nyuuzyou/wikis huggingface dataset:
Linear Combination of Convergent Sequences in Topological Vector Space is Convergent

== Theorem ==
Let $K$ be a topological field. 
Let $X$ be a topological vector space over $K$.
Let $x, y \in X$ and $\lambda, \mu \in K$. 
Let $\sequence {x_n}_{n \in \N}$ and $\sequence {y_n}_{n \in \N}$ be convergent sequences with:
:$x_n \to x$
and:
:$y_n \to y$
Then $\sequence {\lambda x_n + \mu y_n}_{n \in \N}$ converges with:
:$\lambda x_n + \mu y_n \to \lambda x + \mu y$
== Proof ==
From Scalar Multiple of Convergent Sequence in Topological Vector Space is Convergent, we have that:
:$\sequence {\lambda x_n}_{n \in \N}$ converges to $\lambda x$
and:
:$\sequence {\mu y_n}_{n \in \N}$ converges to $\mu y$.
From Sum of Convergent Sequences in Topological Vector Space is Convergent, $\sequence {\lambda x_n + \mu y_n}_{n \in \N}$ converges to $\lambda x + \mu y$.
Category:Topological Vector Spaces
Linear Combination of Convergent Series

== Theorem ==
Let $\sequence {a_n}_{n \mathop \ge 1}$ and $\sequence {b_n}_{n \mathop \ge 1}$ be sequences of real numbers.
Let the two series $\ds\sum_{n \mathop = 1}^\infty a_n$ and $\ds \sum_{n \mathop = 1}^\infty b_n$ converge to $\alpha$ and $\beta$ respectively.
Let $\lambda, \mu \in \R$ be real numbers.
Then the series $\ds \sum_{n \mathop = 1}^\infty \paren {\lambda a_n + \mu b_n}$ converges to $\lambda \alpha + \mu \beta$.
== Proof ==
== Sources ==
* : $\S 6.8$
Category:Series
Linear Combination of Derivatives

== Theorem ==
Let $\map f x, \map g x$ be real functions defined on the open interval $I$.
Let $\xi \in I$ be a point in $I$ at which both $f$ and $g$ are differentiable.
Then:
:$\map D {\lambda f + \mu g} = \lambda D f + \mu D g$
at the point $\xi$ for some $\lambda \in \R$ and $\mu \in \R$.
It follows from the definition of derivative that if $f$ and $g$ are both differentiable on the interval $I$, then:
:$\forall x \in I: \map D {\lambda \map f x + \mu \map g x} = \lambda D \map f x + \mu D \map g x$
== Proof ==
Let $\xi \in I$ be a point in $I$ at which $f$ is differentiable. Then by definition, $f$ is differentiable at the point $\xi$ implies that the limit:
:$\ds \lim_{h \mathop \to 0} \frac {\map f {\xi + h} - \map f \xi} h$
exists.
Let $\xi \in I$ be a point in $I$ at which $g$ is differentiable. Then by definition, $g$ is differentiable at the point $\xi$ implies that the limit:
:$\ds \lim_{h \mathop \to 0} \frac {\map g {\xi + h} - \map g \xi} h$
exists.
So the limit:
:$\ds \lim_{h \mathop \to 0} \lambda \paren {\frac {\map f {\xi + h} - \map f \xi} h} + \mu \paren {\frac {\map g {\xi + h} - \map g \xi} h}$
exists.
Since $f$ and $g$ are real functions that are differentiable at $\xi \in I$, and $\lambda, \mu \in \R$, it follows that:
:$\map{ \paren{\lambda f + \mu g} } {x}$
exists, and is differentiable at $\xi \in I$ real function.
So the limit:
:$\ds \lim_{h \mathop \to 0} \frac { \map{ \paren{\lambda f + \mu g} } {\xi + h} - \map{ \paren{\lambda f + \mu g} } \xi} h$
exists.
We have:
:$\ds \lim_{h \mathop \to 0} \frac { \map{ \paren{\lambda f + \mu g} } {\xi + h} - \map{ \paren{\lambda f + \mu g} } \xi} h = \ds \lim_{h \mathop \to 0} \lambda \paren {\frac {\map f {\xi + h} - \map f \xi} h} + \mu \paren {\frac {\map g {\xi + h} - \map g \xi} h} $
Hence, the result follows from the definition of derivative.
== Sources ==
* : $\S 10.9 \ \text{(i)}$
* : $\S 3.14$: Second Order Linear Equations: Introduction
Category:Differential Calculus
Linear Combination of Direct Sums of Bounded Linear Operators on Hilbert Space

