nyuuzyou/wikis huggingface dataset:
Measure Space from Outer Measure

== Theorem ==
Suppose $\mu^*$ is an outer measure on a set $X$.
Let $\map {\mathfrak M} {\mu^*}$ be the collection of $\mu^*$-measurable sets.
Let $\mu$ be the restriction of $\mu^*$ to $\map {\mathfrak M} {\mu^*}$.
Then $\struct {X, \map {\mathfrak M} {\mu^*}, \mu}$ is a measure space.
== Proof ==
First, note that $\map {\mathfrak M} {\mu^*}$ is a $\sigma$-algebra over $X$.
Next, choose $E_1, E_2 \in \map {\mathfrak M} {\mu^*}$ such that $E_1 \cap E_2 = \O$.
Thus:
Therefore $\mu$ is additive. But because it is constructed from an outer measure, it is also countably subadditive.
Since Additive and Countably Subadditive Function is Countably Additive, it follows that $\mu$ is countably additive.
Finally, also because it is constructed from an outer measure, $\mu$ is nonnegative.
Hence $\struct {X, \map {\mathfrak M} {\mu^*} \mu}$ meets all the criteria of a measure space.
== Consequences ==
It immediately follows from the theorem that the Lebesgue measure forms a measure over the collection of Lebesgue measurable sets of reals.
Category:Outer Measures
Measure Space has Exhausting Sequence of Finite Measure iff Cover by Sets of Finite Measure

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Then there exists an exhausting sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ such that:
:$\forall n \in \N: \map \mu {E_n} < \infty$
 there exists a sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ such that:
:$(1): \quad \ds \bigcup_{n \mathop \in \N} E_n = X$
:$(2): \quad \forall n \in \N: \map \mu {E_n} < +\infty$
== Proof ==
=== Necessary Condition ===
Suppose that there exists an exhausting sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ such that:
::$\forall n \in \N: \map \mu {E_n} < \infty$
Let $\sequence {F_n}_{n \mathop \in \N}$ be an exhausting sequence in $\Sigma$ such that:
:$\forall n \in \N: \map \mu {F_n} < +\infty$
Then as $\sequence {F_n}_{n \mathop \in \N}$ is exhausting, have:
:$\ds \bigcup_{n \mathop \in \N} F_n = X$
It follows that the sequence $\sequence {F_n}_{n \mathop \in \N}$ satisfies $(1)$ and $(2)$.
=== Sufficient Condition ===
Let $\mu$ be any measure.
Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence satisfying $(1)$ and $(2)$.
Define $F_n := \ds \bigcup_{k \mathop = 1}^n E_k$.
Then by Sigma-Algebra Closed under Union:
: $F_n \in \Sigma$ for all $n \in \N$
Also, by Set is Subset of Union:
: $F_{n+1} = F_n \cup E_{n+1}$, hence $F_n \subseteq F_{n + 1}$
The definition of the $F_n$ assures that:
: $X = \ds \bigcup_{n \mathop \in \N} E_n = \bigcup_{n \mathop \in \N} F_n$
Hence $\sequence {F_n}_{n \mathop \in \N}$ is an exhausting sequence in $\Sigma$.
Furthermore, compute, for any $n \in \N$:
Hence there exists an exhausting sequence $\sequence {E_n}_{n \mathop \in \N}$ in $\Sigma$ such that:
:$\forall n \in \N: \map \mu {E_n} < \infty$
== Sources ==
* : $\S 4$: Problem $15$
Category:Sigma-Finite Measures
Measure is Countably Subadditive

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Then $\mu$ is a countably subadditive function.
== Proof ==
Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of sets in $\Sigma$.
It is required to show that:
:$\ds \map \mu {\bigcup_{n \mathop \in \N} E_n} \le \sum_{n \mathop \in \N} \map \mu {E_n}$
Now define the sequence $\sequence {F_n}_{n \mathop \in \N}$ in $\Sigma$ by:
:$F_n := \ds \bigcup_{k \mathop = 0}^n E_n$
By Subset of Union, it follows that, for all $n \in \N$, $F_n \subseteq F_{n + 1}$.
Hence, $\sequence {F_n}_{n \mathop \in \N}$ is increasing.
It is immediate that $F_n \uparrow \ds \bigcup_{n \mathop \in \N} E_n$, where $\uparrow$ signifies the limit of an increasing sequence of sets.
Now reason as follows:
Hence the result.
== Sources ==
* : $4.6$
Category:Measures
Measure is Discrete iff Concentrated on Countable Set

