nyuuzyou/wikis huggingface dataset:
Condition of Tangency to Circle whose Center is Origin

== Theorem ==
Let $\CC$ be a circle embedded in the Cartesian plane of radius $r$ with its center located at the origin.
Let $\LL$ be a straight line in the plane of $\CC$ whose equation is given by:
:$(1): \quad l x + m y + n = 0$
such that $l \ne 0$.
Then $\LL$ is tangent to $\CC$ :
:$\paren {l^2 + m^2} r^2 = n^2$
== Proof ==
From Equation of Circle center Origin, $\CC$ can be described as:
:$(2): \quad x^2 + y^2 = r^2$
Let $\LL$ intersect with $\CC$.
To find where this happens, we find $x$ and $y$ which satisfy both $(1)$ and $(2)$.
So:
This is a quadratic in $y$.
This corresponds to the two points of intersection of $\LL$ with $\CC$.
When $\LL$ is tangent to $\CC$, these two points coincide.
Hence $(3)$ has equal roots.
From Solution to Quadratic Equation, this happens when the discriminant of $(3)$ is zero.
That is:
== Sources ==
* : Chapter $\text {III}$. The Circle: $13$. Condition that a straight line should touch a given circle
Category:Circles
Category:Tangents
Condition on Congruence Relations for Cancellable Monoid to be Group

== Theorem ==
Let $\struct {S, \circ}$ be a cancellable monoid whose identity element is $e$.
Then:
:$\struct {S, \circ}$ is a group
:
:every non-trivial congruence relation on $\struct {S, \circ}$ is induced by a normal subgroup of $\struct {S, \circ}$.
=== Counterexample ===
== Proof ==
=== Necessary Condition ===
Let $\struct {S, \circ}$ be such that every non-trivial congruence relation on $\struct {S, \circ}$ is induced by a normal subgroup of $\struct {S, \circ}$.
Hence, let $\RR$ be an arbitrary non-trivial congruence relation.
From Condition for Subgroup of Monoid to be Normal, there exists a normal subgroup $\struct {H, \circ}$ such that:
:the set of left cosets of $\struct {H, \circ}$ in $\struct {S, \circ}$ form a partition of $S$
and:
:the set of right cosets of $\struct {H, \circ}$ in $\struct {S, \circ}$ form a partition of $S$
such that the equivalence relations induced by those partitions is $\RR$.
By the definition of normal subgroup, the set of left cosets is the same as the set of right cosets.
It remains to be shown that $\struct {S, \circ}$ is a group.
We already have that
:
:
:
are satisfied by the fact that $\struct {S, \circ}$ is a monoid.
Hence it remains to prove .
=== Sufficient Condition ===
Let $\struct {S, \circ}$ be a group.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.18 \ \text {(a)}$
Category:Normal Subgroups
Category:Cancellable Monoids
Category:Condition on Congruence Relations for Cancellable Monoid to be Group
Condition on Congruence Relations for Cancellable Monoid to be Group/Counterexample

