nyuuzyou/wikis huggingface dataset:
Sum of Independent Poisson Random Variables is Poisson

== Theorem ==
Let $X$ and $Y$ be independent discrete random variables with:
:$X \sim \Poisson {\lambda_1}$
and
:$Y \sim \Poisson {\lambda_2}$
for some $\lambda_1, \lambda_2 \in \R_{> 0}$.
Then their sum $X + Y$ is distributed:
:$X + Y \sim \Poisson {\lambda_1 + \lambda_2}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 4.4$: Sums of independent random variables: Exercise $7$
Category:Poisson Distribution
Category:Probability Generating Functions
Category:Sum of Independent Poisson Random Variables is Poisson
Sum of Independent Poisson Random Variables is Poisson/Proof 1

== Theorem ==
== Proof ==
From Probability Generating Function of Poisson Distribution, we have that the probability generating functions of $X$ and $Y$ are given by:
:$\map {\Pi_X} s = e^{-\lambda_1 \paren {1 - s} }$
:$\map {\Pi_Y} s = e^{-\lambda_2 \paren {1 - s} }$
respectively.
Now because of their independence, we have:
This is the probability generating function for a discrete random variable with a Poisson distribution:
:$\Poisson {\lambda_1 + \lambda_2}$
Therefore:
:$X + Y \sim \Poisson {\lambda_1 + \lambda_2}$
Category:Sum of Independent Poisson Random Variables is Poisson
Sum of Independent Poisson Random Variables is Poisson/Proof 2

== Theorem ==
== Proof == 
From Moment Generating Function of Poisson Distribution, the moment generating functions $X$ and $Y$, $M_X$ and $M_Y$, are given by: 
:$\map {M_X} t = e^{\lambda_1 \paren {e^t - 1} }$
and
:$\map {M_Y} t = e^{\lambda_2 \paren {e^t - 1} }$
As $X$ and $Y$ are independent, we may apply Moment Generating Function of Linear Combination of Independent Random Variables, to give: 
This is the moment generating function of a random variable with distribution $\Poisson {\lambda_1 + \lambda_2}$. 
So, $X + Y \sim \Poisson {\lambda_1 + \lambda_2}$.
Category:Sum of Independent Poisson Random Variables is Poisson
Sum of Independent Random Variables with Mean Zero is Martingale

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\sequence {X_n}_{n \mathop \ge 0}$ be a sequence of integrable independent random variables with:
:$\expect {X_n} = 0$ for each $n \in \N$
and: 
:$X_0 = 0$
For $n \ge 0$ define:
:$\ds S_n = \sum_{i \mathop = 0}^n X_i$
Let $\sequence {\FF_n^X}_{n \mathop \ge 0}$ be the natural filtration for $\sequence {X_n}_{n \mathop \ge 0}$. 
Then $\sequence {S_n}_{n \mathop \ge 0}$ is a $\sequence {\FF_n^X}_{n \mathop \ge 0}$-martingale.
== Proof ==
We first show that $\sequence {S_n}_{n \mathop \ge 0}$ is $\sequence {\FF_n^X}_{n \mathop \ge 0}$-adapted.
From the definition of the $\sigma$-algebra generated by a collection of mappings, we have:
:$X_i$ is $\map \sigma {X_0, \ldots, X_n}$-measurable for $0 \le i \le n$.
So from Pointwise Sum of Measurable Functions is Measurable: General Result, we have:
:$S_n$ is $\map \sigma {X_0, \ldots, X_n}$-measurable.
By the definition of the natural filtration, we have:
:$\map \sigma {X_0, \ldots, X_n} = \FF_n^X$
and hence $\sequence {S_n}_{n \mathop \ge 0}$ is $\sequence {\FF_n^X}_{n \mathop \ge 0}$-adapted.
Now let $n \ge 0$. 
We have:
almost surely.
From Random Variable Independent of Sigma-Algebra Generated by Independent Random Variables, we have: 
:$\map \sigma {X_{n + 1} }$ is independent from $\FF_n^X$. 
So, from Conditional Expectation Unchanged on Conditioning on Independent Sigma-Algebra: Corollary we have:
:$\expect {X_{n + 1} \mid \FF_n^X} = \expect {X_{n + 1} } = 0$ almost surely.
So we have:
:$\expect {S_{n + 1} \mid \FF_n^X} = S_n$ almost surely. 
So $\sequence {S_n}_{n \mathop \ge 0}$ is a $\sequence {\FF_n^X}_{n \mathop \ge 0}$-martingale.
== Sources ==
* : $10.4$: Some examples of martingales
Category:Martingales
Sum of Indexed Suprema

