nyuuzyou/wikis huggingface dataset:
Pointwise Multiplication is Associative

== Theorem ==
Let $S$ be a non-empty set.
Let $\mathbb F$ be one of the standard number sets: $\Z, \Q, \R$ or $\C$.
Let $f, g, h: S \to \mathbb F$ be functions.
Let $f \times g: S \to \mathbb F$ denote the pointwise product of $f$ and $g$.
Then:
:$\paren {f \times g} \times h = f \times \paren {g \times h}$
That is, pointwise multiplication is associative.
== Proof ==
== Specific Contexts ==
This result can be applied and proved in the context of the various standard number sets:
=== Pointwise Multiplication on Integer-Valued Functions is Associative ===
=== Pointwise Multiplication on Rational-Valued Functions is Associative ===
=== Pointwise Multiplication on Real-Valued Functions is Associative ===
=== Pointwise Multiplication on Complex-Valued Functions is Associative ===
Category:Pointwise Multiplication
Category:Examples of Associative Operations‎
Category:Pointwise Multiplication is Associative
Pointwise Multiplication is Commutative

== Theorem ==
Let $S$ be a non-empty set.
Let $\mathbb F$ be one of the standard number sets: $\Z, \Q, \R$ or $\C$.
Let $f, g, h: S \to \mathbb F$ be functions.
Let $f \times g: S \to \mathbb F$ denote the pointwise product of $f$ and $g$.
Then:
:$f \times g = g \times f$
That is, pointwise multiplication is commutative.
== Proof ==
== Specific Contexts ==
This result can be applied and proved in the context of the various standard number sets:
=== Pointwise Multiplication on Integer-Valued Functions is Commutative ===
=== Pointwise Multiplication on Rational-Valued Functions is Commutative ===
=== Pointwise Multiplication on Real-Valued Functions is Commutative ===
=== Pointwise Multiplication on Complex-Valued Functions is Commutative ===
Category:Pointwise Multiplication
Category:Examples of Commutative Operations
Category:Pointwise Multiplication is Commutative
Pointwise Multiplication on Complex-Valued Functions is Associative

== Theorem ==
Let $S$ be a non-empty set.
Let $f, g, h: S \to \C$ be complex-valued functions.
Let $f \times g: S \to \C$ denote the pointwise product of $f$ and $g$.
Then:
:$\paren {f \times g} \times h = f \times \paren {g \times h}$
That is, pointwise multiplication on complex-valued functions is associative.
== Proof ==
Category:Pointwise Multiplication is Associative
Category:Complex Multiplication
Pointwise Multiplication on Complex-Valued Functions is Commutative

== Definition ==
Let $S$ be a non-empty set.
Let $f, g: S \to \C$ be complex-valued functions.
Let $f \times g: S \to \C$ denote the pointwise product of $f$ and $g$.
Then:
:$f \times g = g \times f$
That is, pointwise multiplication of complex-valued functions is commutative.
== Proof ==
Category:Pointwise Multiplication is Commutative
Category:Complex Multiplication
Pointwise Multiplication on Integer-Valued Functions is Associative

== Theorem ==
Let $S$ be a non-empty set.
Let $f, g, h: S \to \Z$ be integer-valued functions.
Let $f \times g: S \to \Z$ denote the pointwise product of $f$ and $g$.
Then:
:$\paren {f \times g} \times h = f \times \paren {g \times h}$
That is, pointwise multiplication on integer-valued functions is associative.
== Proof ==
Category:Pointwise Multiplication is Associative
Category:Integer Multiplication
Pointwise Multiplication on Integer-Valued Functions is Commutative

== Definition ==
Let $S$ be a set.
Let $f, g: S \to \Z$ be integer-valued functions.
Let $f \times g: S \to \Z$ denote the pointwise product of $f$ and $g$.
Then:
:$f \times g = g \times f$
That is, pointwise multiplication of integer-valued functions is commutative.
== Proof ==
Category:Pointwise Multiplication is Commutative
Category:Integer Multiplication
Pointwise Multiplication on Rational-Valued Functions is Associative

