nyuuzyou/wikis huggingface dataset:
Mean Value of Convex Real Function

== Theorem ==
Let $f$ be a real function which is continuous on the closed interval $\closedint a b$ and differentiable on the open interval $\openint a b$.
Let $f$ be convex on $\openint a b$.
Then:
:$\forall \xi \in \openint a b: \map f x - \map f \xi \ge \map {f'} \xi \paren {x - \xi}$
== Proof ==
By the Mean Value Theorem:
:$\exists \eta \in \openint x \xi: \map {f'} \eta = \dfrac {\map f x - \map f \xi} {x - \xi}$
From Real Function is Convex iff Derivative is Increasing, the derivative of $f$ is increasing.
Thus:
:$x > \xi \implies \map {f'} \eta \ge \map {f'} \xi$
:$x < \xi \implies \map {f'} \eta \le \map {f'} \xi$
Hence:
:$\map f x - \map f \xi \ge \map {f'} \xi \paren {x - \xi}$
== Also see ==
* Mean Value of Concave Real Function
== Sources ==
* : $\S 12.21 \ (4)$
Category:Convex Real Functions
Category:Differential Calculus
Mean of Grouped Data/Examples/Heights of People

== Examples of Mean of Grouped Data ==
For $200$ people, their exact height $x$ in centimetres is measured.
The observations are grouped as follows:
$\begin {array} {r|l} \hline
\text {Height $x$ (cm)} & \text {Count} \\
\hline
x < 165 & 6 \\
165 \le x < 170 & 39 \\
170 \le x < 175 & 93 \\
175 \le x < 180 & 44 \\
180 \le x < 185 & 15 \\
x \ge 185 & 3 \\
\hline \end {array}$
The mid-interval values for the above data are:
:$162.5, 167.5, 172.5, 177.5, 182.5, 187.5$
Hence the mean for the above data is calculated as:
:$\mu = \dfrac {6 \times 162.5 + 39 \times 167.5 + 93 \times 172.5 + 44 \times 177.5 + 15 \times 182.5 + 3 \times 187.5} {200} = 173.5$
== Sources ==
* 
* 
Category:Examples of Mean of Grouped Data
Mean of Random Sample from Chi-Squared Distribution has Gamma Distribution

== Theorem == 
Let $n$ be a strictly positive integer. 
Let $X_1, X_2, \ldots, X_k$ form a random sample of size $k$ from the chi-squared distribution with $n$ degrees of freedom. 
Then: 
:$\ds \overline X = \frac 1 k \sum_{i \mathop = 1}^k X_i \sim \map \Gamma {\frac {n k} 2, \frac k 2}$
where $\map \Gamma {\dfrac {n k} 2, \dfrac k 2}$ is the gamma distribution with parameters $\dfrac {n k} 2$ and $\dfrac k 2$.
== Proof ==
By Sum of Chi-Squared Random Variables, we have: 
:$\ds \sum_{i \mathop = 1}^k X_i \sim \chi^2_{n k}$
By Multiple of Chi-Squared Random Variable has Gamma Distribution, we then have: 
:$\ds \frac 1 k \sum_{i \mathop = 1}^k X_i \sim \map \Gamma {\frac {n k} 2, \frac 1 {\frac 2 k} } = \map \Gamma {\frac {n k} 2, \frac k 2}$
Category:Chi-Squared Distribution
Category:Gamma Distribution
Mean of Unequal Real Numbers is Between them

== Theorem ==
:$\forall x, y \in \R: x < y \implies x < \dfrac {x + y} 2 < y$
== Proof ==
First note that:
Then:
Similarly:
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $2 \ \text{(k)}$
Category:Real Numbers
Measurability in Trace Sigma-Algebra