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space such that:
:$\set x \in \Sigma$ for all $x \in X$.
Let $\mu$ be a measure. 
Then $\mu$ is discrete :
:there exists a countable set $C \subseteq X$ such that $\map \mu {X \setminus C} = 0$. 
== Proof ==
=== Necessary Condition ===
Suppose that $\mu$ is discrete.
Then there exists a sequence $\sequence {x_n}_{n \mathop \in \N}$ in $X$ and a sequence of non-negative real numbers $\sequence {\lambda_n}_{n \mathop \in \N}$ such that:
:$\ds \mu = \sum_{n \mathop = 1}^\infty \lambda_n \delta_{x_n}$
where $\delta_{x_n}$ is the dirac measure at $x_n$.
Let:
:$C = \set {x_n : n \in \N}$
Then $C$ is countable.
Since $x_n \not \in X \setminus C$ for each $n \in \N$, we have $\map {\delta_{x_n} } {X \setminus C} = 0$.
Hence we have $\map \mu {X \setminus C} = 0$. 
=== Sufficient Condition ===
Suppose that:
:there exists a countable set $C \subseteq X$ such that $\map \mu {X \setminus C} = 0$.
Let $\sequence {x_n}_{n \mathop \in \N}$ be an enumeration of $C$. 
Now let $S \in \Sigma$. 
Let $\lambda_j = \map \mu {\set {x_j} }$ 
From countable additivity, we have that:
:$\ds \map \mu S = \sum_{j : x_j \in S} \map \mu {\set {x_j} } = \sum_{j : x_j \in S} \lambda_j$ 
We have $x_j \in S$ $\map {\delta_{x_j} } S = 1$, and $\map {\delta_{x_j} } S = 0$ otherwise, so we have:
:$\ds \map \mu S = \sum_{j \mathop = 1}^\infty \lambda_j \map {\delta_{x_j} } S$
So we have the result. 
Category:Discrete Measures
Measure is Finitely Additive Function

== Theorem ==
Let $\Sigma$ be a $\sigma$-algebra on a set $X$.
Let $\mu: \Sigma \to \overline {\R}$ be a measure on $\Sigma$.
Then $\mu$ is finitely additive.
== Proof ==
From the definition of a measure, $\mu$ is countably additive.
From Countably Additive Function also Finitely Additive, $\mu$ is finitely additive.
== Sources ==
* : $4.3 \ \text{(i)}$
* : $1.2$: Measures
Category:Finitely Additive Functions
Category:Measures
Measure is Monotone

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Then $\mu$ is monotone, that is:
:$\forall E, F \in \Sigma: E \subseteq F \implies \map \mu E \le \map \mu F$
=== Resolution of the Identity ===
== Proof ==
A direct corollary of Non-Negative Additive Function is Monotone.
== Sources ==
* : $4.3 \ \text{(ii)}$
* : $1.2$: Measures
Category:Measures
Category:Measure is Monotone
Measure is Monotone/Resolution of the Identity

== Theorem ==
Let $X$ be a topological space.
Let $\map \BB X$ be the Borel $\sigma$-algebra of $X$.
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\map B \HH$ be the space of bounded linear transformations on $\HH$.
Let $\le_{\map B \HH}$ be the canonical preordering of $\map B \HH$.
Let $\EE : \map \BB X \to \map B {\HH}$ be a resolution of the identity.
Let $E, F \in \map \BB X$ be such that $E \subseteq F$.
Then:
:$\map \EE E \le_{\map B \HH} \map \EE F$
=== Corollary ===
== Proof ==
Write:
:$F = E \cup \paren {F \setminus E}$
From $(4)$ in the definition of a resolution of the identity, we have:
:$\map \EE F = \map \EE E + \map \EE {F \setminus E}$ 
From Bounds on Projection in Unital C*-Algebra, we have:
:${\mathbf 0}_{\map B \HH} \le_{\map B \HH} \map \EE {F \setminus E}$ 
From $(1)$ in the definition of a preordered vector space, we have:
:$\map \EE E + \map \EE {F \setminus E} \ge_{\map B \HH} \map \EE E$ 
Hence:
:$\map \EE E \le_{\map B \HH} \map \EE F$ 
Category:Resolutions of the Identity
Category:Measure is Monotone
Measure is Monotone/Resolution of the Identity/Corollary

