nyuuzyou/wikis huggingface dataset:
Pi Squared is Irrational/Proof 2

== Theorem ==
== Proof ==
 $\pi^2$ is rational.
Then $\pi^2 = \dfrac p q$ where $p$ and $q$ are integers and $q \ne 0$.
Let us define a polynomial:
:$\ds \map f x = \frac {\paren {1 - x^2}^n} {n!} = \sum_{m \mathop = n}^{2 n} \frac {c_m} {n!} x^m$
for $c_m \in \Z$.
$\map f x = \map f {-x}$ and so we get:
hence they are all integers.
Now let:
:$\ds A_n = \int_{-1}^1 \map f x \map \cos {\dfrac {\pi x} 2} \rd x = 2 \int_0^1 \map f x \map \cos {\dfrac {\pi x} 2} \rd x$
Repeated integration by Parts gives:
Now let:
:$\ds N_n = \frac \pi 4 p^n A_n = \sum_{k \mathop = 0}^n p^{n - k} \paren {-4 q}^k \map {f^{\paren {2 k} } } 1$
Since $\map f x >0$ and $\map \cos {\dfrac {\pi x} 2} > 0$ on the open interval $\openint {-1} 1$:
:$A_n > 0$
Hence $N_n$ is a positive integer.
Nevertheless, for $x \in \closedint {-1} 1$:
:$0 \le \map f x \le \dfrac 1 {n!}$
and:
:$0 \le \map \cos {\dfrac {\pi x} 2} \le 1$
Hence:
:$0 < A_n < \dfrac 2 {n!}$
Therefore:
:$0 < N_n < \dfrac \pi 2 \dfrac {p^n} {n!}$
From Power over Factorial:
:$\ds \lim_{n \mathop \to \infty} \dfrac {p^n} {n!} = 0$
It follows from the Squeeze Theorem that:
:$\ds \lim_{n \mathop \to \infty} N_n = 0$
Hence for sufficiently large $n$, $N_n$ is strictly between $0$ and $1$.
This contradicts the supposition that $N_n$ is an integer.
It follows that $\pi^2$ is irrational.
Category:Pi Squared is Irrational
Pi Squared is Irrational/Proof 3

== Theorem ==
== Proof ==
 $\pi^2$ is rational.
We establish a lemma:
=== Lemma ===
We will use the definition of $A_n$ from the lemma.
Then we will deduce that $A_n$ is an integer for all $n$.
First we confirm by direct integration that $A_0$ and $A_1$ are integers:
and 
Suppose $A_{k - 2}$ and $A_{k - 1}$ are integers. 
By our lemma:
:$A_k = \paren {16 n - 8} q A_{k - 1} - 16 p q A_{k - 2}$
As $n$, $p$ and $q$ are all integers , $A_k$ is also an integer.
So $A_n$ is an integer for all $n$ by Second Principle of Mathematical Induction.
Let $x \in \closedint 0 1$.
From Shape of Cosine Function and Real Cosine Function is Bounded, we have:
:$0 \le \map \cos {\dfrac {\pi x} 2} \le 1$
From Derivative at Maximum or Minimum, we have:
Therefore, plugging $0$ into $\map f x$, we obtain:
:$0 \le \paren {1 - x^2}^n \le 1$
Since $A_n$ is clearly positive and the length of the interval is $1$ and the integrand is bounded at $\paren {1}^n$, from Darboux's Theorem, we have:
:$0 < A_n < 1 \times \dfrac \pi 2 \times \dfrac {p^n} {n!} \times \paren {1}^n$
From Power over Factorial:
:$\ds \lim_{n \mathop \to \infty} \dfrac {p^n} {n!} = 0$
It follows from the Squeeze Theorem that:
:$\ds \lim_{n \mathop \to \infty} A_n = 0$
Hence for sufficiently large $n$, $A_n$ is strictly between $0$ and $1$.
This contradicts the deduction that $A_n$ is an integer.
This depends on our supposition that $\pi^2$ is a rational number,
Hence by Proof by Contradiction it follows that $\pi^2$ is irrational.
Category:Pi Squared is Irrational
Pi Squared is Irrational/Proof 3/Lemma