== Theorem ==
Let $\struct {S, \circ}$ be a monoid which is not cancellable.
Let every non-trivial congruence relation on $\struct {S, \circ}$ be induced by a normal subgroup of $\struct {S, \circ}$.
Then it is not necessarily the case that $\struct {S, \circ}$ is a group.
== Proof ==
Consider the Multiplicative Monoid of Integers Modulo $3$ $\struct {\Z_3, \times_3}$, defined by its Cayley table:
It is noted that $\struct {\Z_3, \times_3}$ is not a cancellable monoid.
Indeed:
:$0 \times_3 1 = 0$
:$0 \times_3 2 = 0$
and so $\times_3$ is not cancellable for $0$.
Hence from Group is Cancellable Monoid:
:$\struct {\Z_3, \times_3}$ is not a group.
We note that $\struct {\Z_3, \times_3}$ has an identity element $\eqclass 1 3$.
We list all the equivalence relations on $\struct {\Z_3, \times_3}$ by the partitions they induce:
:$\set {\set 0, \set 1, \set 2}$ is induced by the diagonal relation
:$\set {\set {0, 1, 2}}$ is induced by the trivial relation
:$\set {\set {0, 1}, \set 2}$ is not a congruence relation because we have $0 \mathrel \RR 1$ and $2 \mathrel \RR 2$ but not $\paren {0 \times 2} \mathrel \RR \paren {1 \times 2}$
:$\set {\set {0, 2}, \set 1}$ is not a congruence relation because we have $0 \mathrel \RR 2$ and $2 \mathrel \RR 2$ but not $\paren {0 \times 2} \mathrel \RR \paren {2 \times 2}$
:$\set {\set 0, \set {1, 2}}$ is induced by the equivalence relation induced by the normal subgroup $\set 0$.
We see that the first four relations are not non-trivial congruence relation on $\struct {S, \circ}$.
Therefore every non-trivial congruence relation on $\struct {S, \circ}$ is induced by a normal subgroup of $\struct {S, \circ}$.
But, , $\struct {\Z_3, \times_3}$ is not a group.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.18 \ \text {(b)}$
Category:Condition on Congruence Relations for Cancellable Monoid to be Group
Condition on Conjugate from Real Product of Complex Numbers

== Theorem ==
Let $z_1, z_2 \in \C$ be complex numbers such that $z_1 z_2 \in \R_{\ne 0}$.
Then:
:$\exists p \in \R: z_1 = p \overline {z_2}$
where $\overline {z_2}$ denotes the complex conjugate of $z_2$.
== Proof ==
Let $z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$.
As $z_1 z_2$ is real:
:$(1): \quad z_1 z_2 = x_1 x_2 - y_1 y_2$
and:
:$(2): \quad x_1 y_2 + y_1 x_2 = 0$
So:
So $z_1 / \overline {z_2} = p$ where $p = \dfrac {x_1 x_2 - y_1 y_2} { {x_2}^2 + {y_2}^2}$, which is real.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $163$
Category:Complex Conjugates
Condition on Equality of Generated Sigma-Algebras

== Theorem ==
Let $X$ be a set.
Let $\GG$, $\HH$ be sets of subsets of $X$.
Suppose that:
:$\GG \subseteq \HH \subseteq \map \sigma \GG$
where $\sigma$ denotes generated $\sigma$-algebra.
Then:
:$\map \sigma \GG = \map \sigma \HH$
== Proof ==
From Generated Sigma-Algebra Preserves Subset, it follows that:
:$\map \sigma \GG \subseteq \map \sigma \HH$
Since $\map \sigma \GG$ is a $\sigma$-algebra containing $\HH$:
:$\map \sigma \HH \subseteq \map \sigma \GG$
from the definition of generated $\sigma$-algebra.
Hence the result, from the definition of set equality.
Category:Sigma-Algebras Generated by Collection of Subsets
Condition on Lower Coefficient for Binomial Coefficient to be Maximum