== Corollary ==
Let $X$ be a topological space.
Let $\map \BB X$ be the Borel $\sigma$-algebra of $X$.
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\map B \HH$ be the space of bounded linear transformations on $\HH$.
Let $\EE : \map \BB X \to \map B {\HH}$ be a resolution of the identity.
Let $E, F \in \map \BB X$ be such that $E \subseteq F$.
Then:
:$\Img {\map \EE E} \subseteq \Img {\map \EE F}$
== Proof ==
Let $\le_{\map B \HH}$ be the canonical preordering of $\map B \HH$.
From Measure is Monotone: Resolution of the Identity, we have:
:$\map \EE E \le_{\map B \HH} \map \EE F$
From the implication $(1) \implies (4)$ in Characterization of Canonical Preordering of Projections on Hilbert Space, we have:
:$\Img {\map \EE E} \subseteq \Img {\map \EE F}$
Category:Measure is Monotone
Measure is Strongly Additive

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Then $\mu$ is strongly additive, that is:
:$\forall E, F \in \Sigma: \map \mu {E \cap F} + \map \mu {E \cup F} = \map \mu E + \map \mu F$
== Proof ==
Combine Measure is Finitely Additive Function with Additive Function is Strongly Additive.
== Sources ==
* : $4.3 \ \text{(iv)}$
Category:Measures
Measure is Subadditive

== Theorem ==
Let $\left({X, \Sigma, \mu}\right)$ be a measure space.
Then $\mu$ is subadditive, that is:
:$\forall E, F \in \Sigma: \mu \left({E \cup F}\right) \le \mu \left({E}\right) + \mu \left({F}\right)$
=== Corollary ===
== Proof ==
A measure is an additive function, and, by definition, nowhere negative.
So Additive Nowhere Negative Function is Subadditive‎ applies.
Hence the result directly:
:$\mu \left({E \cup F}\right) \le \mu \left({E}\right) + \mu \left({F}\right)$
== Sources ==
* : $4.3 \ \text{(v)}$
Category:Measures
Category:Measure is Subadditive
Measure is Subadditive/Corollary

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $E_1, \ldots, E_n \in \Sigma$.
Then:
:$\ds \map \mu {\bigcup_{k \mathop = 1}^n E_k} \le \sum_{k \mathop = 1}^n \map \mu {E_k}$.
== Proof ==
We have Measure is Subadditive.
The result follows by an application of Finite Union of Sets in Subadditive Function.
Category:Measure is Subadditive
Measure is countably subadditive

== Theorem ==
Let $\left({X, \Sigma, \mu}\right)$ be a measure space.
Then $\mu$ is $\sigma$-subadditive, that is:
:$\ds\forall \sequence {A_n} _{n \in \N} \subseteq \Sigma: \map \mu {\bigcup _{n \in \N} A_n} \le \sum _{n \in \N} \map \mu {A_n}$
== Proof ==
For each $n\in\N$, let:
:$\ds A'_n := A_n \setminus \bigcup _{i=0} ^{n-1} A_i$
Then, by (SA3) and (SA2'):
:$\forall n\in\N : A'_n \in \Sigma$
Furthermore, by construction:
:$\bigcup _{n \in \N} A'_n = \bigcup _{n \in \N} A_n$
:$\sequence {A'_n} _{n \in \N}$ are pairwise disjoint
:$\forall n\in\N : A' _n\subseteq A_n$
Therefore:
Category:Measures
Measure of Empty Set is Zero

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Then $\map \mu \O = 0$.
That is, $\O$ is a $\mu$-null set.
== Proof ==
By definition of measure, there exists at least one $E \in \Sigma$ such that $\map \mu E$ is finite.
So, suppose that $E \in \Sigma$ such that $\map \mu E$ is finite.
Let $\map \mu E = x$.
Consider the sequence $\sequence {S_n}_{n \mathop \in \N} \subseteq \Sigma$ defined as:
:$S_n = \begin {cases} E & : n = 1 \\ \O & : n > 1 \end {cases}$
Then:
:$\ds \bigcup_{n \mathop = 1}^\infty S_n = E$
Hence:
It follows directly that $\map \mu \O = 0$.
== Sources ==
* : $\S 4$: Problem $10 \ \text{(i)}$
Category:Measures
Category:Empty Set
Measure of Half-Open Interval as Difference of Distribution Function