== Theorem ==
Let $\struct {A, \Sigma_A}$ and $\struct {C, \Sigma_C}$ be measurable spaces.
Let $B \subseteq C$. 
Let $\Sigma_B$ be the trace $\sigma$-algebra on $B$ induced by $\Sigma_C$. 
Let $f : A \to B$ be $\Sigma_A/\Sigma_B$-measurable.
Then $f$ is $\Sigma_A/\Sigma_C$-measurable. 
== Proof ==
Let $S \in \Sigma_C$.
By Image is Subset of Codomain/Corollary 2, we have:
:$f \sqbrk A \subseteq B$
we have:
:$f^{-1} \sqbrk S = f^{-1} \sqbrk {S \cap B}$
From the definition of the trace $\sigma$-algebra, we have:
:$S \cap B \in \Sigma_B$
Since $f$ is $\Sigma_A/\Sigma_B$-measurable, we have:
:$f^{-1} \sqbrk {S \cap B} \in \Sigma_A$
Thus:
:$f^{-1} \sqbrk S \in \Sigma_A$.
Therefore we have shown that:
:$\forall S \in \Sigma_C$: $f^{-1} \sqbrk S \in \Sigma_A$
Hence $f$ is $\Sigma_A/\Sigma_C$-measurable. 
Category:Measurable Functions
Category:Trace Sigma-Algebras
Measurable Function Zero A.E. iff Absolute Value has Zero Integral

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a $\Sigma$-measurable function.
Then the following are equivalent:
:$(1) \quad$ $f = 0$ almost everywhere
:$(2) \quad$ $\ds \int \size f \rd \mu = 0$
=== Corollary ===
== Proof ==
Let $\EE^+$ be the space of positive simple functions.
=== $(1)$ implies $(2)$ ===
Suppose that: 
:$f = 0$ almost everywhere.
Note that if $\map f x = 0$ for some $x \in X$, then: 
:$\size {\map f x} = 0$ for some $x \in X$. 
So:
:$\size f = 0$ almost everywhere.
That is: 
:there exists a null set $N \subseteq X$ such that if $\size {\map f x} \ne 0$, then $x \in N$. 
From Absolute Value of Measurable Function is Measurable:
:$\size f$ is $\Sigma$-measurable. 
So, its $\mu$-integral is well-defined.
Let $g \in \EE^+$ have $g \le \size f$. 
Then, if $\map g x \ne 0$ for $x \in X$ we must have:
:$\size {\map f x} \ne 0$
So, if $x \in X$ has $\map g x \ne 0$, then $x \in N$. 
From Simple Function has Standard Representation: 
:there exists disjoint $\Sigma$-measurable sets $E_1, E_2, \ldots, E_n$ and non-negative real numbers $a_1, a_2, \ldots, a_n$ such that: 
::$\ds \map f x = \sum_{i \mathop = 1}^n a_i \map {\chi_{E_i} } x$
:for each $x \in X$.
If $a_i \ne 0$, then: 
:$\map f x \ne 0$ for $x \in E_i$
That is: 
:$x \in N$ for all $x \in E_i$
So, we obtain: 
:$E_i \subseteq N$
for each $i$. 
Then, from Measure is Monotone, we have: 
:$\map \mu {E_i} \le \map \mu N = 0$ 
So:
:$\map \mu {E_i} = 0$
for each $i$ such that $a_i \ne 0$
So: 
:$a_i \map \mu {E_i} = 0$
for each $i$. 
Then:
:$\ds \map {I_\mu} g = 0$
for all $g \in \EE^+$ with $g \le \size f$, where: 
:$\map {I_\mu} g$ denotes the $\mu$-integral of the positive simple function $g$.
So, from the definition of the $\mu$-integral:
:$\ds 0 = \sup \set {\map {I_\mu} g: g \le f, g \in \EE^+} = \int \size f \rd \mu$
=== $(2)$ implies $(1)$ ===
Suppose that: 
:$\ds \int \size f \rd \mu = 0$ 
From Markov's Inequality, we have, for each $n \in \N$: 
So:
:$\ds \map \mu {\set {x \in X : \size {\map f x} \ge \frac 1 n} } = 0$
for each $n \in \N$. 
Note that: 
From Measure is Countably Subadditive, we have: 
So:
:$\map \mu {\set {x \in X : \map f x \ne 0} } = 0$
That is: 
:$f = 0$ almost everywhere.
== Sources ==
* : $10.9 \ \text{(i)}$
Category:Measurable Functions
Category:Measurable Function Zero A.E. iff Absolute Value has Zero Integral
Measurable Function Zero A.E. iff Absolute Value has Zero Integral/Corollary