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\sequence {\family {\HH_i, \innerprod \cdot \cdot_i} }_{i \mathop \in I}$ be a $I$-indexed family of Hilbert spaces over $\GF$.
For each $i \in I$, let $T_i : \HH_i \to \HH_i$ and $S_i : \HH_i \to \HH_i$ be bounded linear operators.
Let $\lambda \in \GF$.
Suppose that:
:$\ds \sup_{i \mathop \in I} \norm {T_i}_{\map B {\HH_i} } < \infty$
and:
:$\ds \sup_{i \mathop \in I} \norm {S_i}_{\map B {\HH_i} } < \infty$
where $\norm {\, \cdot \,}_{\map B {\HH_i} }$ is the norm of a bounded linear operator on $\HH_i$.
Let:
:$\ds \HH = \bigoplus_{i \mathop \in I} \HH_i$
be the Hilbert space direct sum of $\sequence {\family {\HH_i, \innerprod \cdot \cdot_i} }_{i \mathop \in I}$ with inner product $\innerprod \cdot \cdot$ and norm $\norm {\, \cdot \,}$.
Let:
:$\ds T = \bigoplus_{i \mathop \in I} T_i$
:$\ds S = \bigoplus_{i \mathop \in I} S_i$
:$\ds \bigoplus_{i \mathop \in I} \paren {T_i + \lambda S_i}$ 
be the direct sum of $\family {T_i}_{i \mathop \in I}$, $\family {S_i}_{i \mathop \in I}$ and $\family {T_i + \lambda S_i}_{i \mathop \in I}$ respectively.
Then we have:
:$\ds T + \lambda S = \bigoplus_{i \mathop \in I} \paren {T_i + \lambda S_i}$ 
That is:
:$\ds \bigoplus_{i \mathop \in I} T_i + \lambda \bigoplus_{i \mathop \in I} S_i = \bigoplus_{i \mathop \in I} \paren {T_i + \lambda S_i}$ 
== Proof ==
First, from and , we have:
Hence:
:$\ds \sup_{i \mathop \in I} \norm {T_i + \lambda S_i}_{\map B {\HH_i} } \le \sup_{i \mathop \in I} \norm {T_i}_{\map B {\HH_i} } + \cmod \lambda \sup_{i \mathop \in I} \norm {S_i}_{\map B {\HH_i} } < \infty$
Hence the direct sum:
:$\ds \bigoplus_{i \mathop \in I} \paren {T_i + \lambda S_i}$ is well-defined.
Let $f \in \HH$ and $i \in I$.
We have:
Hence we have:
:$\ds \paren {\bigoplus_{i \mathop \in I} T_i} f + \lambda \paren {\bigoplus_{i \mathop \in I} S_i} f = \paren {\bigoplus_{i \mathop \in I} \paren {T_i + \lambda S_i} } f$
for each $f \in \HH$.
Hence:
:$\ds \bigoplus_{i \mathop \in I} T_i + \lambda \bigoplus_{i \mathop \in I} S_i = \bigoplus_{i \mathop \in I} \paren {T_i + \lambda S_i}$ 
Category:Hilbert Space Direct Sums
Linear Combination of Functions of Exponential Order

== Theorem ==
Let $f, g: \R \to \F$ be functions, where $\F \in \set {\R, \C}$.
Let $\lambda, \mu$ be complex numbers.
Suppose $f$ is of exponential order $a$ and $g$ is of exponential order $b$.
Then $\map {\paren {\lambda f + \mu g} } t = \lambda \, \map f t + \mu \, \map g t$ is of exponential order $\max \set {a, b}$.
== Proof ==
Follows from:
* Scalar Multiple of Function of Exponential Order
* Sum of Functions of Exponential Order
Category:Exponential Order
Linear Combination of Generating Functions