== Theorem ==
Let $a, b \in \R$. 
Let $\mu$ be a finite Borel measure on $\R$.
Let $F_\mu$ be the distribution function of $\mu$.
Then: 
:$\map \mu {\hointl a b} = \map {F_\mu} b - \map {F_\mu} a$
== Proof ==
Note that: 
:$\map \mu {\hointl {-\infty} a} < \infty$
since $\mu$ is finite.
Then, we have: 
== Sources ==
* : $1.3$: Outer Measures
Category:Distribution Function of Finite Borel Measure
Measure of Horizontal Section of Cartesian Product

== Theorem ==
Let $\struct {X, \Sigma_X, \mu_X}$ and $\struct {Y, \Sigma_Y, \mu_Y}$ be measure spaces.
Let $S_1 \in \Sigma_X$ and $S_2 \in \Sigma_Y$. 
Let $y \in Y$. 
Then: 
:$\map {\mu_X} {\paren {S_1 \times S_2}^y} = \map {\mu_X} {S_1} \map {\chi_{S_2} } y$
where:
:$\paren {S_1 \times S_2}^y$ is the $y$-horizontal section of $S_1 \times S_2$
:$\chi_{S_2}$ is the characteristic function for $S_2$. 
== Proof ==
From Horizontal Section of Cartesian Product, we have: 
:$\paren {S_1 \times S_2}^y = \begin{cases}S_1 & y \in S_2 \\ \O & y \not \in S_2\end{cases}$
So:
:$\map {\mu_X} {\paren {S_1 \times S_2}^y} = \begin{cases}\map {\mu_X} {S_1} & y \in S_2 \\ 0 & y \not \in S_2\end{cases}$
That is: 
:$\map {\mu_X} {\paren {S_1 \times S_2}^y} = \map {\mu_X} {S_1} \times \begin{cases}1 & y \in S_2 \\ 0 & y \not \in S_2\end{cases}$
giving:
:$\map {\mu_X} {\paren {S_1 \times S_2}^y} = \map {\mu_X} {S_1} \map {\chi_{S_2} } y$
from the definition of a characteristic function.
Category:Horizontal Section of Sets
Category:Cartesian Product
Measure of Limit of Decreasing Sequence of Measurable Sets

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space. 
Let $E \in \Sigma$.
Let $\sequence {E_n}_{n \mathop \in \N}$ be an decreasing sequence of $\Sigma$-measurable sets such that: 
:$E_n \downarrow E$ 
where $E_n \downarrow E$ denotes the limit of decreasing sequence of sets.
Suppose also that $\map \mu {E_1} < \infty$. 
Then: 
:$\ds \map \mu E = \lim_{n \mathop \to \infty} \map \mu {E_n}$ 
=== Corollary ===
== Proof ==
From Relative Complement of Decreasing Sequence of Sets is Increasing, we have: 
:$\sequence {E_1 \setminus E_n}_{n \mathop \in \N}$ is increasing.
Further, we have: 
so:
:$E_1 \setminus E_n \uparrow E_1 \setminus E$ 
So, from Measure of Limit of Increasing Sequence of Measurable Sets, we have: 
:$\ds \map \mu {E_1 \setminus E} = \lim_{n \mathop \to \infty} \map \mu {E_1 \setminus E_n}$
From Measure of Set Difference with Subset, we have: 
:$\ds \map \mu {E_1} - \map \mu E = \lim_{n \mathop \to \infty} \paren {\map \mu {E_1} - \map \mu {E_n} }$
since $\map \mu {E_1} < \infty$. 
From the Difference Rule for Real Sequences, we then have: 
:$\ds \map \mu {E_1} - \map \mu E = \map \mu {E_1} - \lim_{n \mathop \to \infty} \map \mu {E_n}$
giving:
:$\ds \map \mu E = \lim_{n \mathop \to \infty} \map \mu {E_n}$ 
== Also see == 
* Signed Measure of Limit of Decreasing Sequence of Measurable Sets
Category:Measures
Category:Decreasing Sequences of Sets
Category:Measure of Limit of Decreasing Sequence of Measurable Sets
Measure of Limit of Decreasing Sequence of Measurable Sets/Corollary