== Theorem ==
Let $n, k \in \Z_{\ge 0}$ be positive integers.
Let $\dbinom n k$ denote the binomial coefficient of $n$ choose $k$.
Then for a given value of $n$, the value of $k$ for which $\dbinom n k$ is a maximum is:
:$k_\max = \floor {\dfrac n 2}$
and also:
:$k_\max = \ceiling {\dfrac n 2}$
== Proof ==
From Condition for Increasing Binomial Coefficients:
:$\forall k \in \Z_{> 0}: \dbinom n k > \dbinom n {k - 1} \iff k \le \dfrac n 2$
Let $n$ be an even integer, such that:
:$n = 2 r$
Then:
:$r = \floor {\dfrac n 2} = \ceiling {\dfrac n 2}$
and:
:$\forall k < r: \dbinom n r > \dbinom n k$
From Symmetry Rule for Binomial Coefficients:
:$\forall k < r: \dbinom n r > \dbinom n {n - k}$
and so:
:$\forall k > r: \dbinom n r > \dbinom n k$
We also have:
:$\dbinom n r = \dbinom n {n - r} = \dbinom n {n / 2}$
Thus for all $k \ne r$:
:$\dbinom n k < \dbinom n {n / 2}$
and so:
:$k_\max = \floor {\dfrac n 2} = \ceiling {\dfrac n 2}$
Let $n$ be an odd integer, such that:
:$n = 2 r + 1$
Then:
:$r = \floor {\dfrac n 2}$
As $r < n$:
:$\forall k < r: \dbinom n r > \dbinom n k$
When $k = r$ we also have:
:$\dbinom n r = \dbinom n {n - r} = \dbinom n {r + 1}$
Then from Symmetry Rule for Binomial Coefficients:
:$\forall k < r: \dbinom n r > \dbinom n {n - k}$
and so:
:$\forall k > r + 1: \dbinom n r > \dbinom n k$
We have that:
:$r = \floor {\dfrac n 2}$
and:
:$r + 1 = \ceiling {\dfrac n 2}$
Hence the result.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $7$
Category:Binomial Coefficients
Condition on Proper Lower Sections for Total Ordering to be Well-Ordering/Mistake

== Source Work ==
:
:Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering:
::$\S 1$ Introduction to well ordering:
:::Exercise $1.2$
== Mistake ==
:Prove that a sufficient condition for a linear ordering $\le$ of $A$ to be a well ordering of $A$ is that, for every proper lower section $L$ of $A$, there is a least element $x$ of $A$ not in $L$.
== Analysis ==
Consider the integers $\Z$ under the usual ordering $\le$.
We have that $\struct {\Z, \le}$ is a linearly ordered set.
Yet let $x \in \Z$ be arbitrary.
We have that $\set {y \in A: y < x}$ is a proper lower section of $\struct {\Z, \le}$.
There is no other way to construct a proper lower section of $\struct {\Z, \le}$.
Then we note that $x$ is then the least element of $A$ not in $L$.
Hence $\struct {\Z, \le}$ fulfils the conditions of the hypothesis, but is not a well ordered set, as can be seen by the fact that there is no smallest integer.
It is suggested that necessary condition was meant, not sufficient condition.
If so, the proof is trivial.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering: Exercise $1.2$
Category:Mistakes/Set Theory and the Continuum Problem
Conditional/Semantics of Conditional/Examples

== Examples of Semantics of Conditional ==
The statement:
:If I pass this course, then (it shows that) I have studied hard for it.
may be rephrased as:
:I will pass this course only if I have studied hard for it.
:To prove that I have studied hard for this course, it is sufficient to know that I passed it.
:For me to pass this course, it is necessary for me to study hard for it.
== Sources ==
* : Appendix $\text{A}.5$: Theorems and Proofs
Category:Conditional
Conditional Dominated Convergence Theorem