== Theorem ==
Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
and $\map H z$ be the generating function for the sequence $\sequence {b_n}$.
Then $\alpha \map G z + \beta \map H z$ is the generating function for the sequence $\sequence {\alpha a_n + \beta b_n}$.
== Proof ==
By definition:
:$\map G z = \ds \sum_{n \mathop \ge 0} a_n z^n$
:$\map H z = \ds \sum_{n \mathop \ge 0} b_n z^n$
Let $\map G z$ and $\map H z$ converge to $x$ and $y$ respectively for some $z_0 \in \R_{>0}$.
Then from Linear Combination of Convergent Series:
:$\ds \sum_{n \mathop \ge 0} \paren {\alpha a_n + \beta b_n} z^n = \alpha \sum_{n \mathop \ge 0} a_n z^n + \beta \sum_{n \mathop \ge 0} b_n z^n$
Hence the result.
== Sources ==
* : $\S 1.2.9$: Generating Functions: $(2)$
Category:Generating Functions
Linear Combination of Indexed Summations

== Theorem ==
Let $\mathbb A$ be one of the standard number systems $\N,\Z,\Q,\R,\C$.
Let $a,b$ be integers.
Let $\closedint a b$ denote the integer interval between $a$ and $b$.
Let $f, g : \closedint a b \to \mathbb A$ be mappings.
Let $\lambda, \mu \in \mathbb A$.
Let $\lambda \cdot f + \mu \cdot g$ be the sum of the product of $f$ with $\lambda$ and the product of $g$ with $\mu$.
Then we have the equality of indexed summations:
:$\ds \sum_{i \mathop = a}^b \paren {\lambda \cdot \map f i + \mu \cdot \map g i} = \lambda \cdot \sum_{i \mathop = a}^b \map f i + \mu \cdot \sum_{i \mathop = a}^b \map g i$
== Proof ==
We have:
== Also see ==
* Linear Combination of Convergent Series
Category:Summations
Linear Combination of Integers is Ideal

== Theorem ==
Let $a, b$ be any integers.
Let $\Bbb S = \set {a x + b y: x, y \in \Z}$.
Then the algebraic structure:
:$\struct {\Bbb S, +, \times}$
is an ideal of $\Z$.
== Proof ==
From the definition of integer combination (or just because it's obvious):
:$\Bbb S \subseteq \Z$
and clearly:
:$\Bbb S \ne \O$
as $a 0 + b 0 = 0 \in \Bbb S$.
Let $w_1, w_2 \in \Bbb S$:
:$w_1 = a x_1 + b y_1, w_2 = a x_2 + b y_2$
Then $w_1 + \paren {-w_2} = a \paren {x_1 - x_2} + b \paren {y_1 - y_2} \in \Bbb S$ as $x_1 - x_2 \in \Z$ and $y_1 - y_2 \in \Z$.
Let $r \in \Z$.
Then:
:$w_1 \times r = a x_1 r + b y_1 r = a \paren {x_1 r} + b \paren {y_1 r} \in \Bbb S$
:$r \times w_1 = r a x_1 + r b y_1 = \paren {r x_1} a + \paren {r y_1} b \in \Bbb S$
The result follows from Test for Ideal.
== Sources==
* : Chapter $5$: Rings: $\S 21$. Ideals: Example $35$
Category:Ideal Theory
Category:Integers
Linear Combination of Integrals

== Theorem ==
Let $f$ and $g$ be real functions which are integrable on the closed interval $\closedint a b$.
Let $\lambda$ and $\mu$ be real numbers.
Then the following results hold:
=== Indefinite Integrals ===
=== Definite Integrals ===
== Sources ==
* : $\S 1.4$
Category:Integral Calculus
Linear Combination of Integrals/Definite

== Theorem ==
Let $f$ and $g$ be real functions which are integrable on the closed interval $\closedint a b$.
Let $\lambda$ and $\mu$ be real numbers.
Then $\lambda f + \mu g$ is integrable on $\closedint a b$ and further:
:$\ds \int_a^b \paren {\lambda \map f t + \mu \map g t} \rd t = \lambda \int_a^b \map f t \rd t + \mu \int_a^b \map g t \rd t$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $6$. Integral Calculus: Appendix: Rules and Techniques of Integration: $1.5$
* 
* 
* : $\S 1.4$
Category:Linear Combination of Definite Integrals
Category:Definite Integrals
Linear Combination of Integrals/Definite/Proof 1

== Theorem ==
== Proof ==
Let $F$ and $G$ be primitives of $f$ and $g$ respectively on $\closedint a b$.
By Linear Combination of Derivatives, $H = \lambda F + \mu G$ is a primitive of $\lambda f + \mu g$ on $\closedint a b$.
Hence by the Fundamental Theorem of Calculus:
== Sources ==
* : $\S 13.20$
* : $\S 1.4$
Category:Linear Combination of Definite Integrals
Linear Combination of Integrals/Definite/Proof 2