== Theorem ==
Let $\sequence {a_i}_{i \mathop \in I}$ be a family of elements of the real numbers $\R$ indexed by $I$.
Let $\sequence {b_j}_{j \mathop \in J}$ be a family of elements of the real numbers $\R$ indexed by $J$.
Let $\map R i$ and $\map S j$ be propositional functions of $i \in I$, $j \in J$.
Let $\ds \sup_{\map R i} a_i$ and $\ds \sup_{\map S j} b_j$ be the indexed suprema on $\sequence {a_i}$ and $\sequence {b_j}$ respectively.
Then:
:$\ds \paren {\sup_{\map R i} a_i} + \paren {\sup_{\map S j} b_j} = \sup_{\map R i} \paren {\sup_{\map S j} \paren {a_i + b_j} }$
== Proof ==
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $35$
Category:Suprema
Sum of Indices of Falling Factorial

== Theorem ==
:$x^{\underline {m + n} } = x^{\underline m} \paren {x - m}^{\underline n}$
where $x^{\underline m}$ denotes $x$ to the $m$ falling.
== Proof ==
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: Exercise $25$
Category:Falling Factorials
Sum of Indices of Real Number

== Theorem ==
Let $r \in \R_{> 0}$ be a (strictly) positive real number.
=== Positive Integers ===
=== Integers ===
=== Rational Numbers ===
== Sources ==
* : Chapter $1$: Real Numbers and Functions of a Real Variable: $1.2$ Operations with Real Numbers: $(6)$
Category:Sum of Indices of Real Number
Category:Powers
Category:Real Numbers
Sum of Indices of Real Number/Integers

== Theorem ==
Let $r \in \R_{> 0}$ be a positive real number.
Let $n, m \in \Z$ be integers.
Let $r^n$ be defined as $r$ to the power of $n$.
Then:
:$r^{n + m} = r^n \times r^m$
== Proof ==
From Sum of Indices of Real Number: Positive Integers, we have that:
:$m \in \Z_{\ge 0}: \forall n \in \Z: r^{n + m} = r^n \times r^m$
It remains to be shown that:
:$\forall m \in \Z_{<0}: \forall n \in \Z: r^{n + m} = r^n \times r^m$
The proof will proceed by induction on $m$.
As $m < 0$ we have that $m = -p$ for some $p \in \Z_{> 0}$.
For all $p \in \Z_{>0}$, let $\map P p$ be the proposition:
:$\forall n \in \Z: r^{n + \paren {-p} } = r^n \times r^{-p}$
that is:
:$\forall n \in \Z: r^{n - p} = r^n \times r^{-p}$
=== Basis for the Induction ===
$\map P 1$ is true, as follows:
When $n > 0$:
When $n \le 0$:
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\forall n \in \Z: r^{n - k} = r^n \times r^{- k}$
Then we need to show:
:$\forall n \in \Z: r^{n - \paren {k + 1} } = r^n \times r^{-\paren {k + 1} }$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n, m \in \Z: r^{n + m} = r^n \times r^m$
== Sources ==
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: $(5)$
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $7$
Category:Sum of Indices of Real Number
Sum of Indices of Real Number/Positive Integers