== Theorem ==
Let $S$ be a non-empty set.
Let $f, g, h: S \to \Q$ be rational-valued functions.
Let $f \times g: S \to \Q$ denote the pointwise product of $f$ and $g$.
Then:
:$\paren {f \times g} \times h = f \times \paren {g \times h}$
That is, pointwise multiplication on rational-valued functions is associative.
== Proof ==
Category:Pointwise Multiplication is Associative
Category:Rational Multiplication
Pointwise Multiplication on Rational-Valued Functions is Commutative

== Definition ==
Let $S$ be a non-empty set.
Let $f, g: S \to \Q$ be rational-valued functions.
Let $f \times g: S \to \Q$ denote the pointwise product of $f$ and $g$.
Then:
:$f \times g = g \times f$
That is, pointwise multiplication of rational-valued functions is commutative.
== Proof ==
Category:Pointwise Multiplication is Commutative
Category:Rational Multiplication
Pointwise Multiplication on Real-Valued Functions is Associative

== Theorem ==
Let $S$ be a non-empty set.
Let $f, g, h: S \to \R$ be real-valued functions.
Let $f \times g: S \to \R$ denote the pointwise product of $f$ and $g$.
Then:
:$\paren {f \times g} \times h = f \times \paren {g \times h}$
That is, pointwise multiplication on real-valued functions is associative.
== Proof ==
Category:Pointwise Multiplication is Associative
Category:Real Multiplication
Pointwise Multiplication on Real-Valued Functions is Commutative

== Definition ==
Let $S$ be a non-empty set.
Let $f, g: S \to \R$ be real-valued functions.
Let $f \times g: S \to \R$ denote the pointwise product of $f$ and $g$.
Then:
:$f \times g = g \times f$
That is, pointwise multiplication of real-valued functions is commutative.
== Proof ==
Category:Pointwise Multiplication is Commutative
Category:Real Multiplication
Pointwise Multiplication on Space of Real-Valued Measurable Functions Identified by A.E. Equality is Well-Defined

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map {\mathcal M} {X, \Sigma, \R}$ be the set of real-valued $\Sigma$-measurable functions on $X$. 
Let $\sim$ be the almost-everywhere equality equivalence relation on $\map {\mathcal M} {X, \Sigma, \R}$.
Let $\map {\mathcal M} {X, \Sigma, \R}/\sim$ be the space of real-valued $\Sigma$-measurable functions identified by $\sim$.
Then pointwise scalar multiplication on $\map {\mathcal M} {X, \Sigma, \R}/\sim$ is well-defined.
== Proof ==
Let $E_1, E_2 \in \map {\mathcal M} {X, \Sigma, \R}/\sim$.
We need to show that $E_1 \cdot E_2$ is independent of the choice of representative for $E_1$ and $E_2$.
Suppose that: 
:$\eqclass f \sim = \eqclass F \sim = E_1$
and:
:$\eqclass g \sim = \eqclass G \sim = E_2$
From Equivalence Class Equivalent Statements, we have:
:$f \sim F$
and:
:$g \sim G$ 
So, from Pointwise Multiplication preserves A.E. Equality, we have: 
:$f \cdot g \sim F \cdot G$
That is, from Equivalence Class Equivalent Statements:
:$\eqclass {f \cdot g} \sim = \eqclass {F \cdot G} \sim$ 
which is what we aimed to show.
Category:Space of Real-Valued Measurable Functions Identified by A.E. Equality
Pointwise Multiplication preserves A.E. Equality