== Theorem ==
== Proof ==
It is clear that for step functions $s$ and $t$:
:$\ds \int_a^b \lambda \map s x + \mu \map t x \rd x = \lambda \int_a^b \map s x \rd x + \mu \int_a^b \map t x \rd x$
Under any partition, the lower Darboux sums and upper Darboux sums of $f$ and $g$ are step functions, so the above formula relates the lower Darboux sums and upper Darboux sums of $f$ and $g$ to the lower Darboux sums and upper Darboux sums of the linear combinations of $f$ and $g$.
Because this identity is preserved for all possible partitions of $\closedint a b$, it is preserved for the supremum and infimum of all possible lower Darboux sums and upper Darboux sums, so the linear combinations of $f$ and $g$ are integrable.
== Sources ==
* : $\S 1.4$
Category:Linear Combination of Definite Integrals
Linear Combination of Integrals/Indefinite

== Theorem ==
Let $f$ and $g$ be real functions which are integrable on the closed interval $\closedint a b$.
Let $\lambda$ and $\mu$ be real numbers.
Then:
:$\ds \int \paren {\lambda \map f x + \mu \map g x} \rd x = \lambda \int \map f x \rd x + \mu \int \map g x \rd x$
== Proof ==
Let $F$ and $G$ be primitives of $f$ and $g$ respectively on $\closedint a b$.
By Linear Combination of Derivatives, $H = \lambda F + \mu G$ is a primitive of $\lambda f + \mu g$ on $\closedint a b$.
Hence:
== Sources ==
* : $\S 1.4$
Category:Integral Calculus
Linear Combination of Laplace Transforms

== Theorem ==
Let $\laptrans f$ denote the Laplace transform of the real function $f$.
Let $f, g$ be functions such that $\laptrans f$ and $\laptrans g$ exist.
Let $\lambda, \mu \in \C$ or $\R$ be constants.
Then:
:$\laptrans {\lambda \, \map f t + \mu \, \map g t} = \lambda \laptrans {\map f t} + \mu \laptrans {\map g t}$
everywhere all the above expressions are defined.
=== General Result ===
== Proof ==
== Examples ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Important Properties of Laplace Transforms: $1$. Linearity property: Theorem $1 \text{-} 2$
* : Chapter $1$: The Laplace Transform: Solved Problems: The Linearity Property: $5$
* : $\S 32$: Table of General Properties of Laplace Transforms: $32.3$
Category:Laplace Transforms
Linear Combination of Laplace Transforms/Examples/Example 1

== Example of Use of Linear Combination of Laplace Transforms ==
Let $\laptrans f$ denote the Laplace transform of the real function $f$.
:$\laptrans {4 t^2 - 3 \cos 2 t + 5 e^{-t} } = \dfrac 8 {s^3} - \dfrac {3 s} {s^2 + 4} + \dfrac 5 {s + 1}$
== Proof ==
== Sources ==
* : Chapter $1$: The Laplace Transform: Some Important Properties of Laplace Transforms: $1$. Linearity property
Category:Examples of Laplace Transforms
Linear Combination of Measurable Functions valued in Topological Vector Space is Measurable

