nyuuzyou/wikis huggingface dataset:
Outer Jordan Content of Right Triangle

== Theorem ==
Let $T \subseteq \R^2$ be defined as:
:$T = \set {\tuple {x, y} \in \R^2 : x \ge 0 \land y \ge 0 \land x + y \le 1}$
Then:
:$\map {m^*} T = \dfrac 1 2$
== Proof ==
Let $\epsilon > 0$ be arbitrary.
By the Axiom of Archimedes, let $n \in \N$ such that:
:$n > 2 \epsilon$
Define $C \subseteq \powerset {\R^2}$ as:
:$C = \set {\closedint {\dfrac p n} {\dfrac {p + 1} n} \times \closedint {\dfrac q n} {\dfrac {q + 1} n} : p, q \in \set {0, 1, \dotsc, n - 1} \land p + q < n}$
By construction, $C$ is a finite set of closed $2$-rectangles.
Let $\tuple {x, y} \in T$ be arbitrary.
Let $p \in \Z_{\ge 0}$ be the smallest non-negative integer such that:
:$\dfrac {p + 1} n \ge x$
Because:
:$\dfrac {\paren {n - 1} + 1} n = 1 \ge x$
we have:
:$p \le n - 1$
If $\dfrac {p'} n < x$ for any $p' \in \Z_{\ge 0}$, then:
:$\dfrac p n < x$
by minimality.
Otherwise, $p = 0$ as the smallest non-negative integer overall, so:
:$\dfrac p n = 0 \le x$
In either case:
:$\dfrac p n \le x$
In like manner, let $q \in \Z_{\ge 0}$ be the smallest non-negative integer such that:
:$\dfrac {q + 1} n \ge y$
By the same argument as before:
:$\dfrac q n \le y$
We also have:
Therefore:
:$\closedint {\dfrac p n} {\dfrac {p + 1} n} \times \closedint {\dfrac q n} {\dfrac {q + 1} n} \in C$
Since, by the previous inequalities:
:$\dfrac p n \le x \le \dfrac {p + 1} n$
:$\dfrac q n \le y \le \dfrac {q + 1} n$
it follows that:
:$\tuple {x, y} \in \closedint {\dfrac p n} {\dfrac {p + 1} n} \times \closedint {\dfrac q n} {\dfrac {q + 1} n}$
As $\tuple {x, y} \in T$ was arbitrary, we have that $C$ is a covering of $T$.
Outer Measure of Limit of Increasing Sequence of Sets

== Theorem ==
Let $\mu^*$ be an outer measure on a set $X$.
Let $\sequence {S_n}$ be an increasing sequence of $\mu^*$-measurable sets, and let $S_n \uparrow S$ (as $n \to \infty$).
Then for any subset $A \subseteq X$:
:$\ds \map {\mu^*} {A \cap S} = \lim_{n \mathop \to \infty} \map {\mu^*} {A \cap S_n}$
== Proof ==
Category:Outer Measures
Category:Outer Measure of Limit of Increasing Sequence of Sets
Outer Measure of Limit of Increasing Sequence of Sets/Proof 1

== Theorem ==
== Proof ==
By the monotonicity of $\mu^*$, it suffices to prove that:
:$\ds \map {\mu^*} {A \cap S} \le \lim_{n \mathop \to \infty} \map {\mu^*} {A \cap S_n}$
Assume that $\map {\mu^*} {A \cap S_n}$ is finite for all $n \in \N$, otherwise the statement is trivial by the monotonicity of $\mu^*$.
Let $S_0 = \O$.
Then $x \in S$ there exists an integer $n \ge 0$ such that $x \in S_{n + 1}$. 
Taking the least possible $n$, it follows that $x \notin S_n$, and so:
:$x \in S_{n + 1} \setminus S_n$
Therefore:
:$\ds S = \bigcup_{n \mathop = 0}^\infty \paren {S_{n + 1} \setminus S_n}$
From Intersection Distributes over Union:
:$\ds A \cap S = A \cap \bigcup_{n \mathop = 0}^\infty \paren {S_{n + 1} \setminus S_n} = \bigcup_{n \mathop = 0}^\infty \paren {A \cap \paren {S_{n + 1} \setminus S_n} }$
Therefore:
Category:Outer Measure of Limit of Increasing Sequence of Sets
Outlier/Examples/Arbitrary Example 1