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g, h : X \to \overline \R$ be functions with: 
:$f = g$ $\mu$-almost everywhere.
Then: 
:$f \times h = g \times h$ $\mu$-almost everywhere
where $f \times h$ and $g \times h$ are the pointwise products of $f$ and $h$, and $g$ and $h$ respectively.
== Proof ==
Since:
:$f = g$ $\mu$-almost everywhere
there exists a $\mu$-null set $N \subseteq X$ such that: 
:if $x \in X$ has $\map f x \ne \map g x$ then $x \in N$.
Note that if $x \in X$ is such that: 
:$\map f x = \map g x$
then: 
:$\map f x \map h x = \map g x \map h x$
By the Rule of Transposition, we therefore have: 
:if $x \in X$ has $\map f x \map h x \ne \map g x \map h x$ then $\map f x \ne \map g x$.
So: 
:if $x \in X$ has $\map f x \map h x \ne \map g x \map h x$ then $x \in N$. 
Since $N$ is $\mu$-null, we therefore have: 
:$f \times h = g \times h$ $\mu$-almost everywhere.
Category:Almost-Everywhere Equality Relation
Pointwise Operation/Examples/Cube and Sine Functions

== Examples of Pointwise Operations ==
Let $f$ and $g$ be the real functions defined as
Then for all $x \in \R$:
Contrast this with:
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces
Category:Examples of Pointwise Operations
Pointwise Operation is Composite of Operation with Mapping to Cartesian Product

== Theorem ==
Let $S$ be a set.
Let $\struct {T, *}$ be an algebraic structure.
Let $T^S$ be the set of all mappings from $S$ to $T$.
Let the algebraic structure $\struct {T^S, \oplus}$ be the algebraic structure on $T^S$ induced by $*$.
Let $f, g \in T^S$, that is, let $f: S \to T$ and $g: S \to T$ be mappings.
Let $f \times g : S \to T \times T$ be the mapping from $S$ to the cartesian product $T \times T$ defined by:
:$\forall x \in S : \map {\paren {f \times g}} x = \tuple {\map f x, \map g x}$
Then:
:$f \oplus g = * \circ \paren {f \times g}$
That is, $f \oplus g$ is the composition of the binary operation $*$ with the mapping $f \times g: S \to T \times T$.
== Proof ==
From Equality of Mappings:
:$* \circ \paren {f \times g} = f \oplus g$
Category:Abstract Algebra
Category:Pointwise Operations
Pointwise Operation on Distributive Structure is Distributive

== Theorem ==
Let $S$ be a set.
Let $\struct {T, +, \circ}$ be an algebraic structure with two operations $+$ and $\circ$.
Let $T^S$ be the set of all mappings from $S$ to $T$.
Let $\struct {T^S, +', \circ'}$ be the structure induced on $T^S$ by $+$ and $\circ$.
Let $\circ$ be distributive over $+$ in $T$.
Then $\circ'$ is distributive over $+'$ in $T$.
== Proof ==
Let $f, g, h: S \to T$ be elements of $T^S$.
Suppose $S$ is the empty set.
Suppose $T^S$ is the set of all mappings from the empty set, $S$, to $T$.
Suppose $\struct {T^S, +', \circ'}$ is the structure induced on $T^S$ by $+$ and $\circ$.
Suppose $\circ$ is distributive over $+$ in $T$.
Then $\circ'$ is distributive over $+'$ in $T$, as required.
Suppose $S$ is non-empty.
Let $x \in S$.
Then:
Similarly:
:$\map {\paren {\paren {g +' h} \circ' f} } x = \map {\paren {\paren {g \circ' f} +' \paren {h \circ' f} } } x$
Hence the result.
Category:Pointwise Operations
Category:Distributive Operations
Pointwise Product of Measurable Functions is Measurable