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be an integrable random variable.
Let $\sequence {X_n}_{n \in \N}$ be an sequence of non-negative integrable random variables converging almost surely to $X$, such that: 
:there exists an integrable random variable $Y$ such that:
::$\size {X_n} \le Y$ almost surely.
Let $\GG \subseteq \Sigma$ be a sub-$\sigma$-algebra.
For each $n \in \N$, let $\expect {X_n \mid \GG}$ be a version of the conditional expectation of $X_n$ conditioned on $\GG$.
Let $\expect {X \mid \GG}$ be a version of the conditional expectation of $X$ conditioned on $\GG$.
Then, we have: 
:$\ds \lim_{n \mathop \to \infty} \expect {X_n \mid \GG} = \expect {X \mid \GG}$ almost surely.
== Proof ==
Let $\expect {Y \mid \GG}$ be a version of the conditional expectation of $Y$ conditioned on $\GG$.
For each $n \in \N$, let $\expect {X_n \mid \GG}$ be a version of the conditional expectation of $X_n$ conditioned on $\GG$.
Since we have: 
:$\size {X_n} \le Y$ almost surely.
and $Y$ is integrable, we have: 
:$\ds \limsup_{n \mathop \to \infty} \expect {X_n \mid \GG} \le \expect {\limsup_{n \mathop \to \infty} X_n \mid \GG} = \expect {X \mid \GG}$ almost surely
by Conditional Reverse Fatou's Lemma. 
From Conditional Fatou's Lemma, we have: 
:$\ds \expect {X \mid \GG} = \expect {\liminf_{n \mathop \to \infty} X_n \mid \GG} \le \liminf_{n \mathop \to \infty} \expect {X_n \mid \GG}$ almost surely.
So almost surely, we have: 
:$\ds \limsup_{n \mathop \to \infty} \expect {X_n \mid \GG} \le \expect {X \mid \GG} \le \liminf_{n \mathop \to \infty} \expect {X_n \mid \GG} \le \limsup_{n \mathop \to \infty} \expect {X_n \mid \GG}$
So almost surely, we have:
:$\ds \liminf_{n \mathop \to \infty} \expect {X_n \mid \GG} = \limsup_{n \mathop \to \infty} \expect {X_n \mid \GG} = \expect {X \mid \GG}$
So, from Convergence of Limsup and Liminf, we have: 
:$\ds \lim_{n \mathop \to \infty} \expect {X_n \mid \GG}$ exists almost surely
and further:
:$\ds \expect {X \mid \GG} = \lim_{n \mathop \to \infty} \expect {X_n \mid \GG}$ almost surely.
== Sources ==
* : $9.7$: Properties of conditional expectation: a list
Category:Conditional Expectation
Category:Lebesgue's Dominated Convergence Theorem
Conditional Entropy Decreases if More Given

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\AA, \CC, \DD \subseteq \Sigma$ be finite sub-$\sigma$-algebras.
Then:
:$\CC \subseteq \DD \implies \map H {\AA \mid \CC} \ge \map H {\AA \mid \DD}$
where:
:$\map H {\cdot \mid \cdot}$ denotes the conditional entropy
== Proof ==
Consider the generated finite partitions:
:$\xi := \map \xi \AA$
:$\eta := \map \xi \CC$
:$\gamma := \map \xi \DD$
By Generating Partition Preserves Order, $\CC \subseteq \DD$ implies:
:$\eta \le \gamma$
where $\le$ denotes the refinement.
By , we shall show:
:$\map H {\xi \mid \eta} \ge \map H {\xi \mid \gamma}$
Recall that $\eta \le \gamma$ implies:
:$\forall B \in \eta, \forall D \in \gamma \implies D \subseteq B \; \text{or} \; B \cap D = \O$
In particular, we have:
:$(1): \quad \forall B \in \eta, \forall D \in \gamma: \map \Pr {B \cap D} = \begin{cases} \map \Pr D &: D \subseteq B \\ 0 &: B \cap D = \O \end{cases}$
Thus, for all $A \in \xi$ and $B \in \gamma$ such that $\map \Pr B > 0$:
The function $\phi$ used in is concave since the second derivative is negative:
:$\forall x > 0 : \map {\phi ' '} x = -\dfrac 1 x < 0$
Note that the concavity holds including $x = 0$, since $\phi$ is continuous there.
So we obtain:
Therefore:
=== Corollary ===
== Sources ==
* $4.3$: Conditional Entropy
Category:Probability Theory
Category:Ergodic Theory
Conditional Entropy Decreases if More Given/Corollary

== Corollary to Conditional Entropy Decreases if More Given ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\AA, \DD \subseteq \Sigma$ be finite sub-$\sigma$-algebras.
Then:
:$\map H \AA \ge \map H {\AA \mid \DD} $
where:
:$\map H \cdot$ denotes the entropy
:$\map H {\cdot \mid \cdot}$ denotes the conditional entropy
== Proof ==
Let $\NN := \set {\O, \Omega}$ be the trivial $\sigma$-algebra.
Then:
== Sources ==
* $4.3$: Conditional Entropy
Category:Probability Theory
Category:Ergodic Theory
Conditional Entropy Given Trivial Sigma-Algebra is Entropy

