nyuuzyou/wikis huggingface dataset:
Characteristic Function Determined by 1-Fiber

== Theorem ==
Let $A \subseteq S$.
Let $f:S \to \left\{{0, 1}\right\}$ be a mapping.
Denote by $\chi_A$ the characteristic function on $A$.
Then the following are equivalent:
:$(1):\quad f = \chi_A$
:$(2):\quad \forall s \in S: f \left({s}\right) = 1 \iff s \in A$
Using the notion of a fiber, $(2)$ may also be expressed as:
:$(2'):\quad f^{-1} \left({1}\right) = A$
== Proof ==
=== $(1)$ implies $(2)$ ===
Follows directly from the definition of characteristic function.
=== $(2)$ implies $(1)$ ===
Let $s \in S$.
Suppose that $s \in A$.
Then by assumption, $f \left({s}\right) = 1$.
Also, by definition of characteristic function, $\chi_A \left({s}\right) = 1$.
Next, suppose that $s \notin A$.
Then $f \left({s}\right) \ne 1$ by assumption.
As $f \left({s}\right) \in \left\{{0, 1}\right\}$, it follows that $f \left({s}\right) = 0$.
Again, by definition of characteristic function, also have $\chi_A \left({s}\right) = 0$.
Hence, for all $s \in S$, have $f \left({s}\right) = \chi_A \left({s}\right)$.
By Equality of Mappings, it follows that $f = \chi_A$.
Category:Characteristic Functions
Characteristic Function Measurable iff Set Measurable

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $E \subseteq X$.
Then the following are equivalent:
:$(1): \quad E \in \Sigma$; that is, $E$ is a $\Sigma$-measurable set
:$(2): \quad \chi_E: X \to \set {0, 1}$, the characteristic function of $E$, is $\Sigma$-measurable
== Proof ==
=== $(1)$ implies $(2)$ ===
Assume that $E \in \Sigma$.
It is clear that $x \notin \set {0, 1}$ implies $\map {\chi_E^{-1} } x = \O$.
Hence Preimage of Union under Mapping and Characteristic Function Determined by 1-Fiber yield, for any $\alpha \in \R$:
:$\set {x \in X: \map {\chi_E} x \ge \alpha} = \begin{cases} \O & \text{if $1 < \alpha$}\\
E & \text{if } 0 < \alpha \le 1 \\
X & \text{if } \alpha \le 0 \end{cases}$
Since $\Sigma$ is a $\sigma$-algebra, $X \in \Sigma$.
By assumption, also $E \in \Sigma$.
Lastly, Sigma-Algebra Contains Empty Set ensures $\O \in \Sigma$.
This establishes $(4)$ of Characterization of Measurable Functions.
Hence $\chi_E$ is $\Sigma$-measurable.
=== $(2)$ implies $(1)$ ===
Assume that $\chi_E$ is $\Sigma$-measurable.
Since for all $x \in X$, it holds that:
:$\map {\chi_E} x > 0 \iff x \in E$
it follows that:
:$E = \map {\chi^{-1} } {\openint 0 {+\infty} }$
and thus $E \in \Sigma$ as $\chi_E$ is $\Sigma$-measurable.
== Sources ==
* : $8.5 \ \text{(i)}$
Category:Characteristic Functions
Category:Measurable Sets
Category:Measurable Functions
Characteristic Function of Disjoint Union