== Example of Outlier ==
Consider the set of observations:
:$\set {7, 9, 3, 5, 4, 202}$
The observation $202$ is an outlier.
It can be speculated that $202$ may have been a typo for $2, 2$, for example.
== Sources ==
* 
* 
Category:Examples of Outliers
Overflow Theorem

== Theorem ==
Let $F$ be a set of first-order formulas which has finite models of arbitrarily large size.
Then $F$ has an infinite model.
=== Corollary ===
== Proof ==
For each $n$, let $\mathbf A_n$ be the formula:
:$\exists x_1 \exists x_2 \ldots \exists x_n: \paren {x_1 \ne x_2 \land x_1 \ne x_3 \land \ldots \land x_{n - 1} \ne x_n}$
Then $\mathbf A_i$ is true in a structure $\AA$ $\AA$ has at least $n$ elements.
Take:
:$\ds \Gamma := A \cup \bigcup_{i \mathop = 1}^\infty A_i$
Since $F$ has models of arbitrarily large size, every finite subset of $\Gamma$ is satisfiable.
From the Compactness Theorem, $\Gamma$ is satisfiable in some model $\MM$.
But since $\MM \models A_i$ for each $i$, $\MM$ must be infinite.
So $A$ has an infinite model.
== Also see ==
* Upward Löwenheim-Skolem Theorem
== Sources ==
* : $\mathrm{II}.7$ First-Order Logic Semantics: Theorem $\mathrm{II.7.17}$
Category:Mathematical Logic
Category:Named Theorems
P-Norm is Norm/Complex Numbers

== Theorem ==
The $p$-norm on the complex numbers is a norm.
== Proof ==
Let $K \in \C^d$, where $d \in \N_{>0}$.
=== ===
Suppose $\sequence {x_n}_{n \mathop \in \set {1, 2, \ldots, d} } \in K$.
By definition of $p$-norm:
:$\ds \norm {\mathbf x}_p = \paren {\sum_{n \mathop = 0}^d \size {x_n}^p}^{1 / p}$
The complex modulus of $x_n$ is real and non-negative.
We have the results:
:Sum of Non-Negative Reals is Non-Negative
:Power of Positive Real Number is Positive
:Zero Raised to Positive Power is Zero
Hence:
:$\norm {\mathbf x}_p \ge 0$
Suppose that $\norm {\mathbf x}_p = 0$.
Then:
Thus is satisfied.
=== ===
Suppose that $\lambda \in K$.
Thus is satisfied.
=== ===
If $\mathbf x = \sequence 0$ and $\mathbf y = \sequence 0$, then by we have equality.
If $\mathbf x + \mathbf y = \sequence 0$ and both $\bf x$ and $\bf y$ nonvanishing, then by we get a strict inequality.
If $\mathbf x + \mathbf y \ne \sequence 0$, then consider p-norm raised to the power of $p$:
Thus is satisfied.
All norm axioms are seen to be satisfied.
Hence the result.
== Sources ==
* : $\S 1.2$: Normed and Banach spaces. Normed Spaces
Category:P-Norms
Category:Examples of Norms
P-Norm is Norm/P-Sequence Space

== Theorem ==
The $p$-norm on the $p$-sequence space is a vector space norm.
== Proof ==
=== ===
Let $x \in \ell^p$ with $1 \le p < \infty$.
By definition of $p$-norm:
:$\ds \norm {\mathbf x}_p = \paren {\sum_{n \mathop = 0}^\infty \size {x_n}^p}^{1/p}$
The complex modulus of $x_n$ is real and non-negative.
We have the results:
:Sum of Non-Negative Reals is Non-Negative
:Power of Positive Real Number is Positive
:Zero Raised to Positive Power is Zero
Hence, $\norm {\mathbf x}_p \ge 0$.
Suppose $\norm {\mathbf x}_p = 0$.
Then:
Thus is satisfied.
=== ===
Suppose $\alpha \in \set {\R, \C}$.
Thus is satisfied.
=== ===
From P-Norm is Norm/Complex Numbers we have that:
:$\ds \paren {\sum_{n \mathop = 0}^d \size {x_n + y_n}^p}^{\frac 1 p} \le \paren {\sum_{n \mathop = 0}^d \size {x_n}^p }^{\frac 1 p} + \paren {\sum_{n \mathop = 0}^d \size {y_n}^p }^{\frac 1 p}$
$\map f z = z^{\frac 1 p}$ is a continuous function for $z \ge 0$ and $p > 0$.
For $\mathbf x \in \ell^p$, changing $d$ is equivalent to changing $z$ in the interval $0 \le z < \infty$. 
Take the composite limit $d \to \infty$.
Then:
:$\norm {\mathbf x + \mathbf y}_p \le \norm {\mathbf x}_p + \norm {\mathbf y}_p$
Thus is satisfied.
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:P-Norms
P-Norm of Real Sequence is Strictly Decreasing Function of P