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f: X \to \overline \R$ be a non-negative integrable function.
Let $A, B \in \Sigma$ have $A \subseteq B$. 
Then: 
:$\ds \int_A f \rd \mu = \int_B f \rd \mu$
: 
:$f \times \chi_{B \setminus A} = 0$ $\mu$-almost everywhere. 
== Proof ==
We can write: 
:$B = A \cup \paren {B \setminus A}$
From Integral of Positive Measurable Function over Disjoint Union, we have: 
:$\ds \int_B f \rd \mu = \int_A f \rd \mu + \int_{B \setminus A} f \rd \mu$
Since: 
:$\ds \int_B f \rd \mu = \int_A f \rd \mu$
we get: 
:$\ds \int_{B \setminus A} f \rd \mu = 0$
From the definition of the $\mu$-integral over $A$, we have: 
:$\ds \int \paren {f \times \chi_{B \setminus A} } \rd \mu$
From Measurable Function Zero A.E. iff Absolute Value has Zero Integral, we have: 
:$\ds \int \paren {f \times \chi_{B \setminus A} } \rd \mu = 0$ $f \times \chi_{B \setminus A} = 0$ $\mu$-almost everywhere. 
So:
:$\ds \int_A f \rd \mu = \int_B f \rd \mu$
:
:$f \times \chi_{B \setminus A} = 0$ $\mu$-almost everywhere. 
Category:Measurable Function Zero A.E. iff Absolute Value has Zero Integral
Measurable Function is Integrable iff A.E. Equal to Real-Valued Integrable Function

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f : X \to \overline \R$ be a $\Sigma$-measurable function.
Then $f$ is $\mu$-integrable :
:there exists a $\mu$-integrable function $g : X \to \R$ such that $g = f$ $\mu$-almost everywhere.
== Proof ==
=== Sufficient Condition ===
Suppose that: 
:there exists a $\mu$-integrable function $g : X \to \R$ such that $g = f$ $\mu$-almost everywhere.
Then, from A.E. Equal Positive Measurable Functions have Equal Integrals: Corollary 1, we have: 
:$f$ is $\mu$-integrable.
=== Necessary Condition ===
Suppose that $f$ is $\mu$-integrable.
From Integrable Function is A.E. Real-Valued, we have: 
:$\map f x \in \R$ for $\mu$-almost all $x \in X$. 
That is, there exists a $\mu$-null set $N \subseteq X$ such that whenever: 
:$\size {\map f x} = \infty$
we have $x \in N$.
Define a function $g : X \to \R$ by:
:$\map g x = \map f x \map {\chi_{X \setminus N} } x$
for each $x \in X$.
Since $N \in \Sigma$, we have: 
:$X \setminus N \in \Sigma$
From Characteristic Function Measurable iff Set Measurable, we have: 
:$\chi_{X \setminus N}$ is $\Sigma$-measurable.
From Pointwise Product of Measurable Functions is Measurable, we have: 
:$g$ is $\Sigma$-measurable.
Note that whenever $x \in X$ has: 
:$\map f x \map {\chi_{X \setminus N} } x \ne \map f x$ 
we have: 
:$\map {\chi_{X \setminus N} } x = 0$
That is, from the definition of set difference: 
:$x \in N$
So:
:$g = f$ $\mu$-almost everywhere. 
From A.E. Equal Positive Measurable Functions have Equal Integrals: Corollary 1, we have: 
:$g$ is $\mu$-integrable.
So:
:there exists a $\mu$-integrable function $g : X \to \R$ such that $g = f$ $\mu$-almost everywhere.
== Sources ==
* : $2.3$: Integral
Category:Measure Theory
Measurable Function is Simple Function iff Finite Image Set