== Theorem ==
Let $X$ be a set. 
Let $\sequence {D_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint subsets of $X$. 
Let: 
:$\ds D = \bigcup_{n \mathop = 1}^\infty D_n$
Then: 
:$\ds \chi_D = \sum_{n \mathop = 1}^\infty \chi_{D_n}$
where:
:$\chi_D$ is the characteristic function of $D$
:$\chi_{D_n}$ is the characteristic function of $D_n$.
=== Corollary ===
== Proof ==
We aim to show that: 
:$\ds \sum_{n \mathop = 1}^\infty \map {\chi_{D_n} } x = \begin{cases}1 & x \in D \\ 0 & x \in X \setminus D\end{cases}$
at which point we will have the demand from the definition of a characteristic function.
Let $x \in D$. 
Then: 
:$\ds x \in \bigcup_{n \mathop = 1}^\infty D_n$ 
From the definition of set union, we have: 
:$x \in D_n$ for some $n \in \N$. 
Since the sets $\sequence {D_n}_{n \mathop \in \N}$ are pairwise disjoint, we have: 
:$x \in D_n$ for precisely one $n \in \N$.
That is, form the definition of a characteristic function:
:$\map {\chi_{D_n} } x = 1$ for precisely one $n \in \N$. 
So that: 
:$\ds \sum_{n \mathop = 1}^\infty \map {\chi_{D_n} } x = 1$
as desired.
Let $x \in X \setminus D$.
Then: 
:$\ds x \in X \setminus \bigcup_{n \mathop = 1}^\infty D_n$ 
so from the set difference:
:$x \in X$ and $\ds x \not \in \bigcup_{n \mathop = 1}^\infty D_n$.
So, from the definition of set union:
:$x \not \in D_n$ for all $n \in \N$. 
That is, form the definition of a characteristic function:
:$\map {\chi_{D_n} } x = 0$ for all $n \in \N$. 
So that: 
:$\ds \sum_{n \mathop = 1}^\infty \map {\chi_{D_n} } x = 0$
as desired.
Category:Characteristic Functions
Category:Set Union
Category:Characteristic Function of Disjoint Union
Characteristic Function of Disjoint Union/Corollary

== Theorem ==
Let $X$ be a set. 
Let $\set {D_1, D_2, \ldots, D_N}$ be a set of pairwise disjoint subsets of $X$. 
Let: 
:$\ds D = \bigcup_{n \mathop = 1}^N D_n$
Then: 
:$\ds \chi_D = \sum_{n \mathop = 1}^N \chi_{D_n}$
where:
:$\chi_D$ is the characteristic function of $D$
:$\chi_{D_n}$ is the characteristic function of $D_n$.
== Proof ==
We can extend $\set {D_1, D_2, \ldots, D_N}$ to a sequence $\sequence {D_n}_{n \mathop \in \N}$ of subsets of $X$ by setting: 
:$D_i = \O$ for $i \ge N + 1$
Clearly, from Intersection with Empty Set, we have: 
:$D_i \cap D_j = \O$ for $i \ge N + 1$ and all $j$. 
So $\sequence {D_n}_{n \mathop \in \N}$ is a sequence of pairwise disjoint subsets of $X$ with: 
:$\ds D = \bigcup_{n \mathop = 1}^\infty D_n$
We therefore have, by Characteristic Function of Disjoint Union:
:$\ds \chi_D = \sum_{n \mathop = 1}^\infty \chi_{D_n}$
We can write:
Category:Characteristic Function of Disjoint Union
Characteristic Function of Intersection/Variant 1

== Theorem ==
Let $A, B \subseteq S$.
Let $\chi_{A \mathop \cap B}$ be the characteristic function of their intersection $A \cap B$.
Then:
:$\chi_{A \mathop \cap B} = \chi_A \chi_B$
== Proof ==
By Characteristic Function Determined by 1-Fiber, it suffices to show that:
:$\map {\chi_A} s \map {\chi_B} s = 1 \iff s \in A \cap B$
Now, both $\chi_A$ and $\chi_B$ are characteristic functions.
It follows that, for any $s \in S$:
:$\map {\chi_A} s \map {\chi_B} s = 1 \iff \map {\chi_A} s = \map {\chi_B} s = 1$
By definition of $\chi_A$ and $\chi_B$, this is equivalent to the statement that both $s \in A$ and $s \in B$.
That is, $s \in A \cap B$, by definition of set intersection.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $7 \ \text{(a)}$
* : $\S 2$: Problem $5 \ \text{(i)}$
Category:Characteristic Function of Intersection
Characteristic Function of Intersection/Variant 2

== Theorem ==
Let $A, B \subseteq S$.
Let $\chi_{A \mathop \cap B}$ be the characteristic function of their intersection $A \cap B$.
Then:
:$\chi_{A \mathop \cap B} = \min \set {\chi_A, \chi_B}$
== Proof ==
By Characteristic Function Determined by 1-Fiber, it suffices to show that:
:$\min \set {\map {\chi_A} s, \map {\chi_B} s} = 1 \iff s \in A \cap B$
By definition of characteristic function, we have:
:$\min \set {\map {\chi_A} s, \map {\chi_B} s} = 1$
:
:$\map {\chi_A} s = \map {\chi_B} s = 1$
because $\chi_A, \chi_B$ take values in $\set {0, 1}$.
By definition of $\chi_A$ and $\chi_B$, this is equivalent to the statement that both $s \in A$ and $s \in B$.
That is:
:$s \in A \cap B$
by definition of set intersection.
== Sources ==
* : $\S 2$: Problem $5 \ \text{(vi)}$
Category:Characteristic Function of Intersection
Characteristic Function of Limit Inferior of Sequence of Sets