== Pi Squared is Irrational: Lemma ==
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let it be supposed that $\pi^2$ is irrational, so that:
:$\pi^2 = \dfrac p q$
where $p$ and $q$ are integers and $q \ne 0$.
Let $A_n$ be defined as:
:$\ds A_n = \frac \pi 2 \frac {p^n} {n!} \int_0^1 \paren {1 - x^2 }^n \map \cos {\dfrac {\pi x} 2} \rd x$
Then:
:$A_n = \paren {16 n - 8} q A_{n - 1} - 16 p q A_{n - 2}$
is a reduction formula for $A_n$.
== Proof ==
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
and let:
Then:
Let:
Now let:
Then:
Category:Pi Squared is Irrational
Pi as Sum of Alternating Sequence of Products of 3 Consecutive Reciprocals

== Theorem ==
:$\dfrac {\pi - 3} 4 = \dfrac 1 {2 \times 3 \times 4} - \dfrac 1 {4 \times 5 \times 6} + \dfrac 1 {6 \times 7 \times 8} \cdots$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
Category:Formulas for Pi
Category:Pi as Sum of Alternating Sequence of Products of 3 Consecutive Reciprocals
Pi as Sum of Alternating Sequence of Products of 3 Consecutive Reciprocals/Lemma

== Theorem ==
:$\ds \iiint \dfrac x {x^2 + 1} \rd x \rd x \rd x = x \map \arctan x + \dfrac {\paren {x^2 - 1} \map \ln {x^2 + 1} - 3 x^2} 4$
with all integration constants at $0$.
== Proof ==
First primitive:
The integration constant is not added due to the series never having a constant during its integration.
Second primitive:
Third primitive:
Category:Pi as Sum of Alternating Sequence of Products of 3 Consecutive Reciprocals
Pi as Sum of Alternating Sequence of Products of 3 Consecutive Reciprocals/Proof 1

== Theorem ==
== Proof ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = x^1 - x^3 + x^5 - x^7 + x^9 - x^{11} + x^{13} - x^{15} \cdots$
We can rewrite this infinite geometric sequence as follows:
Integrating the infinite geometric sequence $3$ times and using Integral of Power, we get:
Integrating the equivalent analytic function $3$ times, we get:
=== Lemma ===
We now have:
:$\ds \sum_{n \mathop = 0}^\infty \dfrac {\paren {-1}^n x^{2 n + 4} } {\paren {2 n + 2} \paren {2 n + 3} \paren {2 n + 4} } = x \map \arctan x + \dfrac {\paren {x^2 - 1} \map \ln {x^2 + 1} - 3 \paren x^2} 4$
Next we confirm that the infinite geometric sequence on the will converge at $x = 1$.
We are guaranteed convergence by the Alternating Series Test:
:$\map f 1 = \dfrac 1 {2 \times 3 \times 4} - \dfrac 1 {4 \times 5 \times 6} + \dfrac 1 {6 \times 7 \times 8} + \cdots + \dfrac {\paren {-1}^n } {\paren {2 n + 2} \paren {2 n + 3} \paren {2 n + 4} }$ 
Finally, we substitute $x = 1$ to obtain our desired result:
Category:Pi as Sum of Alternating Sequence of Products of 3 Consecutive Reciprocals
Pi as Sum of Alternating Sequence of Products of 3 Consecutive Reciprocals/Proof 2

== Theorem ==
== Proof ==
The alternating sum can be written as $\ds \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^{n + 1}} {2 n \paren{2 n + 1} \paren{2 n + 2}}$.
By Partial Fraction Decomposition:
:$\ds \frac 1 {2 n \paren {2 n + 1} \paren {2 n + 2}} = \frac 1 2 \paren{\frac 1 {2 n} - \frac 2 {2 n + 1} + \frac 1 {2 n + 2}}$
Therefore:
Category:Pi as Sum of Alternating Sequence of Products of 3 Consecutive Reciprocals
Pi as Sum of Odd Reciprocals Alternating in Sign in Pairs