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f: X \to \R$ be a measurable function.
Then $f$ is a simple function its image is finite:
:$\card {\Img f} < \infty$
=== Corollary ===
== Proof ==
=== Necessary Condition ===
Suppose that $f$ is a simple function, and that:
:$\ds \forall x \in X: \map f x = \sum_{i \mathop = 1}^n a_i \map {\chi_{S_i} } x$
Since each of the $\chi_{S_i}$ is a characteristic function, it can take only the values $0$ and $1$.
Thus each summand can take two values.
It follows immediately that $f$ can take at most $2^n$ different values.
The conclusion follows from Simple Function is Measurable.
=== Sufficient Condition ===
Suppose that the image of $f$ is finite.
Call the distinct values $f$ attains $y_1, \ldots, y_n$.
For brevity, denote $\set {f = a}$ to mean $\set {x \in X: \map f x = a}$ (compare Set Definition by Predicate).
Define for each $i$ with $1 \le i \le n$:
:$B_i := \set {f = y_i}$
From Characterization of Measurable Functions $(2)$ and $(4)$, and Sigma-Algebra Closed under Intersection we obtain that:
:$\set {f = y_i} = \set {f \ge y_i} \cap \set {f \le y_i} \in \Sigma$
Furthermore, since the $y_i$ are distinct, the $B_i$ are necessarily disjoint.
It follows that:
:$(1): \quad \map f x = \ds \sum_{i \mathop = 1}^n y_j \map {\chi_{B_j} } x$
As the $B_i$ are measurable, $f$ is shown to be a simple function.
== Sources ==
* : $8.7 \ \text{(iii)}$
Category:Measurable Functions
Category:Simple Functions
Measurable Function is Simple Function iff Finite Image Set/Corollary

== Corollary to Measurable Function is Simple Function iff Finite Image Set ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f: X \to \R$ be a measurable function.
Then $f$ has a standard representation.
== Proof ==
Applying the main theorem to a simple function yields the representation $(1)$:
:$\map f x = \ds \sum_{i \mathop = 1}^n y_j \map {\chi_{B_j} } x$
which is of the required form.
== Sources ==
* : $8.7 \ \text{(iii)}$
Category:Measure Theory
Measurable Functions Determine Measurable Sets

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f, g: X \to \overline \R$ be $\Sigma$-measurable functions.
Then the following sets are measurable:
:$\set {f < g}$
:$\set {f \le g}$
:$\set {f = g}$
:$\set {f \ne g}$
where, for example, $\set {f < g}$ is short for $\set {x \in X: \map f x < \map g x}$.
== Proof ==
From Pointwise Difference of Measurable Functions is Measurable, $f - g: X \to \overline \R$ is $\Sigma$-measurable.
Now we have the following evident identities:
:$\set {f < g} = \set {f - g < 0}$
:$\set {f \ge g} = \set {f - g \le 0}$
:$\set {f = g} = \set {f - g = 0}$
:$\set {f \ne g} = \set {f - g \ne 0}$
Subsequently, using the preimage under $f - g$, the latter may respectively be expressed as:
:$\set {f - g < 0} = \paren {f - g}^{-1} \sqbrk {\openint {-\infty} 0}$
:$\set {f - g \ge 0} = \paren {f - g}^{-1} \sqbrk {\hointl {-\infty} 0}$
:$\set {f - g = 0} = \paren {f - g}^{-1} \sqbrk {\set 0}$
:$\set {f - g \ne 0} = \paren {f - g}^{-1} \sqbrk {\R \setminus \set 0}$
Now the sets:
:$\openint {-\infty} 0$
:$\hointl {-\infty} 0$
:$\set 0$
:$\R \setminus \set 0$
are all open or closed in the Euclidean topology.
Hence by definition of Borel $\sigma$-algebra, and by Closed Set Measurable in Borel Sigma-Algebra, they are $\map \BB \R$-measurable.
Since $f - g$ is $\Sigma$-measurable, it follows that:
:$\set {f < g}$
:$\set {f \le g}$
:$\set {f = g}$
:$\set {f \ne g}$
are all $\Sigma$-measurable sets.
== Sources ==
* : $8.12$, $\S 8$: Problem $11$
Category:Measurable Functions
Category:Measurable Sets
Measurable Functions with Equal Integrals on Sub-Sigma-Algebra are A.E. Equal