== Theorem ==
Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of sets.
Let $E := \ds \liminf_{n \mathop \to \infty} E_n$ be the limit inferior of the $E_n$.
Then:
:$\ds \chi_E = \liminf_{n \mathop \to \infty} \chi_{E_n}$
where:
:$\chi$ denotes characteristic function
:$\ds \liminf_{n \mathop \to \infty} \chi_{E_n}$ is the pointwise limit inferior of the $\chi_{E_n}$.
== Proof ==
== Sources ==
* : $\S 9$: Problem $9 \ \text{(i)}$
Category:Characteristic Functions
Characteristic Function of Limit Superior of Sequence of Sets

== Theorem ==
Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence of sets.
Let $E := \ds \limsup_{n \mathop \to \infty} E_n$ be the limit superior of the $E_n$.
Then:
:$\ds \chi_E = \limsup_{n \mathop \to \infty} \chi_{E_n}$
where:
:$\chi$ denotes characteristic function
:$\ds \liminf_{n \mathop \to \infty} \chi_{E_n}$ is the pointwise limit superior of the $\chi_{E_n}$
== Proof ==
== Sources ==
* : $\S 9$: Problem $9 \ \text{(i)}$
Category:Characteristic Functions
Characteristic Function of Normal Distribution

== Theorem ==
The characteristic function of the normal distribution with mean $\mu$ and variance $\sigma^2$ is given by:
:$\map \phi t = e^{i t \mu - \frac 1 2 t^2 \sigma^2}$
=== Corollary ===
== Proof ==
=== ===
=== ===
By :
:$\map \phi t = c \dfrac 1 {\sqrt {2 \pi \sigma^2} } \ds \int_{x \mathop \in \R} e^{-\paren {\frac {x - k} {\sqrt 2 \sigma} }^2} \rd x$
Let $z = \paren {\dfrac {x - k} {\sqrt 2 \sigma} }$.
Then:
Category:Characteristic Function of Normal Distribution
Category:Characteristic Functions of Random Variables
Category:Normal Distribution
Characteristic Function of Normal Distribution/Corollary

== Theorem ==
The characteristic function of the standard normal distribution is:
:$\map \phi t = e^{-\frac 1 2 t^2}$
== Proof ==
Recall Characteristic Function of Normal Distribution:
The standard normal distribution is the normal distribution with $\mu = 0$ and $\sigma = 1$.
Hence:
== Sources ==
* 
* 
Category:Characteristic Function of Normal Distribution
Characteristic Function of Normal Distribution/Lemma 1

== Lemma for Characteristic Function of Normal Distribution ==
Let $\map \phi t$ denote the characteristic function of the normal distribution with mean $\mu$ and variance $\sigma^2$.
Let:
:$k = \mu + i t \sigma^2$
:$c = e^{\mu i t - \frac 1 2 t^2 \sigma^2}$
Then:
:$\map \phi t = c \dfrac 1 {\sqrt {2 \pi \sigma^2} } \ds \int_{x \mathop \in \R} e^{-\paren {\frac {x - k} {\sqrt 2 \sigma} }^2} \rd x$
== Proof ==
The characteristic function is defined as
Begin by verifying that:
:$i t x - \dfrac {\paren {x - \mu}^2} {2 \sigma^2} = -\dfrac {\paren {x - k}^2 + 2 \mu i t \sigma^2 - t^2 \sigma^4} {2 \sigma^2}$
We can then simplify the integral in $(1)$:
Category:Characteristic Function of Normal Distribution
Characteristic Function of Normal Distribution/Lemma 2