== Theorem ==
:$\dfrac {\pi \sqrt 2} 4 = 1 + \dfrac 1 3 - \dfrac 1 5 - \dfrac 1 7 + \dfrac 1 9 + \dfrac 1 {11} - \dfrac 1 {13} - \dfrac 1 {15} \cdots$
== Proof ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \dfrac {x^1} 1 + \dfrac {x^3} 3 - \dfrac {x^5} 5 - \dfrac {x^7} 7 + \dfrac {x^9} 9 + \dfrac {x^{11} } {11} - \dfrac {x^{13} } {13} - \dfrac {x^{15} } {15} \cdots$
We first confirm that the series will converge at $x = 1$.
By grouping the series two terms at a time, we are guaranteed convergence by the Alternating Series Test:
:$ \map f 1 = \paren {1 + \dfrac 1 3 } - \paren { \dfrac 1 5 + \dfrac 1 7 } + \paren { \dfrac 1 9 + \dfrac 1 {11} } - \cdots + \paren {-1}^{\floor {\frac {k - 1} 2} } \paren {\dfrac 1 {2k - 1 } + \dfrac 1 {2k + 1 } } $ 
We now move on to determining the sum of $\map f 1$.
We have:
== Sources ==
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
Category:Formulas for Pi
Pi is Irrational

== Theorem ==
Pi ($\pi$) is irrational.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also see ==
* Pi Squared is Irrational
== Historical Note ==
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 1$: The Language of Set Theory
* : Introduction: Review of Algebra, Geometry, and Trigonometry: $\text{0-1}$: The Real Numbers
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Example $2$
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.1$ Real Numbers
* : Chapter $1$: Numbers: Real Numbers: $4$
* : $\S 1$: Real Numbers: $\S 1.2$: The set of real numbers
* : Chapter $1$: Complex Numbers: The Real Number System: $4$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : $\S 1.2.2$: Numbers, Powers, and Logarithms
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
* 
Category:Pi
Category:Irrationality Proofs
Category:Pi is Irrational
Pi is Irrational/Historical Note

== Historical Note on $\pi$ is Irrational ==
The fact that $\pi$ (pi) is irrational was first proven in $1761$, published in $1768$, by .
== Sources ==
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : Chapter $\text {A}.29$: Liouville ()
* : Chapter $\text {B}.2$: More about Numbers: Irrationals, Perfect Numbers and Mersenne Primes
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
* 
* 
* 
Category:Historical Notes
Pi is Irrational/Historical Note/Mistake

== Source Work ==
:
:pi
== Mistake ==
:Pi was proved to be *irrational by in $1797$, ...
== Correction ==
It was $1761$ when he did this, and it was published in $1768$.
== Also see ==
* Uniform Prism is Uniform Polyhedron
* Uniform Antiprism is Uniform Polyhedron
== Sources ==
* 
* 
Category:Mistakes/The Penguin Dictionary of Mathematics
Pi is Irrational/Proof 1/Historical Note

== Historical Note on Proof 1 of Pi is Irrational ==
This proof was first published by in $1947$, and is considered a classic.
The underlying method was developed by in $1873$, but it was who provided the final details.
== Sources ==
* : Chapter $\text {B}.17$: More About Irrational Numbers. $\pi$ is Irrational
Category:Historical Notes
Pi is Irrational/Proof 2