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\GG$ be a sub-$\sigma$-algebra of $\Sigma$.
Suppose that $\mu \restriction_\GG$, the restriction of $\mu$ to $\GG$, is $\sigma$-finite.
Let $f, g: X \to \overline \R$ be $\GG$-measurable functions.
Suppose that, for all $G \in \GG$:
:$\ds \int_G f \rd \mu = \int_G g \rd \mu$
Then $f = g$ $\mu$-almost everywhere.
== Proof ==
First, assume that $f$ and $g$ are $\mu$-integrable.
Observe:
For each $n \in \N_{>0}$:
Swapping $f$ and $g$, we also have:
Thus:
Therefore:
Hence the result, by definition of almost-everywhere equality.
Now, consider general $f$ and $g$.
Recall that $\mu \restriction_\GG$　is $\sigma$-finite.
That is, there is an exhausting sequence $\sequence {E_n}_{n\in\N} \subseteq \GG$ such that:
:$\map \mu {E_n} < \infty$
We define $\sequence {A_n}_{n\in\N} \subseteq \GG$ by:
:$A_n := E_n \cap \set { \size f \le n} \cap \set {\size g \le n}$
Then $\sequence {A_n}_{n\in\N}$ is also an exhausting sequence.
Let $f_n := f \chi_{A_n}$ and $g_n := g \chi_{A_n}$.
Then $f_n$ and $g_n$ are $\mu$-integrable so that:
:$\forall n \in \N : \map \mu {\set {f_n \ne g_n} } = 0$
On the other hand:
:$\set {f_n \ne g_n} = \set {f \ne g} \cap A_n$
Therefore:
== Also see ==
* Integrable Functions with Equal Integrals on Sub-Sigma-Algebra are A.E. Equal
== Sources ==
* : $10.14 \ \text{(ii)}$
Category:Measure Theory
Measurable Image