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\AA \subseteq \Sigma$ be a finite sub-$\sigma$-algebra.
Let $\NN := \set {\O, \Omega}$ be the trivial $\sigma$-algebra.
Then:
:$\ds \map H {\AA \mid \NN} = \map H \AA$
where:
:$\map H {\cdot \mid \cdot}$ denotes the conditional entropy
:$\map H {\, \cdot \,}$ denotes the entropy
== Proof ==
== Sources ==
* $4.3$: Conditional Entropy
Category:Probability Theory
Category:Ergodic Theory
Conditional Entropy of Join as Sum

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\AA, \CC, \DD \subseteq \Sigma$ be finite sub-$\sigma$-algebras.
Then:
:$\ds \map H {\AA \vee \CC \mid \DD} = \map H {\AA \mid \DD} + \map H {\CC \mid \AA \vee \DD} $
where:
:$\map H {\cdot \mid \cdot}$ denotes the conditional entropy
:$\vee$ denotes the join
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
=== Corollary 4 ===
=== Corollary 5 ===
== Proof ==
Consider the generated finite partitions:
:$\xi := \map \xi \AA$
:$\eta := \map \xi \CC$
:$\gamma := \map \xi \DD$
By , we shall show:
:$\map H {\xi \vee \eta \mid \gamma} = \map H {\xi \mid \gamma} + \map H {\eta \mid \xi \vee \gamma}$
Then:
Now:
and:
== Sources ==
* $4.3$: Conditional Entropy
Category:Probability Theory
Category:Ergodic Theory
Conditional Entropy of Join as Sum/Corollary 1

== Corollary to Conditional Entropy of Join as Sum ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\AA, \CC \subseteq \Sigma$ be finite sub-$\sigma$-algebras.
Then:
:$\map H {\AA \vee \CC} = \map H {\AA} + \map H {\CC \mid \AA} $
where:
:$\map H {\cdot \mid \cdot}$ denotes the conditional entropy
:$\map H \cdot$ denotes the entropy
:$\vee$ denotes the join
== Proof ==
Let $\NN := \set {\O, \Omega}$ be the trivial $\sigma$-algebra.
Then:
== Sources ==
* $4.3$: Conditional Entropy
Category:Probability Theory
Category:Ergodic Theory
Conditional Entropy of Join as Sum/Corollary 2

== Corollary to Conditional Entropy of Join as Sum ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\AA, \CC, \DD \subseteq \Sigma$ be finite sub-$\sigma$-algebras.
Then:
:$\AA \subseteq \CC \implies \map H {\AA \mid \DD} \le \map H {\CC \mid \DD} $
where:
:$\map H {\cdot \mid \cdot}$ denotes the conditional entropy
:$\vee$ denotes the join
== Proof ==
Let $\AA \subseteq \CC$.
Then:
 
== Sources ==
* $4.3$: Conditional Entropy
Category:Probability Theory
Category:Ergodic Theory
Conditional Entropy of Join as Sum/Corollary 3

== Corollary to Conditional Entropy of Join as Sum ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\AA, \CC \subseteq \Sigma$ be finite sub-$\sigma$-algebras.
Then:
:$\AA \subseteq \CC \implies \map H \AA \le \map H \CC $
where:
:$\map H \cdot$ denotes the entropy
== Proof ==
Let $\AA \subseteq \CC$.
Let $\NN := \set {\O, \Omega}$ be the trivial $\sigma$-algebra.
Then:
== Sources ==
* $4.3$: Conditional Entropy
Category:Probability Theory
Category:Ergodic Theory
Conditional Entropy of Join as Sum/Corollary 4