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space. 
Let $F \in \Sigma$.
Let $\sequence {F_n}_{n \mathop \in \N}$ be an decreasing sequence of $\Sigma$-measurable sets such that: 
:$F_n \downarrow F$ 
where $F_n \downarrow F$ denotes the limit of decreasing sequence of sets.
Suppose also that $\map \mu {F_m} < \infty$ for some $m \in \N$. 
Then: 
:$\ds \map \mu F = \lim_{n \mathop \to \infty} \map \mu {F_n}$
== Proof ==
Define the sequence $\sequence {E_n}_{n \mathop \in \N}$ by: 
:$E_n = F_{m + n}$ 
Then from Tail of Decreasing Sequence of Sets is Decreasing:
:$\sequence {E_n}_{\mathop \in \N}$ is an decreasing sequence of $\Sigma$-measurable sets.
From Limit of Tail of Decreasing Sequence of Sets, we have: 
:$E_n \downarrow F$
with: 
:$\map \mu {E_1} = \map \mu {F_{m + 1} }$
Since $\sequence {F_n}_{n \mathop \in \N}$ is decreasing, we have: 
:$F_{m + 1} \subseteq F_m$
So from Measure is Monotone, we have: 
:$\map \mu {F_{m + 1} } \le \map \mu {F_m} < \infty$
so:
:$\map \mu {E_1} < \infty$
Then:
Category:Measure of Limit of Decreasing Sequence of Measurable Sets
Measure of Limit of Increasing Sequence of Measurable Sets

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space. 
Let $E \in \Sigma$.
Let $\sequence {E_n}_{n \mathop \in \N}$ be an increasing sequence of $\Sigma$-measurable sets such that: 
:$E_n \uparrow E$ 
where $E_n \uparrow E$ denotes limit of increasing sequence of sets.
Then: 
:$\ds \map \mu E = \lim_{n \mathop \to \infty} \map \mu {E_n}$ 
== Proof ==
We define a sequence $\sequence {F_n}_{n \mathop \in \N}$ inductively.
Set $F_1 = E_1$. 
For $n > 1$, define: 
:$F_n = E_n \setminus E_{n - 1}$
From the definition of set difference we have: 
:$E_n \setminus E_{n - 1} = E_n \cap \paren {X \setminus E_{n - 1} }$
Since $\Sigma$ is closed under complementation, we have: 
:$X \setminus E_{n - 1} \in \Sigma$
and so, from Sigma-Algebra Closed under Countable Intersection:
:$F_n \in \Sigma$
for $n > 1$. 
Clearly $F_1 \in \Sigma$ also.
Since $\sequence {E_n}_{n \in \N}$ is increasing, we have that: 
:$E_i \subseteq E_j$ for $i \le j$. 
So:
:$F_n$ is disjoint to $E_1, E_2, \ldots, E_{n - 1}$.
Since $F_i \subseteq E_i$ for each $i$, we have:
:$F_n$ is disjoint to $F_1, F_2, \ldots, F_{n - 1}$.
Picking $n = \max \set {i, j}$ we obtain that: 
:$F_i \cap F_j = \O$ for $i \ne j$.
We now show that: 
:$\ds E_k = \bigcup_{i \mathop = 1}^k F_i$
We have: 
:$\ds x \in \bigcup_{i \mathop = 1}^k F_i$
:
:$x \in F_i$ for some $1 \le i \le k$.
This is equivalent to: 
:$x \in E_i$ for some $1 \le i \le k$.
Since $\sequence {E_n}_{n \in \N}$ increasing sequence, this is equivalent to: 
:$x \in E_k$
showing that: 
:$\ds x \in \bigcup_{i \mathop = 1}^k F_i$ $x \in E_k$
giving:
:$\ds E_k = \bigcup_{i \mathop = 1}^k F_i$
So, we have: 
:$\ds \bigcup_{k \mathop = 1}^\infty E_k = \bigcup_{i \mathop = 1}^\infty F_i$
That is: 
:$\ds E = \bigcup_{i \mathop = 1}^\infty F_i$
from the definition of a limit of an increasing sequence of sets.
We can now compute:
Category:Measures
Category:Increasing Sequences of Sets
Measure of Set Difference with Subset