== Theorem ==
Let $p \ge 1$ be a real number.
Let ${\ell^p}_\R$ denote the real $p$-sequence space.
Let $\mathbf x = \sequence {x_n} \in {\ell^p}_\R$.
Suppose $\mathbf x$ is not a sequence of zero elements.
Let $\norm {\mathbf x}_p$ denote the $p$-norm of $\mathbf x$ where $p \ge 1$.
Then the mapping $p \to \norm {\mathbf x}_p$ is strictly decreasing $p$.
== Proof ==
Equality holds only for a sequence of zero elements.
Suppose $\mathbf x$ is not a sequence of zero elements.
Then:
By derivative of $p$-norm $p$:
By $\paren \star$, the term in parenthesis is negative.
By $p$-Norm is Norm:
:$\norm {\mathbf x}_p > 0$ for $\mathbf x \ne \sequence 0$.
Hence:
:$\forall p \ge 1 : \forall \mathbf x \ne \sequence 0 : \dfrac \d {\d p} \norm {\mathbf x}_p < 0$
Category:P-Norms
P-Product Metric Induces Product Topology

== Theorem ==
Let $M_A = \struct{A, d_A}$ and $M_B = \struct{B, d_B}$ be metric spaces.
Let $\tau_A$ and $\tau_B$ be the topologies on $A$ and $B$ induced by $d_A$ and $d_B$, respectively.
Let $p \ge 1$ be an extended real number.
Let $M = \struct{A \times B, d}$ be the $p$-product of $M_A$ and $M_B$.
We have that $M$ is a metric space.
Let $\tau$ be the topology on $A \times B$ induced by $d$.
Then $\struct{A \times B, \tau}$ is the product space of $\struct{A, \tau_A}$ and $\struct{B, \tau_B}$.
== Proof ==
By $p$-Product Metrics are Lipschitz Equivalent and Lipschitz Equivalent Metrics are Topologically Equivalent, it suffices to consider the case $p = \infty$.
Let $\struct{A \times B, \tau'}$ be the product space of $\struct{A, \tau_A}$ and $\struct{B, \tau_B}$.
By the definition of $d$, it follows that an open ball in $M$ is the (Cartesian) product of an open ball in $M_A$ and an open ball in $M_B$.
Since an open ball is an open set, it follows from Equivalence of Definitions of Analytic Basis that $\tau \subseteq \tau'$.
Let $W \in \tau'$, $\tuple {x, y} \in W$.
By the definition of $\tau'$, it follows from Equivalence of Definitions of Analytic Basis that there exists a $U$ open in $M_A$ and a $V$ open in $M_B$ such that $\tuple {x, y} \in U \times V \subseteq W$.
By the definition of an open set:
:There exists a strictly positive real number $\alpha \in \R_{>0}$ such that the open $\alpha$-ball of $x$ in $M_A$ is contained in $U$
:There exists a strictly positive real number $\beta \in \R_{>0}$ such that the open $\beta$-ball of $y$ in $M_B$ is contained in $V$
Let $\epsilon = \min \set {\alpha, \beta}$.
Then $\epsilon \in \R_{>0}$, and the open $\epsilon$-ball of $\tuple {x, y}$ in $M$ is contained in $U \times V$.
Hence, $W$ is open in $M$.
That is, $\tau' \subseteq \tau$.
By definition of set equality:
:$\tau = \tau'$
as desired.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 14$
Category:Metric Spaces
Category:P-Product Metrics
P-Product Metric is Metric