== Theorem ==
Let $r \in \R_{> 0}$ be a positive real number.
Let $n, m \in \Z_{\ge 0}$ be positive integers.
Let $r^n$ be defined as $r$ to the power of $n$.
Then:
: $r^{n + m} = r^n \times r^m$
== Proof ==
Proof by induction on $m$:
For all $m \in \Z_{\ge 0}$, let $\map P m$ be the proposition:
:$\forall n \in \Z_{\ge 0}: r^{n + m} = r^n \times r^m$
$\map P 0$ is true, as this just says:
:$r^{n + 0} = r^n = r^n \times 1 = r^n \times r^0$
=== Basis for the Induction ===
$\map P 1$ is true, by definition of power to an integer:
:$r^{n + 1} = r^n \times r = r^n \times r^1$
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\forall n \in \Z_{\ge 0}: r^{n + k} = r^n \times r^k$
Then we need to show:
:$\forall n \in \Z_{\ge 0}: r^{n + k + 1} = r^n \times r^{k + 1}$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n, m \in \Z_{\ge 0}: r^{n + m} = r^n \times r^m$
== Sources ==
* : $\S 0.1$. Arithmetic: Example $1: \ \text{I}$
Category:Sum of Indices of Real Number
Sum of Indices of Real Number/Rational Numbers

== Theorem ==
Let $r \in \R_{> 0}$ be a (strictly) positive real number.
Let $x, y \in \Q$ be rational numbers.
Let $r^x$ be defined as $r$ to the power of $n$.
Then:
: $r^{x + y} = r^x \times r^y$
== Proof ==
Let $x = \dfrac p q, y = \dfrac u v$.
Then:
== Sources ==
* : $\S 1$: Real Numbers: Exercise $\S 1.12 \ (3) \, \text{(i)}$
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $9$
Category:Sum of Indices of Real Number
Sum of Indices of Rising Factorial

== Theorem ==
:$x^{\overline {m + n} } = x^{\overline m} \paren {x + m}^{\overline n}$
where $x^{\overline m}$ denotes $x$ to the $m$ rising.
== Proof ==
== Sources ==
* : $\S 1.2.5$: Permutations and Factorials: Exercise $25$
Category:Rising Factorials
Sum of Infinite Arithmetic-Geometric Sequence

== Theorem ==
Let $\sequence {a_k}$ be an arithmetic-geometric sequence defined as:
:$a_k = \paren {a + k d} r^k$ for $n = 0, 1, 2, \ldots$
Let:
:$\size r < 1$
where $\size r$ denotes the absolute value of $r$.
Then:
:$\ds \sum_{n \mathop = 0}^\infty \paren {a + k d} r^k = \frac a {1 - r} + \frac {r d} {\paren {1 - r}^2}$
== Proof ==
From Sum of Arithmetic-Geometric Sequence, we have:
:$\ds s_n = \sum_{k \mathop = 0}^{n - 1} \paren {a + k d} r^k = \frac {a \paren {1 - r^n} } {1 - r} + \frac {r d \paren {1 - n r^{n - 1} + \paren {n - 1} r^n} } {\paren {1 - r}^2}$
We have that $\size r < 1$.
So by Sequence of Powers of Number less than One:
:$r^n \to 0$ as $n \to \infty$
and
:$r^{n - 1} \to 0$ as $n - 1 \to \infty$
Hence:
:$s_n \to \dfrac a {1 - r} + \dfrac {r d} {\paren {1 - r}^2}$
as $n \to \infty$.
The result follows.
== Linguistic Note ==
== Sources ==
* : $\S 19$: Arithmetic-Geometric Series: $19.7$
* : $\S 21$: Series of Constants: Arithmetic-Geometric Series: $21.7.$
Category:Sums of Sequences
Category:Arithmetic-Geometric Sequences
Sum of Infinite Geometric Sequence

== Theorem ==
Let $S$ be a standard number field, that is $\Q$, $\R$ or $\C$.
Let $z \in S$.
Let $\size z < 1$, where $\size z$ denotes:
:the absolute value of $z$, for real and rational $z$
:the complex modulus of $z$ for complex $z$.
Then $\ds \sum_{n \mathop = 0}^\infty z^n$ converges absolutely to $\dfrac 1 {1 - z}$.
=== Corollary 1 ===
With the same restriction on $z \in S$:
=== Corollary 2 ===
With the same restriction on $z \in S$:
=== Corollary 3 ===
With the same restriction on $z \in S$:
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Sources ==
* : $\S 4.4$. Power Series: Example $\text {(i)}$
* : $1$
* : Chapter $\text {B}.5$: Fermat's Calculation of $\int_0^b x^n \rd x$ for Positive Rational $n$
* : $1$
Category:Sums of Sequences
Category:Series
Category:Examples of Power Series
Category:Geometric Sequences
Category:Sum of Geometric Sequence
Category:Sum of Infinite Geometric Sequence
Sum of Infinite Geometric Sequence/Corollary 1