== Theorem ==
== Proof ==
 $\pi$ is rational.
We establish a lemma:
=== Lemma ===
We will use the definition of $A_n$ from the lemma.
Then we will deduce that $A_n$ is an integer for all $n$.
First we confirm by direct integration that $A_0$ and $A_1$ are integers:
and 
Suppose $A_{k - 2}$ and $A_{k - 1}$ are integers. 
By our lemma:
:$A_k = \paren {4 n - 2} q A_{k - 1} - p^2 A_{k - 2}$
As $n$, $p$ and $q$ are all integers , $A_k$ is also an integer.
So $A_n$ is an integer for all $n$ by Second Principle of Mathematical Induction.
Let $x \in \closedint 0 \pi$.
From Shape of Sine Function and Real Sine Function is Bounded, we have:
:$0 \le \sin x \le 1$
Let:
From Derivative at Maximum or Minimum, $\map f x$ is a maximum or minimum at $x = \dfrac \pi 2$.
Then also:
:$\map f 0 = \map f \pi = 0$
and:
:$\map f {\dfrac \pi 2} = \dfrac {\pi^2} 4$
So on $\closedint 0 \pi$, $x = \dfrac \pi 2$ is in fact a maximum for $x \paren {\pi - x}$.
So we have:
Hence it follows that:
:$0 \le x \paren {\pi - x} \le \dfrac {\pi^2} 4$
Since $A_n$ is clearly positive and the length of the interval is $\pi$ and the integrand is bounded at $\paren {\pi^2 / 4}^n$, from Darboux's Theorem, we have:
:$0 < A_n < \pi \times \dfrac {q^n} {n!} \times \paren {\pi^2 / 4}^n$
From Power over Factorial:
:$\ds \lim_{n \mathop \to \infty} \frac {\paren {q \pi^2 / 4}^n} {n!} = 0$
It follows from the Squeeze Theorem that:
:$\ds \lim_{n \mathop \to \infty} A_n = 0$
Hence for sufficiently large $n$, $A_n$ is strictly between $0$ and $1$.
This contradicts the deduction that $A_n$ is an integer.
This depends on our supposition that $\pi$ is a rational number,
Hence by Proof by Contradiction it follows that $\pi$ is irrational.
Category:Pi is Irrational
Pi is Irrational/Proof 2/Lemma

== Pi is Irrational: Lemma ==
Let $n \in \Z_{> 0}$ be a positive integer.
Let it be supposed that $\pi$ is irrational, so that:
:$\pi = \dfrac p q$
where $p$ and $q$ are integers and $q \ne 0$.
Let $A_n$ be defined as:
:$\ds A_n = \frac {q^n} {n!} \int_0^\pi \paren {x \paren {\pi - x} }^n \sin x \rd x$
Then:
:$A_n = \paren {4 n - 2} q A_{n - 1} - p^2 A_{n - 2}$
is a reduction formula for $A_n$.
== Proof ==
With a view to expressing the primitive in the form:
:$\ds \int u \frac {\d v} {\d x} \rd x = u v - \int v \frac {\d u} {\d x} \rd x$
let:
and let:
Then:
Let:
Now let:
We note in passing that:
:$(1): \quad \paren {q \pi}^2 = q^2 \paren {\dfrac {p^2} {q^2} } = p^2$
which, , is an integer.
Then:
Category:Pi is Irrational
Pi is Irrational/Proof 3

== Theorem ==
== Proof ==
From Rational Points on Graph of Sine Function, the only rational point on the graph of the sine function in the real Cartesian plane $\R^2$:
:$f := \left\{ {\left({x, y}\right) \in \R^2: y = \sin x}\right\}$
is the point $\left({0, 0}\right)$.
But $\left({\pi, 0}\right)$ is also on $f$.
Hence $\pi$ cannot be rational.
== Sources ==
* : Chapter $\text {B}.17$: More About Irrational Numbers. $\pi$ is Irrational
Category:Pi is Irrational
Pi is Transcendental

== Theorem ==
$\pi$ (pi) is transcendental.
== Proof ==
Proof by Contradiction:
 $\pi$ is not transcendental.
Hence by definition, $\pi$ is algebraic.
Let $\pi$ be the root of a non-zero polynomial with rational coefficients, namely $\map f x$.
Then, $\map g x := \map f {i x} \map f {-i x}$ is also a non-zero polynomial with rational coefficients such that:
:$\map g {i \pi} = 0$
Hence, $i \pi$ is also algebraic.
From the Weaker Hermite-Lindemann-Weierstrass Theorem, $e^{i \pi}$ is transcendental.
However, from Euler's Identity:
:$e^{i \pi} = -1$
which is the root of $\map h z = z + 1$, and so is algebraic.
This contradicts the conclusion that $e^{i \pi}$ is transcendental.
Hence by Proof by Contradiction it must follow that $\pi$ is transcendental.
== Historical Note ==
== Sources ==
* : Chapter $8$: Field Extensions: $\S 38$. Simple Algebraic Extensions: Example $77$
* : $\S 1$: Real Numbers: $\S 1.13$: Irrational Numbers
* : $1$: Complex Numbers: Solved Problems: Miscellaneous Problems: $47$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
* 
* 
* 
* 
* 
Category:Transcendental Number Theory
Category:Pi
Pi is Transcendental/Historical Note