== Lemma for Characteristic Function of Normal Distribution ==
:$\ds \lim_{\alpha \mathop \to \infty} \int_{\frac{-\alpha - \mu} {\sqrt 2 \sigma} - i \frac {t \sigma} {\sqrt 2} }^{\frac {\alpha - \mu} {\sqrt 2 \sigma} - i \frac {t \sigma} {\sqrt 2} } e^{-z^2} \rd z = \sqrt {2 \pi \sigma^2}$
== Proof ==
Let $\Gamma \subset \C$ be the rectangular contour with corners:
Since $e^{-z^2}$ is holomorphic everywhere in the region bounded by $\Gamma$, the Cauchy Integral Theorem states that:
We now evaluate each linear contour integral in the limit as $\alpha$ goes to infinity:
=== Between $c_3$ and $c_4$ ===
=== Between $c_2$ and $c_3$ ===
=== Between $c_4$ and $c_1$ ===
By a similar argument as for between $c_2$ and $c_3$, the integral between $c_4$ and $c_1$ is also $0$.
=== Between $c_1$ and $c_2$ ===
and therefore:
Category:Characteristic Function of Normal Distribution
Characteristic Function of Null Set is A.E. Equal to Zero

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $N$ be a $\mu$-null set. 
Then: 
:$\chi_N = 0$ $\mu$-almost everywhere.
where $\chi_N$ is the characteristic function of $N$. 
=== Corollary ===
== Proof ==
Let $x \in X$ be such that: 
:$\map {\chi_N} x \ne 0$
Then. since $\map {\chi_N} x \in \set {0, 1}$: 
:$\map {\chi_N} x = 1$
which is equivalent to: 
:$x \in N$ 
from the definition of a characteristic function.
So:
:if $x \in X$ is such that $\map {\chi_N} x \ne 0$, then $x \in N$.
Since $N$ is a $\mu$-null set, we have: 
:$\chi_N = 0$ $\mu$-almost everywhere.
Category:Characteristic Functions
Category:Measure Theory
Category:Characteristic Function of Null Set is A.E. Equal to Zero
Characteristic Function of Null Set is A.E. Equal to Zero/Corollary

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $N$ be a $\mu$-null set. 
Then: 
:$\chi_{X \setminus N} = 1$ $\mu$-almost everywhere.
where $\chi_{X \setminus N}$ is the characteristic function of $X \setminus N$. 
== Proof ==
From Characteristic Function of Set Difference, we have: 
:$\chi_{X \setminus N} = \chi_X - \chi_{X \cap N}$
From Intersection with Subset is Subset, we therefore have: 
:$\map {\chi_{X \setminus N} } x = 1 - \map {\chi_N} x$
for each $x \in X$.
From Characteristic Function of Null Set is A.E. Equal to Zero, we have: 
:$\chi_N = 0$ $\mu$-almost everywhere. 
So we have, from Pointwise Addition preserves A.E. Equality:
:$1 - \chi_N = 1$ $\mu$-almost everywhere.
So:
:$\chi_{X \setminus N} = 1$ $\mu$-almost everywhere.
Category:Characteristic Function of Null Set is A.E. Equal to Zero
Characteristic Function of Preimage

== Theorem ==
Let $X$ and $Y$ be sets.
Let $f : X \to Y$ be a function.
Let $A \subseteq Y$.
Then: 
:$\chi_{f^{-1} \sqbrk A} = \chi_A \circ f$
where $\chi_{f^{-1} \sqbrk A}$ denotes the characteristic function of $f^{-1} \sqbrk A$.
== Proof ==
We show that if $x \in f^{-1} \sqbrk A$, then: 
:$\map {\paren {\chi_A \circ f} } x = 1$
and if $x \in X \setminus f^{-1} \sqbrk A$, then: 
:$\map {\paren {\chi_A \circ f} } x = 0$
Let $x \in f^{-1} \sqbrk A$.
Then $\map f x \in A$, and so: 
:$\map {\paren {\chi_A \circ f} } x = \map {\chi_A} {\map f x} = 1$
Now let $x \in X \setminus f^{-1} \sqbrk A$. 
Then $\map f x \in Y \setminus A$, and so: 
:$\map {\paren {\chi_A \circ f} } x = \map {\chi_A} {\map f x} = 0$
Category:Characteristic Functions
Characteristic Function of Random Variable is Well-Defined

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be a real-valued random variable on $\struct {\Omega, \Sigma, \Pr}$.
The characteristic function $\phi: \R \to \C$ of $X$ is well-defined.
== Proof ==
Let $t \in \R$.
Recall:
Thus we need to show that the last integral exists.
By Modulus of Exponential of Imaginary Number is One:
:$\cmod {e^{i t X } } = 1$
since $t \map X \omega \in \R$ for all $\omega \in \Omega$.
In particular:
:$\ds \int \cmod {e^{i t X} } \rd \Pr = \int 1 \rd \Pr = 1$
Thus, in view of Characterization of Integrable Functions:
:$\ds \int e^{i t X} \rd \Pr$
exists.
Category:Probability Theory
Characteristic Function of Set Difference