== Corollary to Conditional Entropy of Join as Sum ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\AA, \CC, \DD \subseteq \Sigma$ be finite sub-$\sigma$-algebras.
Then:
:$\map H {\AA \vee \CC \mid \DD} \le \map H {\AA \mid \DD} + \map H {\CC \mid \DD}$
where:
:$\map H {\cdot \mid \cdot}$ denotes the conditional entropy
:$\vee$ denotes the join
== Proof ==
== Sources ==
* $4.3$: Conditional Entropy
Category:Probability Theory
Category:Ergodic Theory
Conditional Entropy of Join as Sum/Corollary 5

== Corollary to Conditional Entropy of Join as Sum ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\AA, \CC \subseteq \Sigma$ be finite sub-$\sigma$-algebras.
Then:
:$\map H {\AA \vee \CC} \le \map H \AA + \map H \CC $
where:
:$\map H \cdot$ denotes the conditional entropy
:$\vee$ denotes the join
== Proof ==
Let $\NN := \set {\O, \Omega}$ be the trivial $\sigma$-algebra.
Then:
== Sources ==
* $4.3$: Conditional Entropy
Category:Probability Theory
Category:Ergodic Theory
Conditional Equation/Examples/Arbitrary Example 1

== Example of Conditional Equation ==
The equation:
:$3 x + y = 7$
is a conditional equation as its truth value depends on the values of $x$ and $y$.
For example:
:if $x = 2$ and $y = 1$, the equation is true
:if $x = 1$ and $y = 2$, the equation is false.
== Sources ==
* 
* 
Category:Examples of Conditional Equations
Conditional Expectation Conditioned on Event of Non-Zero Probability

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be an integrable random variable on $\struct {\Omega, \Sigma, \Pr}$.
Let $B \in \Sigma$ be an event with:
:$\map \Pr B > 0$
Let:
:$\GG = \map \sigma B = \set {\O, B, B^c, \Omega}$
where $\map \sigma B$ is the $\sigma$-algebra generated by $B$.
Let: 
:$\ds \alpha = \frac {\expect {X \cdot 1_B} } {\map \Pr B}$
and:
:$\ds \beta = \frac {\expect {X \cdot 1_{B^c} } } {\map \Pr {B^c} }$
Let $\expect {X \mid \GG}$ be a version of the conditional expectation of $X$ given $\GG$.
Then:
:$\ds \expect {X \mid \GG} = \alpha \cdot 1_B + \beta \cdot 1_{B^c}$ almost everywhere. 
== Proof ==
We show that:
:$\ds Z = \alpha \cdot 1_B + \beta \cdot 1_{B^c}$ is a version of the conditional expectation of $X$ given $\GG$.
We will then be done by Existence and Essential Uniqueness of Conditional Expectation Conditioned on Sigma-Algebra.
From Characteristic Function Measurable iff Set Measurable, $1_B$ and $1_{B^c}$ are both measurable functions.
From Pointwise Product of Measurable Functions is Measurable and Pointwise Sum of Measurable Functions is Measurable, we therefore have that $Z$ is a random variable. 
So we now verify that the expectations take the correct values. 
From Integral of Integrable Function over Null Set, we have: 
:$\expect {Z \cdot 1_\O} = 0 = \expect {X \cdot 1_\O}$
We also have: 
and:
Finally, we have: 
Category:Conditional Expectation
Conditional Expectation Conditioned on Trivial Sigma-Algebra