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $S, T \in \Sigma$ be such that $S \subseteq T$, and suppose that $\mu \paren S < +\infty$.
Then:
:$\mu \paren {T \setminus S} = \mu \paren T - \mu \paren S$
where $T \setminus S$ denotes set difference.
=== Signed Measure ===
== Proof ==
== Sources ==
* : $1.2$: Measures
Category:Measures
Category:Set Difference
Category:Measure of Set Difference with Subset
Measure of Set Difference with Subset/Signed Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$.
Let $S, T \in \Sigma$ be such that $S \subseteq T$ with $\size {\map \mu S} < \infty$.
Then: 
:$\map \mu {T \setminus S} = \map \mu T - \map \mu S$
== Proof ==
Let $\struct {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$.
Then, since: 
:$\size {\map \mu S} < \infty$
We have: 
:$\map {\mu^+} S < \infty$ and $\map {\mu^-} S < \infty$
Then, we have: 
Category:Measure of Set Difference with Subset
Category:Signed Measures
Measure of Stieltjes Function of Measure

== Theorem ==
Let $\mu$ be a measure on $\map \BB \R$, the Borel $\sigma$-algebra on $\R$.
Suppose that for all $n \in \N$, $\mu$ satisfies:
:$\map \mu {\hointr {-n} n} < +\infty$
Let $f_\mu$ be the Stieltjes function of $\mu$.
Let $\mu_{f_\mu}$ be the measure of $f_\mu$.
Then $\mu_{f_\mu} = \mu$.
== Proof ==
From Pre-Measure of Finite Stieltjes Function Extends to Unique Measure, it suffices to verify that:
:$\map {\mu_{f_\mu} } {\hointr a b} = \map \mu {\hointr a b}$
for all half-open intervals $\hointr a b$.
Now we have:
If either $a = 0$ or $b = 0$, the result follows immediately from the definition of $f_\mu$.
Now suppose that $a < b < 0$.
Then:
Finally, let $0 < a < b$.
Then:
The final case $a < 0 < b$ is a trivial consequence of Measure is Finitely Additive Function.
Hence it must be that $\mu_{f_\mu} = \mu$.
== Sources ==
* : $\S 7$: Problem $9 \ \text{(iii)}$
Category:Measure Theory
Measure of Total Solid Angle around Point

== Theorem ==
The total solid angle around a point has a measure of $4 \pi$ steradians.
== Proof ==
Let $Q$ denote this total solid angle
Hence $Q$ is equal to the total surface area of $S$ divided by $r^2$.
That is:
Hence the result.
== Sources ==
* 
* 
Category:Solid Angles
Measure of Trihedral Angle formed by 3 Mutually Perpendicular Half-Lines

== Theorem ==
Let $Q$ be the trihedral angle formed by $3$ half-lines each of which is perpendicular to the other two half-lines.
Then $Q$ measures $\dfrac \pi 2$ steradians.
== Proof ==
== Sources ==
* 
* 
Category:Trihedral Angles
Category:Solid Angles
Measure of Vertical Section of Cartesian Product

== Theorem ==
Let $\struct {X, \Sigma_X, \mu}$ and $\struct {Y, \Sigma_Y, \nu}$ be measure spaces.
Let $S_1 \in \Sigma_X$ and $S_2 \in \Sigma_Y$. 
Let $x \in X$. 
Then: 
:$\map {\nu} {\paren {S_1 \times S_2}_x} = \map {\nu} {S_2} \map {\chi_{S_1} } x$
where:
:$\paren {S_1 \times S_2}_x$ is the $x$-vertical section of $S_1 \times S_2$
:$\chi_{S_1}$ is the characteristic function for $S_1$. 
== Proof ==
From Vertical Section of Cartesian Product, we have: 
:$\paren {S_1 \times S_2}_x = \begin{cases}S_2 & x \in S_1 \\ \O & x \not \in S_1\end{cases}$
So:
:$\map {\nu} {\paren {S_1 \times S_2}_x} = \begin{cases}\map {\nu} {S_2} & x \in S_1 \\ 0 & x \not \in S_1\end{cases}$
That is: 
:$\map {\nu} {\paren {S_1 \times S_2}_x} = \map {\nu} {S_2} \times \begin{cases}1 & x \in S_1 \\ 0 & x \not \in S_1\end{cases}$
giving:
:$\map {\nu} {\paren {S_1 \times S_2}_x} = \map {\nu} {S_2} \map {\chi_{S_1} } x$
from the definition of a characteristic function.
Category:Vertical Section of Sets
Category:Cartesian Product
Measure with Density is Measure