== Theorem ==
Let $A, B \subseteq S$.
Then:
:$\chi_{A \mathop \setminus B} = \chi_A - \chi_{A \cap B}$ 
where:
:$A \setminus B$ denotes set difference
:$\chi$ denotes characteristic function.
== Proof ==
Suppose that $\map {\chi_{A \mathop \setminus B} } s = 1$.
Then by definition of characteristic function:
:$s \in A \setminus B$
That is, by definition of set difference, $s \in A$ and $s \notin B$; in particular, $s \notin A \cap B$.
Hence:
:$\map {\chi_A} s = 1$ and $\map {\chi_{A \cap B} } s = 0$
Conclude that:
:$\map {\chi_{A \mathop \setminus B} } s = 1 \implies \map {\chi_A} s - \map {\chi_{A \cap B} } s = 1$
Now conversely suppose $\map {\chi_A} s - \map {\chi_{A \cap B} } s = 1$.
Considering that $\map {\chi_{A \cap B} } s \ge 0$, it must be that $\map {\chi_A} s = 1$ and $\map {\chi_{A \cap B} } s = 0$.
Hence, $s \in A$, and $s \notin A \cap B$.
By definition of set intersection, this means $s \notin B$.
Hence $s \in A \setminus B$ by definition of set difference.
Thus:
:$\map {\chi_{A \mathop \setminus B} } s = 1$
Hence, it has been established that:
:$\map {\chi_{A \mathop \setminus B} } s = 1 \iff \map {\chi_A} s - \map {\chi_{A \cap B} } s = 1$
which, by application of Characteristic Function Determined by 1-Fiber, gives the result.
== Sources ==
* : $\S 2$: Problem $5 \ \text{(iii)}$
Category:Characteristic Functions
Characteristic Function of Square-Free Integers is Multiplicative

== Theorem ==
Let $S \subseteq \Z$ be the set of positive integers defined as:
:$S := \set {n \in \Z: \forall k \in \Z_{>1}: k^2 \nmid n}$
That is, let $S$ be the set of all square-free positive integers.
Let $\chi_S: \N \to \Z$ denote the characteristic function of $S$:
:$\forall n \in \Z: \map {\chi_S} n = \sqbrk {n \in S}$
where $\sqbrk {n \in S}$ is Iverson's convention.
Then $\chi_S$ is multiplicative.
== Proof ==
Let $r, s \in \Z$ such that $r \perp s$.
=== Case 1: Either factor is not square-free ===
Let either $r \notin S$ or $s \notin S$ or both.
Then either:
:$\exists k \in \Z_{>1}: k^2 \divides r$
or:
:$\exists k \in \Z_{>1}: k^2 \divides s$
Thus either:
:$\map {\chi_S} r = 0$
or:
:$\map {\chi_S} s = 0$
and so:
:$\map {\chi_S} r \, \map {\chi_S} s = 0$
But then:
:$\exists k \in \Z_{>1}: k^2 \divides r s$
and so:
:$\map {\chi_S} {r s} = 0$
demonstrating that:
:$\map {\chi_S} r \, \map {\chi_S} s = \map {\chi_S} {r s}$
=== Case 1: Both factors are square-free ===
Let $r \in S$ and $s \in S$.
Thus:
:$\nexists k \in \Z_{>1}: k^2 \divides r$
and:
:$\nexists k \in \Z_{>1}: k^2 \divides s$
Hence:
:$\map {\chi_S} r = 1$
and:
:$\map {\chi_S} s = 1$
and so:
:$\map {\chi_S} r \, \map {\chi_S} s = 1$
Because $r \perp s$:
:$\nexists k \in \Z_{>1}: k \divides r, k \divides s$
Hence there can be no $k \in \Z_{>1}$ whose multiplicity in $r s$ is greater than $1$.
Thus:
:$\nexists k \in \Z_{>1}: k^2 \divides {r s}$
and so:
:$\map {\chi_S} {r s} = 1$
once more demonstrating that:
:$\map {\chi_S} r \, \map {\chi_S} s = \map {\chi_S} {r s}$
In both cases:
:$\map {\chi_S} r \, \map {\chi_S} s = \map {\chi_S} {r s}$
Hence the result, by definition of multiplicative function.
== Also see ==
* Möbius Function is Multiplicative, a similar result.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $29 \ \text{(b)}$
Category:Multiplicative Functions
Category:Number Theory
Category:Square-Free Integers
Characteristic Function of Subset