== Theorem ==
Let $\mathfrak M$ be the set of measurable sets of $\R$.
For any extended real-valued function $f: \R \to \R \cup \set {-\infty, +\infty}$ whose domain is measurable, the following statements are equivalent:
:$(1): \quad \forall \alpha \in \R: \set {x: \map f x > \alpha} \in \mathfrak M$
:$(2): \quad \forall \alpha \in \R: \set {x: \map f x \ge \alpha} \in \mathfrak M$
:$(3): \quad \forall \alpha \in \R: \set {x: \map f x < \alpha} \in \mathfrak M$
:$(4): \quad \forall \alpha \in \R: \set {x: \map f x \le \alpha} \in \mathfrak M$
These statements imply:
:$(5): \quad \forall \alpha \in \R \cup \set {-\infty, +\infty}: \set {x: \map f x = \alpha} \in \mathfrak M$
== Proof ==
Let the domain of $f$ be $D$.
We have that Measurable Sets form Algebra of Sets.
First we note that, from Properties of Algebras of Sets, the difference of two measurable sets is measurable.
So:
:$\set {x: \map f x \le \alpha} = D - \set {x: \map f x > \alpha}$
and so $(1) \iff (4)$.
Similarly, $(2) \iff (3)$.
Next we note that, also from Properties of Algebras of Sets, the intersection of a sequence of measurable sets is measurable.
Now:
:$\ds \set {x: \map f x \ge \alpha} = \bigcap_{n \mathop = 1}^\infty \set {x: \map f x > \alpha -\dfrac 1 n}$
because if:
:$x \in \set {x: \map f x \ge \alpha}$
that is:
:$\map f x \ge \alpha$
and since:
:$\forall n \in \N: n > 0: \alpha > \alpha - \dfrac 1 n$
then:
:$\forall n \in \N: n > 0: \map f x > \alpha - \dfrac 1 n$
That is:
:$\forall n \in \N: n > 0: x \in \set {x: \map f x > \alpha - \dfrac 1 n}$
Hence:
:$\ds x \in \bigcap_{n \mathop = 1}^\infty \set {x: \map f x > \alpha - \dfrac 1 n}$
Conversely, suppose:
:$\ds x \in \bigcap_{n \mathop = 1}^\infty \set {x: \map f x > \alpha - \dfrac 1 n}$
that is:
:$\forall n \in \N: n > 0: x \in \set {x: \map f x > \alpha - \dfrac 1 n}$
Claim $\map f x \ge \alpha $.
Otherwise $\map f x < \alpha$, say for example $\map f x = \alpha - \size \epsilon$.
Choose $N = \ceiling {\dfrac 1 {\size \epsilon} } + 1 \in \N$.
Therefore:
:$N > \ceiling {\dfrac 1 {\size \epsilon} } \ge \dfrac 1 {\size \epsilon}$
and so:
:$\alpha - \dfrac 1 N > \alpha - \size \epsilon$
By hypothesis:
:$\forall N \in \N: \map f x > \alpha - \dfrac 1 N$
and therefore by the just previous:
:$\map f x > \alpha - \size \epsilon$
But we had $\map f x = \alpha - \size \epsilon$, a contradiction.
Therefore:
:$\map f x \ge \alpha $
that is:
$x \in \set {x: \map f x \ge \alpha}$
which was to be shown.
So $(1) \implies (2)$.
Similarly:
:$\ds \set {x: \map f x > \alpha} = \bigcup_{n \mathop = 1}^\infty \set {x: \map f x \ge \alpha + \dfrac 1 n}$
and so $(2) \implies (1)$.
This shows that $(1) \iff (2) \iff (3) \iff (4)$.
For the fifth statement, we have:
:$\set {x: \map f x = \alpha} = \set {x: \map f x \ge \alpha} \cap \set {x: \map f x \le \alpha}$
and so $(3) \land (4) \implies (5)$ for $\alpha \in \R$.
Since:
:$\ds \set {x: \map f x = +\infty} = \bigcap_{n \mathop = 1}^\infty \set {x: \map f x \ge n}$
we have that $(2) \implies (5)$ for $\alpha = +\infty$.
Similarly $(4) \implies (5)$ for $\alpha = - \infty$.
Category:Measure Theory
Measurable Mappings from Product Measurable Space

== Theorem ==
Let $\left({X, \Sigma}\right)$, $\left({X_1, \Sigma_1}\right)$ and $\left({X_2, \Sigma_2}\right)$ be measurable spaces.
Let $\Sigma_1 \otimes \Sigma_2$ be the product $\sigma$-algebra on $X_1 \times X_2$.
Let $f: X_1 \times X_2 \to X$ be a $\Sigma_1 \otimes \Sigma_2 \, / \, \Sigma$-measurable mapping.
Then:
:$\forall x_1 \in X_1: f \left({x_1, \cdot}\right): X_2 \to X$ is $\Sigma_2 \, / \, \Sigma$-measurable
:$\forall x_2 \in X_2: f \left({\cdot, x_2}\right): X_1 \to X$ is $\Sigma_1 \, / \, \Sigma$-measurable
== Proof ==
== Sources ==
* : $13.10 \ \text{(iii)}$
Category:Measure Theory
Measurable Mappings to Product Measurable Space

