.
But then $V \in \VV_A$, so that:
:$a \in U_A$
Hence $A \subseteq U_A$, and similarly $B \subseteq U_B$.
 that $x \in U_A \cap U_B$.
Then there exist $V_A \in \VV_A, V_B \in \VV_B$ such that:
:$x \in V_A, x \in V_B$
By definition of $\VV_A$ and $\VV_B$, there exist:
:$a \in A \cap V_A$
:$b \in B \cap V_B$
Now recall the definition of the star $x^*$ of $x$:
:$x^* := \bigcup \set {V \in \VV: x \in V}$
Since $T$ is fully $T_4$, either $x^* \subseteq \map \complement A$ or $x^* \subseteq \map \complement B$.
But $a \in V_A \subseteq x^*$, so $x^* \nsubseteq \map \complement A$.
Similarly $b \in V_B \subseteq x^*$, so $x^* \nsubseteq \map \complement B$.
From this contradiction it is inferred that $x \in U_A \cap U_B$ cannot exist.
That is:
:$U_A \cap U_B = \O$
and it follows that $T$ is a $T_4$ space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $3$: Compactness: Paracompactness
Category:Fully T4 Spaces
Category:T4 Spaces
Function A.E. Equal to Measurable Function in Complete Measure Space is Measurable

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a complete measure space.
Let $f : X \to \overline \R$ be a $\Sigma$-measurable function.
Let $g : X \to \overline \R$ be a function such that: 
:$f = g$ $\mu$-almost everywhere.
Then $g$ is $\Sigma$-measurable.
== Proof ==
We aim to show that: 
:$\set {x \in X : \map g x \le \alpha} \in \Sigma$
for each $\alpha \in \R$. 
Let $\alpha \in \R$. 
Since $f = g$ $\mu$-almost everywhere there exists a $\mu$-null set such that: 
:whenever $x \in X$ has $\map f x \ne \map g x$, we have $x \in N$.
We have: 
For $x \in X \setminus N$, we have: 
:$\map g x = \map f x$
so:
:$\set {x \in X \setminus N : \map g x \le \alpha} \cap \paren {X \setminus N} = \set {x \in X \setminus N : \map f x \le \alpha} \cap \paren {X \setminus N}$
Since $f$ is $\Sigma$-measurable, we have: 
:$\set {x \in X : \map f x \le \alpha} \in \Sigma$
Since $\sigma$-algebras are closed under complement, we have: 
:$X \setminus N \in \Sigma$ 
From Sigma-Algebra Closed under Countable Intersection, we therefore have: 
:$\set {x \in X \setminus N : \map f x \le \alpha} \cap \paren {X \setminus N} \in \Sigma$
Now, from Intersection is Subset, we have: 
:$\set {x \in X : \map g x \le \alpha} \cap N \subseteq N$
Since $N$ is $\mu$-null, we have: 
:$\set {x \in X : \map g x \le \alpha}$ is $\mu$-null
since $\struct {X, \Sigma, \mu}$ is complete. 
In particular: 
:$\set {x \in X : \map g x \le \alpha} \cap N \in \Sigma$
So, since $\sigma$-algebras are closed under countable union, we have: 
:$\paren {\set {x \in X : \map g x \le \alpha} \cap N} \cup \paren {\set {x \in X : \map g x \le \alpha} \cap \paren {X \setminus N} } \in \Sigma$
that is: 
:$\set {x \in X : \map g x \le \alpha} \in \Sigma$
Since $\alpha \in \R$ was arbitrary, we have: 
:$g$ is $\Sigma$-measurable.
Category:Complete Measure Spaces
Function Measurable iff Positive and Negative Parts Measurable

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f: X \to \overline \R$ be an extended real-valued function.
Let $f^+, f^-: X \to \overline \R$ be the positive and negative parts of $f$.
Then $f$ is $\Sigma$-measurable both $f^+$ and $f^-$ are $\Sigma$-measurable.
== Proof ==
=== Necessary Condition ===
Suppose $f$ is measurable.
By definition, its positive part $f^+$ equals the pointwise maximum:
:$f^+ = \max \set {f, 0}$
where $0$ denotes the zero function.
By Constant Function is Measurable, $0$ is a measurable function.
Thus, by Pointwise Maximum of Measurable Functions is Measurable, $f^+$ is measurable.
Subsequently, the negative part $f^-$ of $f$ is defined by means of a pointwise minimum:
:$f^- = - \min \set {f, 0}$
By Pointwise Maximum of Measurable Functions is Measurable and Negative of Measurable Function is Measurable, $f^-$ is measurable as well.
=== Sufficient Condition ===
Suppose $f^+$ and $f^-$ are both measurable.
By Difference of Positive and Negative Parts:
:$f = f^+ - f^-$
and hence, by Pointwise Difference of Measurable Functions is Measurable, $f$ is also measurable.
== Sources ==
* : $8.11$, $\S 8$: Problem $11$
Category:Measurable Functions
Category:Positive Parts
Category:Negative Parts
Function Measurable with respect to Power Set

== Theorem ==
Let $\struct {X, \map \PP X}$ be a measurable space, where $\map \PP X$ is the power set of $X$. 
Let $f : X \to \overline \R$ be a function.
Then $f$ is $\map \PP X$-measurable function.
== Proof ==
For each $\alpha \in \R$, we have: 
:$\set {x \in X : \map f x \le \alpha} \subseteq X$
That is, from the definition of power set:
:$\set {x \in X : \map f x \le \alpha} \in \map \PP X$
So for each $\alpha \in \R$:
:the set $\set {x \in X : \map f x \le \alpha}$ is $\map \PP X$-measurable.
So:
:$f$ is $\map \PP X$-measurable.
Category:Measurable Functions
Function Obtained by Minimization from URM Computable Relations

== Theorem ==
Let $\RR$ be a URM computable $k+1$-ary relation on $\N^{k + 1}$.
Let the function $f: \N^{k + 1} \to \N$ be a URM computable function.
Let $g: \N^k \to \N$ be the function obtained by minimization from $f$ thus:
:$\map g {n_1, n_2, \ldots, n_k} \approx \mu y \map \RR {n_1, n_2, \ldots, n_k, y}$
Then $g$ is also URM computable.
== Proof ==
From Minimization on Relation Equivalent to Minimization on Function, minimization on $\RR$ is equivalent to minimization on $\overline{\operatorname{sgn} } \circ \chi_\RR$.
We have that a Primitive Recursive Function is URM Computable.
By definition, if $\RR$ is URM computable then so is its characteristic function $\chi_\RR$.
We have that $\overline{\operatorname{sgn} }$ is primitive recursive and thus URM computable.
Thus, from Function Obtained by Substitution from URM Computable Functions, $\overline{\operatorname{sgn} } \circ \chi_\RR$ is URM computable.
Category:URM Programs