== Historical Note on $\pi$ is Transcendental ==
The transcendental nature of $\pi$ (pi) was investigated without success by in $1844$, at around the time he conjectured that Euler's number $e$ was likewise transcendental.
His ideas contributed towards work done by , who proved in $1873$ that Euler's number $e$ is transcendental, but had not noticed that it was a short step from there, via Euler's Identity $e^{i \pi} + 1 = 0$, that $\pi$ is transcendental:
:I shall risk nothing on an attempt to prove the transcendence of the number $\pi$. If others undertake this enterprise, no one will be happier than I at their success, but believe me, my dear friend, it will not fail to cost them some effort.
:::: -- , in a letter to a friend
That final step was made by , who finally achieved this proof in $1882$.
Many people believed that was a grossly inferior mathematician to , and that he achieved this result by pure luck, and that it should have been who gained the credit for it.
However, be that as it may, it was indeed and not who made that actual step of reasoning, and the result falls fair and square at his feet.
== Sources ==
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : Chapter $\text {A}.29$: Liouville ()
* : Chapter $\text {A}.30$: Hermite ()
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
* 
* 
Category:Historical Notes
Picard's Existence Theorem

== Theorem ==
Let $\map f {x, y} : \R^2 \to \R$ be continuous in a region $D \subseteq \R^2$.
Let $\exists M \in \R: \forall x, y \in D: \size {\map f {x, y} } < M$.
Let $\map f {x, y}$ satisfy in $D$ the Lipschitz condition in $y$:
:$\size{\map f {x, y_1} - \map f {x, y_2} } \le A \size {y_1 - y_2}$
where $A$ is independent of $x, y_1, y_2$.
Let the rectangle $R$ be defined as $\set {\tuple {x, y} \in \R^2: \size {x - a} \le h, \size {y - b} \le k}$ such that $M h \le k$.
Let $R \subseteq D$.
Then $\forall x \in \R: \size {x - a} \le h$, the first order ordinary differential equation:
:$y' = \map f {x, y}$
has one and only one solution $y = \map y x$ for which $b = \map y a$.
== Proof 1 ==
== Proof 2 ==
== Also known as ==
== Sources ==
* : Chapter $\text {I}$: Equations of the First Order and Degree: $2$. Integration: footnote ${}^*$
* : $1$: The Nature of Differential Equations: $\S 2$: General Remarks on Solutions
* : Chapter $1$: Normed and Banach spaces
Category:First Order ODEs
Category:Picard's Existence Theorem
Picard's Existence Theorem/Also known as

== Picard's Existence Theorem: Also known as ==
Picard's Existence Theorem is also known as:
:the Picard-Lindelöf Theorem, after and 
:the Cauchy-Lipschitz Theorem, after and .
Some sources give this as Picard's Theorem but there are other theorems with this appellation so it is better to disambiguate.
Category:Picard's Existence Theorem
Pick's Theorem

== Theorem ==
Let $P$ be a polygon whose vertices are the lattice points of an integer lattice embedded in the plane.
Then the area of $P$ is given by the formula:
:$A = l + \dfrac b 2 - 1$
where:
:$l$ denotes the number of lattice points in the interior of $P$
:$b$ denotes the number of lattice points on the boundary of $P$, including the vertices themselves.
== Proof ==
== Examples ==
== Also known as ==
== Historical Note ==
== Sources ==
* 
Category:Pick's Theorem
Category:Integer Lattices
Category:Area Formulas
Pick's Theorem/Examples/Arbitrary Example 1