== Corollary to Sum of Infinite Geometric Sequence ==
Let $S$ be a standard number field, i.e. $\Q$, $\R$ or $\C$.
Let $z \in S$.
Let $\size z < 1$, where $\size z$ denotes:
:the absolute value of $z$, for real and rational $z$
:the complex modulus of $z$ for complex $z$.
Then:
:$\ds \sum_{n \mathop = 1}^\infty z^n = \frac z {1 - z}$
== Proof 1 ==
== Proof 2 ==
Category:Sum of Infinite Geometric Sequence
Sum of Infinite Geometric Sequence/Corollary 2

== Corollary to Sum of Infinite Geometric Sequence ==
Let $S$ be a standard number field, that is $\Q$, $\R$ or $\C$.
Let $z \in S$.
Let $\size z < 1$, where $\size z$ denotes:
:the absolute value of $z$, for real and rational $z$
:the complex modulus of $z$ for complex $z$.
Then:
:$\ds \sum_{n \mathop = 0}^\infty a z^n = \frac a {1 - z}$
== Proof ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.1$ Binomial Theorem etc.: Sum of Geometric Progression to $n$ Terms: $3.1.10$
* : $\S 19$: Geometric Series: $19.5$
* 
* 
* 
* : $\S 21$: Series of Constants: Geometric Series: $21.5.$
* 
Category:Sum of Infinite Geometric Sequence
Sum of Infinite Geometric Sequence/Corollary 2/Mistake

== Source Work ==
:
:geometric series
== Mistake ==
:The sum of the finite initial segments of the series is:
::$\dfrac {a \paren {r^n - 1} } {r - 1}$,
:whence, if the infinite series converges, its sum is $a / \paren {r - 1}$.
== Correction ==
This should read:
:whence, if the infinite series converges, its sum is $a / \paren {1 - r}$.
This is demonstrated in Sum of Infinite Geometric Sequence: Corollary 2.
== Sources ==
* 
Category:Mistakes/Dictionary of Mathematics
Sum of Infinite Geometric Sequence/Corollary 3

== Corollary to Sum of Infinite Geometric Sequence ==
Let $S$ be a standard number field, that is $\Q$, $\R$ or $\C$.
Let $z \in S$.
Let $\size z < 1$, where $\size z$ denotes:
:the absolute value of $z$, for real and rational $z$
:the complex modulus of $z$ for complex $z$.
Then:
:$\ds \sum_{n \mathop = N}^\infty z^n = \frac {z^N } {1 - z}$
== Proof ==
From Sum of Geometric Sequence, we have:
:$\ds\sum_{n \mathop = 0}^{\paren {N - 1} } z^n = \frac {1 - z^{\paren {N - 1} + 1} } {1 - z}$
From Sum of Infinite Geometric Sequence, we have:
:$\ds \sum_{n \mathop = 0}^\infty z^n = \dfrac 1 {1 - z}$
Therefore:
Category:Sum of Infinite Geometric Sequence
Sum of Infinite Geometric Sequence/Proof 1

== Theorem ==
== Proof ==
From Sum of Geometric Sequence, we have:
:$\ds s_N = \sum_{n \mathop = 0}^N z^n = \frac {1 - z^{N + 1} } {1 - z}$
We have that $\size z < 1$.
So by Sequence of Powers of Number less than One:
:$z^{N + 1} \to 0$ as $N \to \infty$
Hence $s_N \to \dfrac 1 {1 - z}$ as $N \to \infty$.
The result follows.
It remains to demonstrate absolute convergence:
The absolute value of $\size z$ is just $\size z$.
By assumption:
:$\size z < 1$
So $\size z$ fulfils the same condition for convergence as $z$.
Hence:
:$\ds \sum_{n \mathop = 0}^\infty \size z^n = \frac 1 {1 - \size z}$
== Sources ==
* : $\S 6.2$
* : $\S 1.2.1$: The Geometric Series: $(1.10)$
Category:Sum of Infinite Geometric Sequence
Sum of Infinite Geometric Sequence/Proof 2