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f, g: X \to \overline \R$ be $\Sigma$-measurable functions.
Then the pointwise product $f \cdot g: X \to \overline \R$ is also $\Sigma$-measurable.
== Proof ==
By Measurable Function is Pointwise Limit of Simple Functions, we find sequences $\sequence {f_n}_{n \mathop \in \N}, \sequence {g_n}_{n \mathop \in \N}$ such that:
:$\ds f = \lim_{n \mathop \to \infty} f_n$
:$\ds g = \lim_{n \mathop \to \infty} g_n$
where the limits are pointwise.
It follows that for all $x \in X$:
:$\map f x \cdot \map g x = \ds \lim_{n \mathop \to \infty} \map {f_n} x \cdot \map {g_n} x$
so that we have the pointwise limit:
:$\ds f \cdot g = \lim_{n \mathop \to \infty} f_n \cdot g_n$
By Pointwise Product of Simple Functions is Simple Function, $f \cdot g$ is a pointwise limit of simple functions.
By Simple Function is Measurable, $f \cdot g$ is a pointwise limit of measurable functions.
Hence $f \cdot g$ is measurable, by Pointwise Limit of Measurable Functions is Measurable.
== Sources ==
* : $8.10$
Category:Measurable Functions
Pointwise Product of Simple Functions is Simple Function

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f, g : X \to \R$ be simple functions.
Then $f \cdot g: X \to \R, \map {\paren {f \cdot g} } x := \map f x \cdot \map g x$ is also a simple function.
== Proof ==
From Measurable Function is Simple Function iff Finite Image Set, it follows that there exist $x_1, \ldots, x_n$ and $y_1, \ldots y_m$ comprising the image of $f$ and $g$, respectively.
But then it immediately follows that any value attained by $f \cdot g$ is of the form $x_i \cdot y_j$.
Hence, there are at most $n \times m$ distinct values $f \cdot g$ can take.
From Pointwise Product of Measurable Functions is Measurable, $f \cdot g$ is also measurable.
Hence by Measurable Function is Simple Function iff Finite Image Set, $f \cdot g$ is a simple function.
== Also see ==
* Definition:Pointwise Operation, of which the definition of $f \cdot g$ is an instantiation
* Scalar Multiple of Simple Function is Simple Function
* Pointwise Sum of Simple Functions is Simple Function
* Space of Simple Functions forms Ring
== Sources ==
* : $8.7 \ \text{(iv)}$
Category:Simple Functions
Category:Pointwise Operations
Category:Pointwise Product of Simple Functions is Simple Function
Pointwise Scalar Multiple of Measurable Function is Measurable

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f : X \to \overline \R$ be a $\Sigma$-measurable function.
Let $\alpha \in \overline \R$ be an extended real number. 
Then the pointwise scalar multiple $\alpha f$ is $\Sigma$-measurable.
== Proof ==
We want to show that: 
:$\set {x \in X : \alpha \map f x \le t}$ is $\Sigma$-measurable for each $t \in \R$.
in each of the cases: 
:$(1): \quad$ $\alpha = 0$
:$(2): \quad$ $0 < \alpha < \infty$
:$(3): \quad$ $-\infty < \alpha < 0$
:$(4): \quad$ $\alpha = \infty$
:$(5): \quad$ $\alpha = -\infty$
If $\alpha = 0$, then $\alpha f = 0$, and $\alpha f$ is $\Sigma$-measurable from Constant Function is Measurable, giving case $(1)$. 
Suppose that $0 < \alpha < \infty$. 
Since $f$ is $\Sigma$-measurable, we have that: 
:$\ds \set {x \in X : \map f x \le \frac t \alpha}$ is $\Sigma$-measurable for each $t \in \R$. 
Since: 
:$\ds \set {x \in X : \alpha \map f x \le t} = \set {x \in X : \map f x \le \frac t \alpha}$
we have the theorem in the case $\alpha > 0$, giving case $(2)$. 
Suppose that $-\infty < \alpha < 0$.
Since $f$ is $\Sigma$-measurable, we have from Characterization of Measurable Functions:
:$\ds \set {x \in X :\map f x \ge \frac t \alpha}$ is $\Sigma$-measurable for each $t \in \R$.
Since:
:$\ds \set {x \in X : \alpha \map f x \le t} = \set {x \in X :\map f x \ge \frac t \alpha}$
we have the theorem in the case $\alpha < 0$, giving case $(3)$. 
Suppose that $\alpha = \infty$.
Then from the definition of extended real multiplication, we have: 
:$\map {\paren {\alpha f} } x = \begin{cases}\infty & \map f x \ne 0 \\ 0 & \map f x = 0\end{cases}$ 
If $t < 0$, we have: 
:$\set {x \in X : \alpha \map f x \le t} = \O$
which is $\Sigma$-measurable.
If $t \ge 0$, we have: 
:$\set {x \in X : \alpha \map f x \le t} = \set {x \in X : \map f x = 0}$
From Measurable Functions Determine Measurable Sets and Constant Function is Measurable, we have that: 
:$\set {x \in X : \map f x = 0} \in \Sigma$
So:
:$\set {x \in X : \alpha \map f x \le t}$ is $\Sigma$-measurable for each $t \in \R$
giving case $(4)$. 
Suppose that $\alpha = -\infty$.
From the rules of extended real multiplication, we have: 
:$\paren {-\infty} \times f = -\paren {\infty \times f}$ 
From case $(4)$, we have that: 
:$\infty \times f$ is $\Sigma$-measurable. 
From case $(2)$, we then have: 
:$\paren {-1} \times \paren {\infty \times f} = \paren {-\infty} \times f$ is $\Sigma$-measurable
hence showing case $(5)$.
Category:Measurable Functions
Pointwise Scalar Multiplication on Space of Real-Valued Measurable Functions Identified by A.E. Equality is Well-Defined