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\struct {K, +_K, \circ_K, \tau_K}$ be a topological field be a topological field.
Let $\struct {Y, \tau}$ be a topological vector space over $K$.
Let $\map \BB Y$ be the Borel $\sigma$-algebra of $\struct {Y, \tau}$.
Let $f : X \to Y$ and $g : X \to Y$ be $\map \BB X/\map \BB Y$-measurable mappings.
Let $\lambda \in K$.
Then $f + \lambda g : X \to Y$ is $\map \BB X/\map \BB Y$-measurable.
== Proof ==
Define $m_\lambda : Y \to Y$ by:
:$\map {m_\lambda} y = \lambda y$
for each $y \in Y$. 
By the definition of a topological vector space, $m_\lambda$ is continuous.
Hence from Continuous Mapping is Measurable, $m_\lambda$ is $\map \BB Y/\map \BB Y$-measurable.
Equip $Y^2$ with the product topology.
Define also $s : Y^2 \to Y$ by:
:$\map s {x, y} = x + y$
for each $\tuple {x, y} \in Y^2$.
Again by the definition of a topological vector space, $s$ is continuous.
Hence from Continuous Mapping is Measurable, $m_\lambda$ is $\map \BB {Y^2}/\map \BB Y$-measurable.
Define $F : Y^2 \to Y$ by:
:$\map F {x, y} = \tuple {x, \lambda y}$ 
for each $\tuple {x, y} \in Y^2$. 
Let $\pr_1: Y^2 \to Y$ and $\pr_2: Y^2 \to Y$ be the first and second projections, respectively.
Hence we have:
:$\map {\pr_1 \circ F} {x, y} = x$
and:
:$\map {\pr_2 \circ F} {x, y} = \lambda y = \map {m_\lambda} y$
From Identity Mapping is Continuous, Continuous Mapping is Measurable and Measurable Mappings from Product Measurable Space, we have that $\pr_1 \circ F$ is $\map \BB {Y^2}/\map \BB Y$-measurable.
Similarly from Measurable Mappings from Product Measurable Space we have that $\pr_2 \circ F$ is $\map \BB {Y^2}/\map \BB Y$-measurable.
Hence from Measurable Mappings to Product Measurable Space, we obtain that $F$ is $\map \BB {Y^2}/\map \BB {Y^2}$-measurable.
Hence from Composition of Measurable Mappings is Measurable, $s \circ F$ is $\map \BB {Y^2}/\map \BB Y$-measurable.
Finally, define $G : X \to Y^2$ by:
:$\map G x = \tuple {\map f x, \map g x}$
for each $x \in X$.
Since $f$ and $g$ are $\map \BB X/\map \BB Y$-measurable mappings, we have that $G$ is $\map \BB X/\map \BB {Y^2}$-measurable from Measurable Mappings to Product Measurable Space.
Finally from Composition of Measurable Mappings is Measurable, $s \circ F \circ G$ is $\map \BB X/\map \BB Y$-measurable with:
for each $x \in X$.
Hence $f + \lambda g : X \to Y$ is $\map \BB X/\map \BB Y$-measurable.
Category:Measurable Functions
Linear Combination of Measures

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu, \nu$ be measures on $\struct {X, \Sigma}$.
Then for all positive real numbers $a, b \in \R_{\ge 0}$, the pointwise sum:
:$a \mu + b \nu: \Sigma \to \overline \R, \ \map {\paren {a \mu + b \nu} } E := a \map \mu E + b \map \nu E$
is also a measure on $\struct {X, \Sigma}$.
== Proof ==
Verifying the axioms $(1)$, $(2)$ and $(3')$ for a measure in turn:
=== Axiom $(1)$ ===
The statement of axiom $(1)$ for $a \mu + b \nu$ is:
:$\forall E \in \Sigma: \map {\paren {a \mu + b \nu} } E \ge 0$
Let $E \in \Sigma$.
Then $\map \mu E, \map \nu E \ge 0$ as $\mu$ and $\nu$ are measures.
Hence, $a \map \mu E \ge 0$ as $a \ge 0$.
Also, $b \map \nu E \ge 0$ since $b \ge 0$.
Therefore it follows that:
:$a \map \mu E + b \map \nu E \ge 0$
as desired.
=== Axiom $(2)$ ===
Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma$.
The statement of axiom $(2)$ for $a \mu + b \nu$ is:
:$\ds \map {\paren {a \mu + b \nu} } {\bigcup_{n \mathop \in \N} E_n} = \sum_{n \mathop \in \N} \map {\paren {a \mu + b \nu} } {E_n}$
So let us do a direct computation:
which establishes $a \mu + b \nu$ satisfies $(2)$.
=== Axiom $(3')$ ===
The statement of axiom $(3')$ for $a \mu + b \nu$ is:
:$\map {\paren {a \mu + b \nu} } \O = 0$
This is verified by the following:
Thus, $a \mu + b \nu$ satisfies $(3')$.
Having verified an appropriate set of axioms, it follows that $a \mu + b \nu$ is a measure.
== Sources ==
* : $\S 4$: Problem $6 \ \text{(i)}$
Category:Measures
Linear Combination of Mellin Transforms