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R_{\ge 0}$ be a positive $\Sigma$-measurable function.
Then the $f \mu$, the measure with density $f$ with respect to $\mu$ is a measure.
== Proof ==
Note that for each $A \in \Sigma$, we have: 
:$\ds \map {\paren {f \mu} } A = \int_A f \rd \mu$ 
We verify each of the three conditions for a measure.
=== Proof of $(1)$ ===
We have: 
:$\map {\paren {f \mu} } A \ge 0$
for each $A \in \Sigma$ from the definition of the $\mu$-integral of a positive measurable function. 
=== Proof of $(2)$ ===
Let $\sequence {D_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets. 
Let: 
:$\ds D = \bigcup_{n \mathop = 1}^\infty D_n$
We then have: 
which shows $(2)$.
=== Proof of $(3')$ ===
We have: 
verifying $(3')$.
Since all three conditions have been verified, we have that $\mu f$ is a measure.
== Sources ==
* : $\S 9$: Problem $5$
* : $10.8$
Category:Measures
Category:Measure with Density
Measurement of Terrestrial Latitude/Pole Star

== Theorem ==
An observer's (terrestrial) latitude can be identified by measuring the (celestial) altitude of the Pole Star.
== Proof ==
From Altitude of North Celestial Pole equals Latitude of Observer, all the observer needs to do is to measure the (celestial) altitude of the north celestial pole.
The star Polaris is situated at practically exactly that point.
Hence the result.
Category:Measurement of Terrestrial Latitude
Measurement of Terrestrial Longitude

== Theorem ==
To measure the (terrestrial) longitude of a position relative to a given principal meridian:
:$(1): \quad$ Identify the exact moment of noon at the location in question
:$(2): \quad$ Work out the precise time of day $t$ that this happens.
Let $N$ be the time of noon at the principal meridian on the day in question.
Let $x$ be the number of hours before $N$ at the principal meridian.
If $t$ is after $N$, then $x$ is treated as negative.
Then the (terrestrial) longitude is $15 x \degrees$.
If $x$ is positive, the (terrestrial) longitude is east.
If $x$ is negative, the (terrestrial) longitude is west.
== Proof ==
== Sources ==
* : Chapter $6$: Curves and Coordinates: Coordinate geometry today
Category:Navigation
Measurements of Common Angles/Acute Angle

== Theorem ==
An acute angle measures $\theta$, where:
:$0 \degrees < \theta < 90 \degrees$
or:
:$0 < \theta < \dfrac \pi 2$
== Proof ==
An acute angle is defined to be an angle whose measure is between that of a zero angle and a right angle.
A zero angle measures $0$ by definition, and a right angle measures $90 \degrees$ or $\dfrac \pi 2$.
Hence the result.
== Sources ==
* 
* 
Category:Acute Angles
Category:Measurements of Common Angles
Measurements of Common Angles/Full Angle

== Theorem ==
A full angle is equal to $360 \degrees$ or $2 \pi$ radians.
== Proof ==
By definition, $1$ radian is the angle which sweeps out an arc on a circle whose length is the radius $r$ of the circle.
From Perimeter of Circle, the length of the circumference of a circle of radius $r$ is equal to $2 \pi r$.
Therefore, $1$ radian sweeps out $\dfrac 1 {2 \pi}$ of a circle.
It follows that $2 \pi$ radians sweeps out the entire circle, or one full angle.
By definition of degree of angle, a full rotation is $360 \degrees$.
Therefore, a full angle is $360 \degrees$ or $2 \pi$.
== Sources ==
* : $1$. Functions: $1.5$ Trigonometric or Circular Functions: $1.5.1$ Unit Circle
* : $6 \cdotp 283 \, 185 \ldots$
* : $6 \cdotp 28318 \, 5 \ldots$
* 
* 
* 
* 
Category:Full Angles
Category:Measurements of Common Angles