== Example of Use of Pick's Theorem ==
Consider the following polygon $P$ embedded in an integer lattice:
:
The area of $P$ is $4$.
== Proof ==
The number $l$ of interior lattice points of $P$ is $2$.
The number $b$ of boundary lattice points of $P$ is $6$.
Hence the area $A$ of $P$ is given by Pick's Theorem as:
== Sources ==
* 
Category:Examples of Use of Pick's Theorem
Piecewise Combination of Measurable Mappings is Measurable/Binary Case

== Theorem ==
Let $\struct {X, \Sigma}$ and $\struct {X', \Sigma'}$ be measurable spaces.
Let $f, g: X \to X'$ be $\Sigma \, / \, \Sigma'$-measurable mappings.
Let $E \in \Sigma$ be a measurable set.
Define $h: X \to X'$ by:
:$\forall x \in X: \map h x := \begin{cases}
\map f x & : \text {if $x \in E$} \\
\map g x & : \text {if $x \notin E$}
\end{cases}$
Then $h$ is also a $\Sigma \, / \, \Sigma'$-measurable mapping.
== Proof ==
Let $E' \in \Sigma'$ be a $\Sigma'$-measurable set.
Then by definition of preimage:
:$\map {h^{-1} } {E'} = \set {x \in X: \map h x \in E'}$
Expanding the definition of $h$, this translates into:
:$\map {h^{-1} } {E'} = \set {x \in E: \map f x \in E'} \cup \set {x \in \relcomp X E: \map g x \in E'}$
where $\complement$ denotes set complement.
That is, we have:
:$\map {h^{-1} } {E'} = \paren {E \cap \map {f^{-1} } {E'} } \cup \paren {\relcomp X E \cap \map {g^{-1} } {E'} }$
All sets on the are $\Sigma$-measurable.
By Sigma-Algebra Closed under Intersection and Sigma-Algebra Closed under Union, so is $\map {h^{-1} } {E'}$.
Since $E' \in \Sigma'$ was arbitrary, $h$ is a $\Sigma \, / \, \Sigma'$-measurable mapping.
== Sources ==
* : $\S 8$: Problem $3 \ \text{(i)}$
Category:Piecewise Combination of Measurable Mappings is Measurable
Piecewise Combination of Measurable Mappings is Measurable/General Case

== Theorem ==
Let $\struct {X, \Sigma}$ and $\struct {X', \Sigma'}$ be measurable spaces.
Let $\sequence {E_n}_{n \mathop \in \N} \in \Sigma, \ds \bigcup_{n \mathop \in \N} E_n = X$ be a countable cover of $X$ by $\Sigma$-measurable sets.
For each $n \in \N$, let $f_n: E_n \to X'$ be a $\Sigma_{E_n} \, / \, \Sigma'$-measurable mapping.
Here, $\Sigma_{E_n}$ is the trace $\sigma$-algebra of $E_n$ in $\Sigma$.
Suppose that for every $m, n \in \N$, $f_m$ and $f_n$ satisfy:
:$(1): \quad f_m \restriction_{E_m \cap E_n} = f_n \restriction_{E_m \cap E_n}$
that is, $f_m$ and $f_n$ coincide whenever both are defined; here $\restriction$ denotes restriction.
Define $f: X \to X'$ by:
:$\ds \forall n \in \N, x \in E_n: \map f x := \map {f_n} x$
Then $f$ is a $\Sigma \, / \, \Sigma'$-measurable mapping.
== Proof ==
First, note that $f$ is well-defined, since if $x \in E_n$ and $x \in E_m$, we have that:
:$\map {f_n} x = \map f x = \map {f_m} x$
by $(1)$, since $x \in E_n \cap E_m$.
Let $E' \in \Sigma'$.
Then by definition of preimage, $f^{-1} \sqbrk {E'} \subseteq X$, and hence:
The last step we derive as follows.
Whenever $x \in E_n$, we have by definition of $f$ that $\map f x = \map {f_n} x$.
Hence, $x \in E_n \cap f^{-1} \sqbrk {E'}$ implies $x \in {f_n}^{-1} \sqbrk {E'}$.
Conversely, if $x \in {f_n}^{-1} \sqbrk {E'}$, that is $\map {f_n} x \in E'$, then $x \in E_n$ since $E_n$ is the domain of $f_n$.
By definition of $f$ then also $\map f x = \map {f_n} x \in E'$.
Hence we have deduced that $E_n \cap f^{-1} \sqbrk {E'} = {f_n}^{-1} \sqbrk {E'}$.
Since all $f_n$ are $\Sigma_{E_n} \, / \, \Sigma'$-measurable, it follows that for all $n \in \N$:
:${f_n}^{-1} \sqbrk {E'} \in \Sigma$
By $\sigma$-algebra axiom $(3)$, it follows that:
:$f^{-1} \sqbrk {E'} = \ds \bigcup_{n \mathop \in \N} {f_n}^{-1} \sqbrk {E'} \in \Sigma$
Since $E' \in \Sigma'$ was arbitrary, it follows that $f$ is $\Sigma \, / \, \Sigma'$-measurable.
== Sources ==
* : $\S 8$: Problem $3 \ \text{(ii)}$
Category:Piecewise Combination of Measurable Mappings is Measurable
Piecewise Continuous Function does not necessarily have Improper Integrals