== Theorem ==
Let $A \subseteq B \subseteq S$.
Then:
:$\forall s \in S: \chi_A \left({s}\right) \le \chi_B \left({s}\right)$
where $\chi$ denotes characteristic function.
== Proof ==
Both $\chi_A$ and $\chi_B$ take values in $\left\{{0, 1}\right\}$ as they are characteristic functions.
So if $\chi_A \left({s}\right) = 0$, then the statement of the theorem is automatically satisfied (since both $0 \le 0$ and $0 \le 1$).
Now assume $\chi_A \left({s}\right) = 1$.
By definition of $\chi_A$, this happens $s \in A$.
But as $A \subseteq B$, this implies $s \in B$ as well.
Hence, $\chi_B \left({s}\right) = 1$, and so $\chi_A \left({s}\right) \le \chi_B \left({s}\right)$.
The theorem follows from Proof by Cases.
Category:Characteristic Functions
Characteristic Function of Symmetric Difference

== Theorem ==
Let $A, B \subseteq S$.
Then:
:$\chi_{A \symdif B} = \chi_A + \chi_B - 2 \chi_{A \cap B}$ 
where:
:$\chi$ denotes characteristic function
:$\symdif$ denotes symmetric difference.
== Proof ==
By definition of symmetric difference:
:$A \symdif B = \paren {A \cup B} \setminus \paren {A \cap B}$
Thus:
:$\chi_{A \symdif B} = \chi_{A \mathop \cup B} - \chi_{\paren {A \mathop \cup B} \mathop \cap \paren {A \mathop \cap B} }$
by Characteristic Function of Set Difference.
But by Intersection is Subset of Union and Intersection with Subset is Subset:
:$\paren {A \cup B} \cap \paren {A \cap B} = A \cap B$
Hence it follows that:
:$\chi_{A \symdif B} = \chi_{A \mathop \cup B} - \chi_{A \mathop \cap B}$
which by Characteristic Function of Union: Variant 2 becomes:
:$\chi_{A \symdif B} = \chi_A + \chi_B - 2 \chi_{A \mathop \cap B}$
as desired.
== Sources ==
* : $\S 2$: Problem $6 \ \text{(i)}$
Category:Characteristic Functions
Characteristic Function of Union/Variant 1

== Theorem ==
Let $A, B \subseteq S$.
Let $\chi_{A \mathop \cup B}$ be the characteristic function of their union $A \cup B$.
Then:
:$\chi_{A \mathop \cup B} = \min \set {\chi_A + \chi_B, 1}$
where $\min$ denotes the min operation.
== Proof ==
By Characteristic Function Determined by 1-Fiber, it suffices to show:
:$\min \set {\map {\chi_A} s + \map {\chi_B} s, 1} = 1 \iff s \in A \cup B$
By the nature of the min operation, this amounts to showing that:
:$\map {\chi_A} s + \map {\chi_B} s \ge 1 \iff s \in A \cup B$
As $\chi_A, \chi_B$ are characteristic functions, the amounts to:
:$s \in A \lor s \in B$
which is precisely the definition of $s \in A \cup B$.
== Sources ==
* : $\S 2$: Problem $5 \ \text{(ii)}$
Category:Characteristic Function of Union
Characteristic Function of Union/Variant 2