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map {\mathcal M} {X, \Sigma, \R}$ be the set of real-valued $\Sigma$-measurable functions on $X$. 
Let $\sim$ be the $\mu$-almost-everywhere equality relation on $\map {\mathcal M} {X, \Sigma, \R}$.
Let $\map {\mathcal M} {X, \Sigma, \R}/\sim$ be the space of real-valued $\Sigma$-measurable functions identified by $\sim$.
Then pointwise scalar multiplication on $\map {\mathcal M} {X, \Sigma, \R}/\sim$ is well-defined.
== Proof ==
Let $\lambda \in \R$.
Let $E \in \map {\mathcal M} {X, \Sigma, \R}/\sim$.
First, we show that if $E = \eqclass f \sim$, then $\eqclass {\lambda f} \sim$ is well-understood.
This follows from Pointwise Scalar Multiple of Measurable Function is Measurable.
We need to show that $\lambda \cdot E$ is independent of the choice of representative for $E$.
Suppose that: 
:$\eqclass f \sim = \eqclass g \sim = E$ 
From Equivalence Class Equivalent Statements, we have:
:$f \sim g$
So, from Pointwise Scalar Multiplication preserves A.E. Equality, we have: 
:$\lambda \cdot f \sim \lambda \cdot g$
That is, from Equivalence Class Equivalent Statements:
:$\eqclass {\lambda \cdot f} \sim = \eqclass {\lambda \cdot g} \sim$ 
which is what we aimed to show.
Category:Space of Real-Valued Measurable Functions Identified by A.E. Equality
Pointwise Scalar Multiplication preserves A.E. Equality

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g : X \to \overline \R$ be extended real-valued functions such that: 
:$f = g$ $\mu$-almost everywhere.
Let $\lambda \in \overline \R$.
Then: 
:$\lambda \cdot f = \lambda \cdot g$ $\mu$-almost everywhere
where $\lambda \cdot f$ denotes pointwise scalar multiplication.
== Proof ==
Since:
:$f = g$ $\mu$-almost everywhere
there exists a $\mu$-null set $N \subseteq X$ such that: 
:if $\map f x \ne \map g x$ then $x \in N$. 
Note that if $\map f x = \map g x$ then $\lambda \map f x = \lambda \map g x$. 
So, from the Rule of Transposition we have:
:if $\lambda \map f x \ne \lambda \map g x$ then $\map f x \ne \map g x$
so:
:if $\lambda \map f x \ne \lambda \map g x$ then $x \in N$. 
So, we have: 
:$\lambda \cdot f = \lambda \cdot g$ $\mu$-almost everywhere.
Category:Almost-Everywhere Equality Relation
Pointwise Sum of Integrable Functions is Integrable Function