== Theorem ==
== Proof ==
By the Chain Rule for Derivatives and the corollary to $n$th Derivative of Reciprocal of $m$th Power:
:$\dfrac {\d^n} {\d z^n} \dfrac 1 {1 - z} = \dfrac {n!} {\paren {1 - z}^{n + 1} }$
Thus the Maclaurin series expansion of $\dfrac 1 {1 - z}$ is:
:$\ds \sum_{n \mathop = 0}^\infty \frac {z^n} {n!} \dfrac {n!} {\paren {1 - 0}^{n + 1} }$
whence the result.
== Sources ==
* : $\S 1.3.2$: Power series: $(1.44)$
Category:Sum of Infinite Geometric Sequence
Sum of Infinite Series of Product of Power and Cosine

== Theorem ==
Let $r \in \R$ such that $\size r < 1$.
Then:
== Proof ==
From Euler's Formula:
:$e^{i \theta} = \cos \theta + i \sin \theta$
Hence:
== Also see ==
* Sum of Infinite Series of Product of Power and Sine
* Sum of Series of Product of Power and Cosine
* Sum of Series of Product of Power and Sine
== Sources ==
* : $\S 4.4$. Power Series
* : $\S 19$: Miscellaneous Series: $19.41$
* : $2$: Miscellaneous Problems: $46 \ \text{(a)}$
* : $\S 21$: Series of Constants: Miscellaneous Series: $21.41.$
Category:Cosine Function
Category:Trigonometric Series
Sum of Infinite Series of Product of Power and Sine

== Theorem ==
Let $r \in \R$ such that $\size r < 1$.
Then:
== Proof ==
From Euler's Formula:
:$e^{i \theta} = \cos \theta + i \sin \theta$
Hence:
== Also see ==
* Sum of Infinite Series of Product of Power and Cosine
* Sum of Series of Product of Power and Sine
* Sum of Series of Product of Power and Cosine
== Sources ==
* : Chapter One: $\S 1$. Trigonometrical Series
* : $\S 19$: Miscellaneous Series: $19.42$
* : $2$: Miscellaneous Problems: $46 \ \text{(b)}$
* : $\S 21$: Series of Constants: Miscellaneous Series: $21.42.$
Category:Sine Function
Category:Trigonometric Series
Sum of Infinite Series of Product of nth Power of Cosine by nth Multiple of Cosine

== Theorem ==
Let $0 < \theta < \dfrac \pi 2$.
Then:
== Proof ==
Let $0 < \theta < \dfrac \pi 2$.
Then $0 < \cos \theta < 1$.
== Sources ==
* : $\S 4.4$. Power Series
Category:Cosine Function
Sum of Infinite Series of Product of nth Power of cos 2 theta by 2n+1th Multiple of Sine

== Theorem ==
Let $\theta \in \R$ such that $\theta \ne m \pi$ for any $m \in \Z$.
Then:
== Proof ==
Let $\theta \ne \dfrac {m \pi} 2$ for any $m \in \Z$.
Then $\size {\cos 2 \theta} < 1$.
== Sources ==
* : $\S 4$. Elementary Functions of a Complex Variable: Exercise $10$
Category:Cosine Function
Sum of Integer Combinations is Integer Combination

== Lemma ==
Let $a, b \in \Z$ be integers.
Let $S = \set {a x + b y: x, y \in \Z}$ be the set of integer combinations of $a$ and $b$.
Let $u \in S$ and $v \in S$.
Then $u + v \in S$.
== Proof ==
As both $u, v \in S$, $u$ and $v$ can be expressed as:
where $x_1, x_2, y_1, y_2$ are integers.
Then:
As Integer Addition is Closed, both $x_1 + x_2$ and $y_1 + y_2$ are integers.
Hence the result.
== Sources ==
* : Appendix $1$: Some Results from the Elementary Theory of Numbers: Lemma $1$
Category:Integer Combinations
Sum of Integer Ideals is Greatest Common Divisor