== Theorem ==
Let $A, B \subseteq S$.
Let $\chi_{A \mathop \cup B}$ be the characteristic function of their union $A \cup B$.
Then:
:$\chi_{A \mathop \cup B} = \chi_A + \chi_B - \chi_{A \mathop \cap B}$
== Proof ==
From Subset of Union:
:$A, B \subseteq A \cup B$
From Intersection is Subset of Union:
:$A \cap B \subseteq A \cup B$
Thus from Characteristic Function of Subset:
:$\map {\chi_{A \mathop \cup B} } s = 0 \implies \map {\chi_A} s = \map {\chi_B} s = \map {\chi_{A \mathop \cap B} } s = 0$
Now suppose that $\map {\chi_A} s + \map {\chi_B} s - \map {\chi_{A \mathop \cap B} } s = 0$.
That is:
:$\map {\chi_A} s + \map {\chi_B} s = \map {\chi_{A \mathop \cap B} } s$
Suppose the latter equals $1$, that is $s \in A \cap B$.
Then both $s \in A$ and $s \in B$, so by definition of characteristic function:
:$\map {\chi_A} s + \map {\chi_B} s = 1 + 1 = 2$
Since $2 \ne 1$, it follows that $\map {\chi_{A \mathop \cap B} } s \ne 1$, that is it equals $0$.
Thence, it follows that:
:$\map {\chi_A} s + \map {\chi_B} s = 0$.
This only happens when $\map {\chi_A} s = \map {\chi_B} s = 0$.
Thus, $s \notin A$ and $s \notin B$, so $s \notin A \cup B$.
It finally follows that $\map {\chi_{A \mathop \cup B} } s = 0$.
It is now established that:
:$\map {\chi_{A \mathop \cup B} } s = 0 \iff \map {\chi_A} s + \map {\chi_B} s - \map {\chi_{A \mathop \cap B} } s = 0$
and from Characteristic Function Determined by 0-Fiber, the result follows.
== Sources ==
* : Chapter $1$: Theory of Sets: $\S 6$: Functions: Exercise $7 \ \text{(b)}$
* : $\S 2$: Problem $5 \ \text{(iv)}$
Category:Characteristic Function of Union
Characteristic Function of Union/Variant 3

== Theorem ==
Let $A, B \subseteq S$.
Let $\chi_{A \mathop \cup B}$ be the characteristic function of their union $A \cup B$.
Then:
:$\chi_{A \mathop \cup B} = \max \set {\chi_A, \chi_B}$
where $\max$ denotes the max operation.
== Proof ==
Suppose $\map {\chi_{A \mathop \cup B} } s = 0$.
Then $s \notin A \cup B$, so $s \notin A$ and $s \notin B$.
Hence:
:$\map {\chi_A} s = \map {\chi_B} s = 0$
and by definition of max operation:
:$\max \set {\map {\chi_A} s, \map {\chi_B} s} = 0$
Conversely, suppose:
:$\max \set {\map {\chi_A} s, \map {\chi_B} s} = 0$
Then it follows that:
:$\map {\chi_A} s = \map {\chi_B} s = 0$
because characteristic functions are $0$ or $1$.
Hence $s \notin A$ and $s \notin B$, so $s \notin A \cup B$.
That is:
:$\map {\chi_{A \mathop \cup B} } s = 0$
Above considerations give:
:$\map {\chi_{A \mathop \cup B} } s = 0 \iff \max \set {\map {\chi_A} s, \map {\chi_B} s} = 0$
and applying Characteristic Function Determined by 0-Fiber, the result follows.
== Sources ==
* : $\S 2$: Problem $5 \ \text{(v)}$
Category:Characteristic Function of Union
Characteristic Function of Universe

== Theorem ==
Let $S$ be a set.
Let $\chi_S: S \to \set {0, 1}$ be its characteristic function (in itself).
Then:
:$\chi_S = f_1$
where $f_1: S \to \set {0, 1}$ is the constant mapping with value $1$.
== Proof ==
From Characteristic Function Determined by 1-Fiber, $\chi_S$ is the mapping determined by:
:$\forall s \in S: \map {\chi_S} s = 1 \iff s \in S$
Thus:
:$\forall s \in S: \map {\chi_S} s = 1$
By definition of constant mapping:
:$\chi_S = f_1$
Category:Characteristic Functions
Characteristic Function on Event is Discrete Random Variable

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $E \in \Sigma$ be any event of $\struct {\Omega, \Sigma, \Pr}$.
Let $\chi_E: \Omega \to \set {0, 1}$ be the characteristic function of $E$.
Then $\chi_E$ is a discrete random variable on $\struct {\Omega, \Sigma, \Pr}$.
== Proof ==
By definition of characteristic function:
:$\forall \omega \in \Omega: \chi_E = \begin{cases}
1 & : \omega \in E \\
0 & : \omega \notin E \\
\end{cases}$
Then clearly:
:$\forall x \in \R: \map { {\chi_E}^{-1} } x = \begin{cases}
E & : x = 1 \\
\Omega \setminus E & : x = 0 \\
\O & : x \notin \set {0, 1}
\end{cases}$
So whatever the value of $x \in \R$, its preimage is in $\Sigma$.
Hence the result.
== Sources ==
* : $\S 2.1$: Probability mass functions: Exercise $3$
Category:Probability Theory