== Theorem ==
Let $\MM$ be the Mellin transform.
Let $\map f t$, $g \left({t}\right)$ be functions such that $\MM \left\{ {\map f t}\right\} \left({s}\right)$ and $\MM \left\{ {\map f t}\right\} \left({s}\right)$ exist.
Let $\lambda \in \C$ be a constant.
Then:
:$\map {\MM \set {\lambda \map f t + \map g t} } s = \lambda \map {\MM \set {\map f t} } s + \map {\MM \set {\map g t} } s$
everywhere all the above expressions are defined.
== Proof ==
Category:Mellin Transforms
Linear Combination of Non-Parallel Complex Numbers is Zero if Factors are Both Zero

== Theorem ==
Let $z_1$ and $z_2$ be complex numbers expressed as vectors such taht $z_1$ is not parallel to $z_2$.
Let $a, b \in \R$ be real numbers such that:
:$a z_1 + b z_2 = 0$
Then $a = 0$ and $b = 0$.
== Proof ==
Suppose it is not the case that $a = b = 0$.
Then:
and $z_1$ and $z_2$ are parallel.
== Sources ==
* : $1$: Solved Problems: Graphical Representations of Complex Numbers. Vectors: $9$
Category:Geometry of Complex Plane
Linear Combination of Normal Random Variables

== Theorem ==
Let $X_1, X_2, X_3, \ldots, X_n$ be independent random variables. 
Let $\sequence {\alpha_i}_{1 \mathop \le i \mathop \le n}$ and $\sequence {\mu_i}_{1 \mathop \le i \mathop \le n}$ be sequences of real numbers.
Let $\sequence {\sigma_i}_{1 \mathop \le i \mathop \le n}$ be a sequence of positive real numbers.
Let $X_i \sim \Gaussian {\mu_i} {\sigma^2_i}$ for $1 \le i \le n$, where $\Gaussian {\mu_i} {\sigma^2_i}$ is the normal distribution with parameters $\mu_i$ and $\sigma^2_i$.
Then: 
:$\ds \sum_{i \mathop = 1}^n \alpha_i X_i \sim \Gaussian {\sum_{i \mathop = 1}^n \alpha_i \mu_i} {\sum_{i \mathop = 1}^n \alpha^2_i \sigma^2_i}$
 
== Proof ==
Let: 
:$\ds Z = \sum_{i \mathop = 1}^n \alpha_i X_i$ 
Let $M_Z$ be the moment generating function of $Z$. 
We aim to show that: 
:$\ds Z \sim \Gaussian {\sum_{i \mathop = 1}^n \alpha_i \mu_i} {\sum_{i \mathop = 1}^n \alpha^2_i \sigma^2_i}$
By Moment Generating Function of Normal Distribution and Moment Generating Function is Unique, it is therefore sufficient to show that: 
:$\ds \map {M_Z} t = \map \exp {\paren {\sum_{i \mathop = 1}^n \alpha_i \mu_i} t + \frac 1 2 \paren {\sum_{i \mathop = 1}^n \alpha^2_i \sigma^2_i} t^2}$
We also have, by Moment Generating Function of Normal Distribution, that the moment generating function of $X_i$, $M_{X_i}$, is given by: 
:$\map {M_{X_i}} t = \map \exp {\mu_i t + \dfrac 1 2 \sigma^2_i t^2}$
We have:
Category:Normal Distribution
Linear Combination of Real-Valued Random Variables is Real-Valued Random Variable

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ and $Y$ be real-valued random variables on $\struct {\Omega, \Sigma, \Pr}$.
Let $\alpha$ and $\beta$ be real numbers.
Then: 
:$\alpha X + \beta Y$ is a real-valued random variable.
=== General Result ===
== Proof ==
Since $X$ and $Y$ are real-valued random variables, we have: 
:$X$ and $Y$ are $\Sigma$-measurable functions.
From Pointwise Scalar Multiple of Measurable Function is Measurable, we have: 
:$\alpha X$ and $\beta Y$ are $\Sigma$-measurable.
From Pointwise Sum of Measurable Functions is Measurable, we have: 
:$\alpha X + \beta Y$ is $\Sigma$-measurable.
So:
:$\alpha X + \beta Y$ is a real-valued random variable.
Category:Random Variables
Category:Linear Combination of Real-Valued Random Variables is Real-Valued Random Variable
Linear Combination of Real-Valued Random Variables is Real-Valued Random Variable/General Result