== Theorem ==
Let $\struct {X, \Sigma}$, $\struct {X_1, \Sigma_1}$ and $\struct {X_2, \Sigma_2}$ be measurable spaces.
Let $\Sigma_1 \otimes \Sigma_2$ be the product $\sigma$-algebra on $X_1 \times X_2$.
Let $\pr_1: X_1 \times X_2 \to X_1$ and $\pr_2: X_1 \times X_2 \to X_2$ be the first and second projections, respectively.
A mapping $f: X \to X_1 \times X_2$ is $\Sigma \, / \, \Sigma_1 \otimes \Sigma_2$-measurable :
:$\pr_i \circ f: X \to X_i$ is $\Sigma \, / \, \Sigma_i$-measurable, for $i = 1, 2$
== Proof ==
== Sources ==
* : $13.10 \ \text{(ii)}$
Category:Measure Theory
Measurable Sets form Algebra of Sets

== Theorem ==
Let $\mu^*$ be an outer measure on a set $X$.
Then the set of $\mu^*$-measurable sets is an algebra of sets.
== Proof ==
For a subset $S \subseteq X$, let $\relcomp X S$ denote the relative complement of $S$ in $X$.
We first prove the second property of an algebra of sets, as described on that page.
Let $S$ be $\mu^*$-measurable. For any subset $A \subseteq X$:
as desired.
Now we prove the first property.
Suppose that $S_1$ and $S_2$ are $\mu^*$-measurable sets.
Let $A$ be any subset of $X$. 
Since:
By subadditivity of an outer measure:
:$\map {\mu^*} {A \cap \paren {S_1 \cup S_2} } \le \map {\mu^*} {A \cap S_1} + \map {\mu^*} {\paren {A \setminus S_1} \cap S_2}$
Thus:
The result follows by the subadditivity of an outer measure.
Alternatively, one could use the equality
to prove the result directly without the use of subadditivity.
Category:Measurable Sets
Category:Algebras of Sets
Measurable Sets form Sigma-Algebra

== Theorem ==
Let $\mu^*$ be an outer measure on a set $X$.
Then the set $\map {\mathfrak M} {\mu^*}$ of all $\mu^*$-measurable subsets of $X$ is a $\sigma$-algebra.
== Proof ==
First, note that $\map {\mathfrak M} {\mu^*}$ is an algebra (of sets).
It remains to be shown that $\map {\mathfrak M} {\mu^*}$ is closed under countable union.
Because $\map {\mathfrak M} {\mu^*}$ is an algebra (of sets), the union of any two $\mu^*$-measurable sets is $\mu^*$-measurable.
Using mathematical induction, it directly follows that the finite union of $\mu^*$-measurable sets is $\mu^*$-measurable.
Let $\sequence {S_n}$ be a sequence of $\mu^*$-measurable subsets of $X$.
Define $\ds S = \bigcup_{n \mathop = 1}^\infty S_n$.
We wish to prove that $S$ is $\mu^*$-measurable.
For all $n \in \N$, the set $T_n = S_1 \cup S_2 \cup \cdots \cup S_n$ is $\mu^*$-measurable.
By Subset of Union, the sequence $\sequence {T_n}$ is increasing.
Also, $T_n \uparrow S$ (as $n \to \infty$) where $\uparrow$ denotes the limit of an increasing sequence of sets.
Let $A$ be any subset of $X$.
By Set Difference Union Intersection:
:$A = \paren {A \cap S} \cup \paren {A \setminus S}$
So by the subadditivity of $\mu^*$, it suffices to prove that $\map {\mu^*} A \ge \map {\mu^*} {A \cap S} + \map {\mu^*} {A \setminus S}$ for any subset $A \subseteq X$.
Then:
Letting $n \to \infty$, the result follows by Outer Measure of Limit of Increasing Sequence of Sets.
== Sources ==
* 
Category:Analysis
Category:Measurable Sets
Category:Sigma-Algebras
Measure-Preserving Transformation Preserves Conditional Entropy