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be an integrable random variable.
Let $\expect {X \mid \set {\O, \Omega} }$ be a version of the conditional expectation of $X$ given $\set {\O, \Omega}$.
Then:
:$\expect {X \mid \set {\O, \Omega} } = \expect X$ almost everywhere.
== Proof ==
We check that $\expect X$ is a version of the conditional expectation of $X$ given $\set {\O, \Omega}$, so that we get:
:$\expect {X \mid \set {\O, \Omega} } = \expect X$ almost everywhere.
from Existence and Essential Uniqueness of Conditional Expectation Conditioned on Sigma-Algebra.
Note that $\expect X$ is $\GG$-measurable from Constant Function is Measurable.
We certainly have:
:$\ds \int_\O \expect X \rd \Pr = \int_\O X \rd \Pr = 0$
from Integral of Integrable Function over Null Set.
We also have:
:$\ds \int_\Omega \expect X \rd \Pr = \expect X = \int_\Omega X \rd \Pr$
from Integral of Characteristic Function.
So $\expect X$ is a version of the conditional expectation of $X$ given $\set {\O, \Omega}$, and we have the result.
Category:Conditional Expectation
Conditional Expectation Unchanged on Conditioning on Independent Sigma-Algebra/Corollary

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\HH \subseteq \Sigma$ be a sub-$\sigma$-algebra.
Let $X$ be a integrable random variable such that: 
:$\map \sigma X$ is independent of $\HH$
where $\map \sigma X$ is the $\sigma$-algebra generated by $X$. 
Let $\expect {X \mid \HH}$ be a version of the conditional expectation of $X$ given $\GG$.
Then: 
:$\expect {X \mid \HH} = \expect X$ almost surely. 
== Proof ==
Note that: 
:$\map \sigma X = \map \sigma {\set {\O, \Omega}, \map \sigma X}$ 
So by Conditional Expectation Unchanged on Conditioning on Independent Sigma-Algebra, we have:
:$\expect {X \mid \map \sigma {\set {\O, \Omega}, \HH} } = \expect {X \mid \set {\O, \Omega} }$
From Conditional Expectation Conditioned on Trivial Sigma-Algebra, we have:
:$\expect {X \mid \set {\O, \Omega} } = \expect X$ almost surely.
Also, since: 
:$\set {\O, \Omega} \subseteq \HH$
we have that:
:$\map \sigma {\set {\O, \Omega}, \HH} = \HH$ 
giving:
:$\expect {X \mid \HH} = \expect X$ almost surely.
== Sources ==
* : $9.7$: Properties of conditional expectation: a list
Category:Conditional Expectation
Category:Conditional Expectation Unchanged on Conditioning on Independent Sigma-Algebra
Conditional Expectation is Linear

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $\GG \subseteq \Sigma$ be a sub-$\sigma$-algebra.
Let $X$ and $Y$ be integrable random variables.
Let $\alpha, \beta \in \R$. 
Let $\expect {X \mid \GG}$ be a version of the conditional expectation of $X$ given $\GG$.
Let $\expect {Y \mid \GG}$ be a version of the conditional expectation of $Y$ given $\GG$.
Let $\expect {\alpha X + \beta Y \mid \GG}$ be a version of the conditional expectation of $\alpha x + \beta Y$ given $\GG$.
Then:
:$\expect {\alpha X + \beta Y \mid \GG} = \alpha \expect {X \mid \GG} + \beta \expect {Y \mid \GG}$ almost everywhere.
== Proof ==
We show that $\alpha \expect {X \mid \GG} + \beta \expect {Y \mid \GG}$ is a version of the conditional expectation of $\alpha x + \beta Y$ given $\GG$.
From Linear Combination of Real-Valued Random Variables is Real-Valued Random Variable, we have that $\alpha \expect {X \mid \GG} + \beta \expect {Y \mid \GG}$ is a $\GG$-measurable random variable.
Let $A \in \GG$.
Then:
So $\alpha \expect {X \mid \GG} + \beta \expect {Y \mid \GG}$ is a version of the conditional expectation of $\alpha x + \beta Y$ given $\GG$, and from Existence and Essential Uniqueness of Conditional Expectation Conditioned on Sigma-Algebra we have:
:$\expect {\alpha X + \beta Y \mid \GG} = \alpha \expect {X \mid \GG} + \beta \expect {Y \mid \GG}$ almost everywhere.
== Sources ==
* : $9.7$: Properties of conditional expectation: a list
Category:Conditional Expectation