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $n \in \N$. 
Let $\sequence {X_i}_{i \mathop \in \N}$ be a sequence of real-valued random variables on $\struct {\Omega, \Sigma, \Pr}$.
Let $\sequence {\alpha_i}_{i \mathop \in \N}$ be a sequence of real numbers. 
Then: 
:$\ds \sum_{i \mathop = 1}^n \alpha_i X_i$ is a real-valued random variable.
== Proof ==
We proceed by induction.
For all $n \in \N$ let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 1}^n \alpha_i X_i$ is $\Sigma$-measurable.
=== Basis for Induction ===
From the construction of $\sequence {X_i}_{i \mathop \in \N}$ we have: 
:$X_1$ is $\Sigma$-measurable.
From Pointwise Scalar Multiple of Measurable Function is Measurable, we have: 
:$\alpha_1 X_1$ 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:
:$\ds \sum_{i \mathop = 1}^n \alpha_i X_i$ is $\Sigma$-measurable.
We aim to show that: 
:$\ds \sum_{i \mathop = 1}^{n + 1} \alpha_i X_i$ is $\Sigma$-measurable.
=== Induction Step ===
This is our induction step.
We have: 
:$\ds \sum_{i \mathop = 1}^{n + 1} \alpha_i X_i = \alpha_{n + 1} X_{n + 1} + \sum_{i \mathop = 1}^N \alpha_i X_i$
From our induction hypothesis, we have: 
:$\ds \sum_{i \mathop = 1}^n \alpha_i X_i$ is $\Sigma$-measurable.
From the construction of $\sequence {X_i}_{i \mathop \in \N}$ we have: 
:$X_{n + 1}$ is $\Sigma$-measurable.
So from Pointwise Scalar Multiple of Measurable Function is Measurable, we have: 
:$\alpha_{n + 1} X_{n + 1}$ is $\Sigma$-measurable.
So, from Pointwise Sum of Measurable Functions is Measurable, we have:
:$\ds \alpha_{n + 1} X_{n + 1} + \sum_{i \mathop = 1}^n \alpha_i X_i = \sum_{i \mathop = 1}^{n + 1} \alpha_i X_i$ is $\Sigma$-measurable.
Category:Linear Combination of Real-Valued Random Variables is Real-Valued Random Variable
Linear Combination of Riemann-Stieltjes Integrals/Integrand

== Theorem ==
Let $f, g, \alpha$ be real functions that are bounded on $\closedint a b$.
Let $c_1, c_2 \in \R$.
Suppose that $f$ and $g$ are Riemann-Stieltjes integrable with respect to $\alpha$ on $\closedint a b$ and:
:$\ds \int_a^b f \rd \alpha = A$
:$\ds \int_a^b g \rd \alpha = B$
Then, the real function $h : \closedint a b \to \R$ defined as:
:$\map h x = c_1 \map f x + c_2 \map g x$
is Riemann-Stieltjes integrable with respect to $\alpha$ on $\closedint a b$ and:
:$\ds \int_a^b h \rd \alpha = c_1 A + c_2 B$
== Proof ==
Let $\epsilon > 0$ be arbitrary.
By definition of the Riemann-Stieltjes integral, let $P'_\epsilon, P' '_\epsilon$ be subdivisions of $\closedint a b$ such that:
:For every $P$ finer than $P'_\epsilon$, $\size {\map S {P, f, \alpha} - A} < \dfrac \epsilon {\size {c_1} + \size {c_2} + 1}$
:For every $P$ finer than $P' '_\epsilon$, $\size {\map S {P, g, \alpha} - B} < \dfrac \epsilon {\size {c_1} + \size {c_2} + 1}$
Define $P_\epsilon = P'_\epsilon \cup P' '_\epsilon$ to be a subdivision that is finer than both $P'_\epsilon$ and $P' '_\epsilon$.
Let $P = {x_0, \dotsc, x_n} \supseteq P_\epsilon$ be an arbitrary subdivision that is finer than $P_\epsilon$.
Clearly, $P$ is also finer than $P'_\epsilon$ and $P' '_\epsilon$.
Finally, fix the $t_1, \dotsc, t_k$ for the following Riemann-Stieltjes sum.
We have:
And so:
As $\epsilon > 0$ was arbitrary, by the definition of the Riemann-Stieltjes integral:
:$\ds \int_a^b h \rd \alpha = c_1 A + c_2 B$
== Sources ==
* : Chapter $7$ The Riemann-Stieltjes Integral: $\S 7.4$: Theorem $7.2$
Category:Riemann-Stieltjes Integral
Category:Integral Calculus