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f, g: X \to \overline \R$ be $\mu$-integrable functions.
Suppose that their pointwise sum $f + g$ is well-defined.
Then $f + g$ is also a $\mu$-integrable function.
That is, the space of $\mu$-integrable functions $\LL^1_{\overline \R}$ is closed under pointwise addition.
== Proof ==
We are given $f, g: X \to \overline \R$ are $\mu$-integrable functions:
where $f^+$ and $f^-$ are the positive and negative parts of $f$ respectively.
Also:
We are given that pointwise sum $f + g$ is well-defined.
So by Bound for Positive Part of Pointwise Sum of Functions, there exists no $x \in X$ such that:
:$\set {\map f x, \map g x} = \set {\infty, -\infty}$
and:
:$\paren {f + g}^+ \le f^+ + g^+$
By Bound for Negative Part of Pointwise Sum of Functions, there exists no $x \in X$ such that:
:$\set {\map f x, \map g x} = \set {\infty, -\infty}$ 
and:
:$\paren {f + g}^- \le f^- + g^-$
Therefore:
Hence $f + g$ is also a $\mu$-integrable function.
== Sources ==
* : $10.4 \, \text{(ii)}$
Category:Integrable Functions (Measure Space)‎
Category:Pointwise Operations
Pointwise Sum of Measurable Functions is Measurable

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f, g: X \to \overline \R$ be $\Sigma$-measurable functions.
Assume that the pointwise sum $f + g: X \to \overline \R$ is well-defined.
Then $f + g$ is a $\Sigma$-measurable function.
=== General Result ===
== Proof ==
By Measurable Function is Pointwise Limit of Simple Functions, we find sequences $\sequence {f_n}_{n \mathop \in \N}, \sequence {g_n}_{n \mathop \in \N}$ such that:
:$\ds f = \lim_{n \mathop \to \infty} f_n$
:$\ds g = \lim_{n \mathop \to \infty} g_n$
where the limits are pointwise.
It follows that for all $x \in X$:
:$\map f x + \map g x = \ds \lim_{n \mathop \to \infty} \map {f_n} x + \map {g_n} x$
so that we have the pointwise limit:
:$\ds f + g = \lim_{n \mathop \to \infty} f_n + g_n$
By Pointwise Sum of Simple Functions is Simple Function, $f + g$ is a pointwise limit of simple functions.
By Simple Function is Measurable, $f + g$ is a pointwise limit of measurable functions.
Hence $f + g$ is measurable, by Pointwise Limit of Measurable Functions is Measurable.
== Sources ==
* : $8.10$
Category:Measurable Functions
Category:Pointwise Sum of Measurable Functions is Measurable
Pointwise Sum of Measurable Functions is Measurable/General Result