== Theorem ==
Let $M_{1'} = \struct {A_{1'}, d_{1'} }, M_{2'} = \struct {A_{2'}, d_{2'} }, \ldots, M_{n'} = \struct {A_{n'}, d_{n'} }$ be metric spaces.
Let $\ds \AA = \prod_{i \mathop = 1}^n A_{i'}$ be the cartesian product of $A_{1'}, A_{2'}, \ldots, A_{n'}$.
Let $p \in \R_{\ge 1}$.
Let $d_p: \AA \times \AA \to \R$ be the $p$-product metric on $\AA$:
:$\ds \map {d_p} {x, y} := \paren {\sum_{i \mathop = 1}^n \paren {\map {d_{i'} } {x_i, y_i} }^p}^{\frac 1 p}$
where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$.
Then $d_p$ is a metric.
== Proof ==
=== Proof of ===
So holds for $d_p$.
=== Proof of ===
Let:
:$(1): \quad z = \tuple {z_1, z_2, \ldots, z_n}$
:$(2): \quad$ all summations be over $i = 1, 2, \ldots, n$
:$(3): \quad \map {d_{i'} } {x_i, y_i} = r_i$
:$(4): \quad \map {d_{i'} } {y_i, z_i} = s_i$.
Thus we need to show that:
:$\ds \paren {\sum \paren {\map {d_{i'} } {x_i, y_i} }^p}^{\frac 1 p} + \paren {\sum \paren {\map {d_{i'} } {y_i, z_i} }^p}^{\frac 1 p} \ge \paren {\sum \paren {\map {d_{i'} } {x_i, z_i} }^p}^{\frac 1 p}$
We have:
So holds for $d_p$.
=== Proof of ===
So holds for $d_p$.
=== Proof of ===
So holds for $d_p$.
== Also see ==
* Taxicab Metric is Metric
* Chebyshev Distance is Metric
* Chebyshev Distance is Limit of P-Product Metric, that is: $\ds \map {d_\infty} {x, y} = \lim_{p \mathop \to \infty} \map {d_p} {x, y}$
Category:P-Product Metrics
P-Product Metric on Real Vector Space is Metric

== Theorem ==
Let $\R^n$ be an $n$-dimensional real vector space.
Let $p \in \R_{\ge 1}$.
Let $d_p: \R^n \times \R^n \to \R$ be the $p$-product metric on $\R^n$:
:$\ds \map {d_p} {x, y} := \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^p}^{\frac 1 p}$
where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \R^n$.
Then $d_p$ is a metric.
== Proof 1 ==
== Proof 2 ==
== Comment on notation ==
It can be shown that:
:$\ds \map {d_\infty} {x, y} = \lim_{p \mathop \to \infty} \map {d_p} {x, y}$
That is:
:$\ds \lim_{p \mathop \to \infty} \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^p}^{\frac 1 p} = \max_{i \mathop = 1}^n \set {\size {x_i - y_i} }$
Hence the notation.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.18$
Category:P-Product Metrics
Category:P-Product Metric on Real Vector Space is Metric
P-Product Metric on Real Vector Space is Metric/Proof 1

== Theorem ==
== Proof ==
=== Proof of ===
So holds for $d_p$.
=== Proof of ===
It is required to be shown:
:$\map {d_p} {x, y} + \map {d_p} {y, z} \ge \map {d_p} {x, z}$
for all $x, y, z \in \R^n$.
Let:
:$(1): \quad z = \tuple {z_1, z_2, \ldots, z_n}$
:$(2): \quad$ all summations be over $i = 1, 2, \ldots, n$
:$(3): \quad \size {x_i - y_i} = r_i$
:$(4): \quad \size {y_i - z_i} = s_i$.
Thus we need to show that:
:$\ds \paren {\sum \size {x_i - y_i}^p}^{\frac 1 p} + \paren {\sum \size {y_i - z_i}^p}^{\frac 1 p} \ge \paren {\sum \size {x_i - z_i}^p}^{\frac 1 p}$
We have:
So holds for $d_p$.
=== Proof of ===
So holds for $d_p$.
=== Proof of ===
So holds for $d_p$.
Category:P-Product Metric on Real Vector Space is Metric
P-Product Metrics are Lipschitz Equivalent