== Theorem ==
Let $\ideal m$ and $\ideal n$ be ideals of the integers $\Z$.
Let $\ideal d = \ideal m + \ideal n$.
Then $d = \gcd \set {m, n}$.
== Proof ==
By Sum of Ideals is Ideal we have that $\ideal d = \ideal m + \ideal n$ is an ideal of $\Z$.
By Ring of Integers is Principal Ideal Domain we have that $\ideal m$, $\ideal n$ and $\ideal d$ are all necessarily principal ideals.
By Subrings of Integers are Sets of Integer Multiples we have that:
:$\ideal m = m \Z, \ideal n = n \Z$
Thus:
:$\ideal d = \ideal m + \ideal n = \set {x \in \Z: \exists a, b \in \Z: x = a m + b n}$
That is, $\ideal d$ is the set of all integer combinations of $m$ and $n$.
The result follows by Bézout's Identity.
== Sources ==
* : Chapter $5$: Rings: $\S 21$. Ideals: Example $38$
Category:Ideal Theory
Category:Integers
Category:Greatest Common Divisor
Sum of Integrals on Adjacent Intervals for Continuous Functions

== Theorem ==
Let $f$ be a real function which is continuous on any closed interval $I$.
Let $a, b, c \in I$.
Then:
:$\ds \int_a^c \map f t \rd t + \int_c^b \map f t \rd t = \int_a^b \map f t \rd t$
== Proof ==
By Continuous Real Function is Darboux Integrable, $f$ is integrable on $I$.
The result follows by application of Sum of Integrals on Adjacent Intervals for Integrable Functions.
== Comment ==
This proof would be very simple if we were to use the Fundamental Theorem of Calculus:
but such a proof would be circular.
== Sources ==
* : $\S 13.8$
* : $\S 4.3$
Category:Integral Calculus
Sum of Integrals on Adjacent Intervals for Integrable Functions/Corollary

== Corollary to Sum of Integrals on Adjacent Intervals for Integrable Functions ==
Let $f$ be a real function which is Darboux integrable on any closed interval $\mathbb I$.
Let $a_0, a_1, \ldots, a_n$ be real numbers, where $n \in \N$ and $n \ge 2$.
Then:
:$\ds \int_{a_0}^{a_n} \map f t \rd t = \sum_{i \mathop = 0}^{n - 1} \int_{a_i}^{a_{i + 1} } \map f t \rd t$
== Proof ==
Proof by induction:
=== Basis for the Induction ===
According to Sum of Integrals on Adjacent Intervals for Integrable Functions, $n = 2$ holds.
This is the basis for the induction.
=== Induction Hypothesis ===
This is our induction hypothesis:
:$\ds \int_{a_0}^{a_k} \map f t \rd x = \sum_{i \mathop = 0}^{k - 1} \int_{a_i}^{a_{i + 1} } \map f t \rd t$
Now we need to show true for $n = k + 1$:
:$\ds \int_{a_0}^{a_{k + 1} } \map f t \rd t = \sum_{i \mathop = 0}^k \int_{a_i}^{a_{i + 1} } \map f t \rd t$
=== Induction Step ===
This is our induction step:
The result follows by induction.
== Sources ==
* : Chapter Two: $\S 1$. Piecewise-Continuous Functions: $(5)$
* : $\S 15$: General Formulas involving Definite Integrals: $15.7$
* : $\S 18$: Definite Integrals: General Formulas involving Definite Integrals: $18.7$
Category:Proofs by Induction
Category:Sum of Integrals on Adjacent Intervals for Integrable Functions
Sum of Integrals on Adjacent Intervals for Integrable Functions/Lemma

== Theorem ==
Let $\closedint a b$ be a closed real interval.
Let $c$ be a real number.
Let $a < c < b$.
Let $f$ be a real function defined on $\closedint a b$.
Let $\map L S$ be the lower Darboux sum of $f$ on $\closedint a b$ where $S$ is a subdivision of $\closedint a b$.
Let $P$ and $Q$ be finite subdivisions of $\closedint a b$.
Let:
:$Q = P \cup \set c$.
Then:
:$\map L Q \ge \map L P$
== Proof 1 ==
== Proof 2 ==
Category:Sum of Integrals on Adjacent Intervals for Integrable Functions
Sum of Integrals on Adjacent Intervals for Integrable Functions/Lemma/Proof 2