== Theorem ==
Let $\struct {X, \BB, \mu}$ be a probability space.
Let $T: X \to X$ be a $\mu$-preserving transformation.
Let $\AA, \DD \subseteq \Sigma$ be finite sub-$\sigma$-algebras.
Then:
:$\map H {T^{-1} \AA \mid T^{-1} \DD} = \map H {\AA \mid \DD}$
where:
:$\map H {\cdot \mid \cdot}$ denotes the conditional entropy
:$T^{-1} \AA$ is the pre-image $\sigma$-algebra of $\AA$ on the domain of $T$
:$T^{-1} \DD$ is the pre-image $\sigma$-algebra of $\DD$ on the domain of $T$
=== Corollary ===
== Proof ==
By , we have:
:$\map \xi {T^{-1} \AA} = T^{-1} {\map \xi \AA}$
for each finite sub-$\sigma$-algebras $\AA \subseteq \Sigma$.
Thus it suffices to show that the entropy of finite partition satisfies:
:$\map H {T^{-1} \xi \mid T^{-1} \eta} = \map H {\xi \mid \eta}$
for all $\xi, \eta$ be finite partitions of $\Omega$.
Now:
Category:Probability Theory
Category:Ergodic Theory
Measure-Preserving Transformation Preserves Conditional Entropy/Corollary

== Corollary to Measure-Preserving Transformation Preserves Conditional Entropy ==
Let $\struct {X, \BB, \mu}$ be a probability space.
Let $T: X \to X$ be a $\mu$-preserving transformation.
Let $\AA \subseteq \Sigma$ be finite sub-$\sigma$-algebra.
Then:
:$\map H {T^{-1} \AA} = \map H \AA$
where:
:$\map H \cdot$ denotes the entropy
:$T^{-1} \AA$ is the pre-image $\sigma$-algebra of $\AA$ on the domain of $T$
== Proof ==
Let $\NN := \set {\O, \Omega}$ be the trivial $\sigma$-algebra.
Since:
:$T^{-1} \sqbrk \O = \O$
and:
:$T^{-1} \sqbrk \Omega = \Omega$
it follows from the definition of pre-image $\sigma$-algebra that:
:$T^{-1} \NN = \NN$
Therefore:
Category:Probability Theory
Category:Ergodic Theory
Measure Invariant on Generator is Invariant

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\theta: X \to X$ be an $\Sigma / \Sigma$-measurable mapping.
Suppose that $\Sigma$ is generated by $\GG \subseteq \powerset X$.
Also, let $\GG$ satisfy the following:
:$(1): \quad \forall G, H \in \GG: G \cap H \in \GG$
:$(2): \quad$ There exists an exhausting sequence $\sequence {G_n}_{n \mathop \in \N} \uparrow X$ in $\GG$ such that:
:::$\quad \forall n \in \N: \map \mu {G_n} < +\infty$
Suppose furthermore that, for all $G \in \GG$, $\mu$ satisfies:
:$(3): \quad \map \mu {\theta^{-1} \sqbrk G} = \map \mu G$
Then $\mu$ is a $\theta$-invariant measure.
== Proof ==
Consider the pushforward measure $\theta_* \mu$ on $\struct {X, \Sigma}$.
By definition, this makes equation $(3)$ come down to:
:$\theta_* \mu \restriction_\GG = \mu \restriction_\GG$
where $\restriction$ denotes restriction.
The suppositions $(1)$, $(2)$ and $(3)$ together constitute precisely the prerequisites to Uniqueness of Measures.
Hence $\theta_* \mu = \mu$, that is, $\mu$ is $\theta$-invariant.
== Sources ==
* : $\S 5$: Problem $9$
Category:Invariant Measures