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space. 
Let $\sequence {\alpha_n}_{n \mathop \in \N}$ be a sequence of real numbers.
Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence of $\Sigma$-measurable functions $f_n : X \to \overline \R$ such that: 
:for each $N \in \N$ and $x \in X$, the sum $\ds \sum_{n \mathop = 1}^N \alpha_n \map {f_n} x$ is well-defined.
Then: 
:$\ds \sum_{n \mathop = 1}^N \alpha_n f_n$ is $\Sigma$-measurable.
== Proof ==
We proceed by induction.
For all $N \in \N$ let $\map P N$ be the proposition:
:$\ds \sum_{n \mathop = 1}^N \alpha_n f_n$ is $\Sigma$-measurable.
=== Basis for Induction ===
From the construction of $\sequence {f_n}_{n \mathop \in \N}$ we have: 
:$f_1$ is $\Sigma$-measurable.
From Pointwise Scalar Multiple of Measurable Function is Measurable, we have: 
:$\alpha_1 f_1$ is is $\Sigma$-measurable.
This is precisely $\map P 1$. 
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P N$ is true, where $N \ge 1$, then it logically follows that $\map P {N + 1}$ is true.
Our induction hypothesis is:
:$\ds \sum_{n \mathop = 1}^N \alpha_n f_n$ is $\Sigma$-measurable.
We aim to show that: 
:$\ds \sum_{n \mathop = 1}^{N + 1} \alpha_n f_n$ is $\Sigma$-measurable.
=== Induction Step ===
This is our induction step.
We have: 
:$\ds \sum_{n \mathop = 1}^{N + 1} \alpha_n f_n = \alpha_{N + 1} f_{N + 1} + \sum_{n \mathop = 1}^N \alpha_n f_n$
From our induction hypothesis, we have: 
:$\ds \sum_{n \mathop = 1}^N \alpha_n f_n$ is $\Sigma$-measurable.
From the construction of $\sequence {f_n}_{n \mathop \in \N}$ we have: 
:$f_{N + 1}$ is $\Sigma$-measurable.
So from Pointwise Scalar Multiple of Measurable Function is Measurable, we have: 
:$\alpha_{N + 1} f_{N + 1}$ is $\Sigma$-measurable.
So, from Pointwise Sum of Measurable Functions is Measurable, we have:
:$\ds \alpha_{N + 1} f_{N + 1} + \sum_{n \mathop = 1}^N \alpha_n f_n = \sum_{n \mathop = 1}^{N + 1} \alpha_n f_n$ is $\Sigma$-measurable.
Category:Pointwise Sum of Measurable Functions is Measurable
Pointwise Sum of Simple Functions is Simple Function

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f, g : X \to \R$ be simple functions.
Then the pointwise sum $f + g: X \to \R$ of $f$ and $g$:
:$\forall x, y \in X: \map {\paren {f + g} } x := \map f x + \map g x$
is also a simple function.
== Proof ==
We have $f + g = + \circ \innerprod f g \circ \Delta_X$, where:
:$\Delta_X: X \to X \times X$ is the diagonal mapping on $X$
:$\innerprod f g: X \times X \to \R \times \R, \map {\innerprod f g} {x, y} := \tuple {\map f x, \map g y}$
:$+: \R \times \R \to \R$ is real addition.
From this, we see that $+$ may be restricted to $\Img {\innerprod f g}$, the image of $\sequence {f, g}$.
It is immediate that this image equals $\Img f \times \Img g$.
By Measurable Function is Simple Function iff Finite Image Set, $\Img f$ and $\Img g$ are finite.
Therefore, by Cardinality of Cartesian Product of Finite Sets, $\Img {\sequence {f, g} }$ is also finite.
Next, from Corollary to Image of Composite Mapping:
:$\Img {\innerprod f g \circ \Delta_X} \subseteq \Img {\innerprod f g}$
whence the former is finite.
Now $+$ is a surjection:
:$+: \Img {\innerprod f g \circ \Delta_X} \to \Img {f + g}$
by Restriction of Mapping to Image is Surjection.
By Cardinality of Codomain of Surjection, this implies that $\Img {f + g}$ is finite.
Whence Measurable Function is Simple Function iff Finite Image Set grants that $f + g$ is a simple function.
== Also see ==
* Definition:Pointwise Operation, of which the definition of $f + g$ is an instantiation
* Scalar Multiple of Simple Function is Simple Function
* Pointwise Product of Simple Functions is Simple Function
* Space of Simple Functions forms Algebra
== Sources ==
* : $8.7 \ \text{(iv)}$
Category:Simple Functions
Category:Pointwise Operations