== Theorem ==
== Proof ==
Let $P = \set {x_0, x_1, \ldots, x_n}$.
Suppose that:
:$c \in P$
Then:
:$Q = P$
We have:
:$\map L P \ge \map L P$
:$\leadsto \map L Q \ge \map L P$ as $Q = P$
This finishes the proof for this case.
The only other possibility for $c$ is:
:$x_{j-1} < c < x_j$
where $1 \le j \le n$.
Let $m_i$ be the infimum of $f$ on the interval $\closedint {x_{i - 1}} {x_i}$ for all $i \in \set {1, 2, \ldots, n}$.
We have:
Let $r$ be the infimum of $f$ on the interval $\closedint {x_{j - 1}} c$.
Let $s$ be the infimum of $f$ on the interval $\closedint c {x_j}$.
The interval $\closedint {x_{j - 1}} c$ is a subset of $\closedint {x_{j - 1}} {x_j}$.
Therefore, a lower bound of $f$ on the interval $\closedint {x_{j - 1}} {x_j}$ is a lower bound of $f$ on the interval $\closedint {x_{j - 1}} c$ as well.
The infimum of $f$ on the interval $\closedint {x_{j - 1}} {x_j}$ is a lower bound for $f$ on this interval.
Accordingly, the infimum of $f$ on the interval $\closedint {x_{j - 1}} {x_j}$ is a lower bound of $f$ on the interval $\closedint {x_{j - 1}} c$ as well.
Therefore, the infimum of $f$ on the interval $\closedint {x_{j - 1}} c$ is greater than or equal to the infimum of $f$ on the interval $\closedint {x_{j - 1}} {x_j}$.
In other words:
:$r \ge m_j$
A similar analysis of $\closedint c {x_j}$ gives:
:$s \ge m_j$
We find:
This finishes the proof.
Category:Sum of Integrals on Adjacent Intervals for Integrable Functions
Sum of Integrals on Complementary Sets

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $A, E \in \Sigma$ with $A \subseteq E$.
Let $f$ be a $\mu$-integrable function on $X$.
Then 
:$\ds \int_E f \rd \mu = \int_A f \rd \mu + \int_{E \mathop \setminus A} f \rd \mu$
== Proof ==
Let $\chi_E$ be the characteristic function of $E$.
Because $A$ and $E \setminus A$ are disjoint:
:$A \cap \paren {E \setminus A} = \O$
By Characteristic Function of Union:
:$\chi_E = \chi_A + \chi_{E \mathop \setminus A}$ 
Integrating $f$ over $E$ gives:
== Also see ==
* Sum of Integrals on Adjacent Intervals for Integrable Functions
== Sources ==
* : Chapter $4$, section $4$, proposition $15 \ \text{(iv)}$
Category:Integral of Integrable Function over Measurable Set
Sum of Internal Angles of Polygon

== Theorem ==
The sum $S$ of all internal angles of a polygon with $n$ sides is given by the formula $S = \paren {n - 2} 180 \degrees$.
=== Corollary ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.9$: Corollary
* : $\S 1.3.1$
* 
* 
Category:Sum of Internal Angles of Polygon
Category:Polygons
Sum of Internal Angles of Polygon/Proof 1

== Theorem ==
== Proof ==
The Polygon Triangulation Theorem shows that there exists a triangulation of the polygon that consists of $n - 2$ triangles.
The sides of these triangles are sides and chords of the polygon, where the chords lie completely in the interior of $P$.
Hence the vertices of the triangles are vertices of the polygon.
Sum of Angles of Triangle equals Two Right Angles shows that the sum of the internal angles of a triangle is $180 \degrees$.
As the triangulation covers the polygon, the sum of the internal angles of the vertices of the triangles in the triangulation is equal to $S$.
So:
:$S = \paren {n - 2} 180 \degrees$
== Sources ==
* : $\S 1.3.1$
Category:Sum of Internal Angles of Polygon