== Theorem ==
Let $f$ be a real function defined on a closed interval $\closedint a b$, $a < b$. 
Let $f$ be a piecewise continuous function:
Then it is not necessarily the case that $f$ is a piecewise continuous function with improper integrals:
== Proof ==
Consider the function:
:$\map f x = \begin{cases}
0 & : x = a \\
\dfrac 1 {x - a} & : x \in \hointl a b
\end{cases}$
Since $\dfrac 1 {x - a}$ is continuous on $\openint a b$, $f$ is continuous on $\openint a b$.
Therefore, $f$ is a piecewise continuous function for the (finite) subdivision $\set {a, b}$ of $\closedint a b$.
We now consider whether the improper integral $\ds \int_{a^+}^{b^-} \dfrac 1 {x - a} \rd x$ exists.
Let $c$ be a point in $\openint a b$.
From the definition of improper integral, the existence of $\ds \int_{a^+}^{b^-} \dfrac 1 {x - a} \rd x$ requires that $\ds \lim_{\gamma \mathop \to a^+} \int_\gamma^c \dfrac 1 {x - a} \rd x$ exists.
We have
The last approaches $\infty$ as $\gamma$ approaches $a$ from above.
So $\ds \int_{a^+}^c \dfrac 1 {x - a} \rd x$ does not exist.
Therefore, $\ds \int_{a^+}^{b^-} \dfrac 1 {x - a} \rd x$ does not exist either.
Accordingly, $f$ is not a piecewise continuous function with improper integrals.
Category:Piecewise Continuous Functions
Piecewise Continuous Function with Improper Integrals may not be Bounded

== Theorem ==
Let $f$ be a real function defined on a closed interval $\closedint a b$, $a < b$. 
Let $f$ be a piecewise continuous function with improper integrals.
Then $f$ may not be piecewise continuous and bounded on $\closedint a b$.
== Proof ==
Consider the function:
:$\map f x = \begin{cases}
0 & : x = a \\
\dfrac 1 {\sqrt{x - a} } & : x \in \hointl a b
\end{cases}$
Since $\dfrac 1 {\sqrt{x - a} }$ is continuous on $\openint a b$, $f$ is continuous on $\openint a b$.
Therefore, $f$ satisfies $(1)$ in the requirements of a piecewise continuous function with improper integrals for the subdivision $\set {a, b}$ of $\closedint a b$.
Also: 
Hence $\ds \int_{a+}^{b-} \map f x \rd x$ exists.
Thus $f$ is a piecewise continuous function with improper integrals.
However, we have that $\map f x$ approaches $\infty$ as $x$ approaches $a$ from above.
Thus $f$ is not bounded.
Therefore $f$ is not piecewise continuous and bounded.
Hence the result.
== Also see ==
* Bounded Piecewise Continuous Function has Improper Integrals
Category:Piecewise Continuous Functions