== Theorem ==
Let $M_A = \left({A, d_A}\right)$ and $M_B = \left({B, d_B}\right)$ be metric spaces.
Let $\tau_A$ and $\tau_B$ be the topologies on $A$ and $B$ induced by $d_A$ and $d_B$, respectively.
For all extended real numbers $p \ge 1$, let $M_p = \left({A \times B, d_p}\right)$ be the $p$-product of $M_A$ and $M_B$.
Then all the metrics $\left\{{d_p: p \ge 1}\right\}$ are Lipschitz equivalent.
== Proof ==
From $p$-Product Metric is Metric, $M_p$ is a metric space.
For all real numbers $p \ge 1$, it follows from the definition of $d_p$ that:
:$\forall x, y \in A \times B: d_{\infty} \left({x, y}\right) \le d_p \left({x, y}\right) \le 2^{1/p} d_{\infty} \left({x, y}\right)$
Hence, $d_p$ and $d_{\infty}$ are Lipschitz equivalent.
The result follows from Lipschitz Equivalence is Equivalence Relation.
Category:Lipschitz Equivalence
Category:P-Product Metrics
P-Product Metrics on Real Vector Space are Topologically Equivalent

== Theorem ==
For $n \in \N$, let $\R^n$ be an Euclidean space.
Let $p \in \R_{\ge 1}$.
Let $d_p$ be the $p$-product metric on $\R^n$.
Let $d_\infty$ be the Chebyshev distance on $\R^n$.
Then $d_p$ and $d_\infty$ are topologically equivalent.
== Proof ==
Let $r, t \in \R_{\ge 1}$.
, assume that $r \le t$.
For all $x, y \in \R^n$, we are going to show that:
:$\map {d_r} {x, y} \ge \map {d_\infty} {x, y} \ge n^{-1} \map {d_r} {x, y}$
Then we can demonstrate Lipschitz equivalence between all of these metrics, from which topological equivalence follows.
Let $d_r$ be the metric defined as:
:$\ds \map {d_r} {x, y} = \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^r}^{1/r}$
=== Inequalities for Chebyshev Distance ===
=== Inequality for General Case ===
When we combine the inequalities, we have:
:$\map {d_r} {x, y} \ge \map {d_\infty} {x, y} \ge n^{-1} \map {d_1} {x, y} \ge n^{-1} \map {d_r} {x, y}$
Therefore, $d_r$ and $d_\infty$ are Lipschitz equivalent for all $r \in \R_{\ge 1}$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.4$: Equivalent metrics: Example $2.4.5$
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 18$
Category:P-Product Metrics
Category:P-Product Metrics on Real Vector Space are Topologically Equivalent
P-Product Metrics on Real Vector Space are Topologically Equivalent/Inequality for General Case

== Theorem ==
For $n \in \N$, let $\R^n$ be a real vector Space.
Let $r, t \in \R_{\ge 1}$.
Let $d_r$ and $d_t$ be $p$-product metrics on $\R^n$.
Then $d_r$ and $d_t$ are topologically equivalent.
== Proof ==
, assume that $r \le t$.
We show that $\ds \map {d_r} {x, y} \ge \map {d_t} {x, y}$, which is equivalent to proving that:
:$\ds \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^r}^{1/r} \ge \paren {\sum_{i \mathop = 1}^n \size {x_i - y_i}^t}^{1/t}$
Let $\forall i \in \closedint 1 n: s_i = \size {x_i - y_i}$.
Suppose $s_k = 0$ for some $k \in \closedint 1 n$.
Then the problem reduces to the equivalent one of showing that:
:$\ds \paren {\sum_{i \mathop = 1}^{n - 1} \size {x_i - y_i}^r}^{1/r} \ge \paren {\sum_{i \mathop = 1}^{n - 1} \size {x_i - y_i}^t}^{1/t}$
that is, of reducing the index by $1$.
Note that when $n = 1$, from simple algebra $\map {d_r} {x, y} = \map {d_t} {x, y}$.
So, let us start with the assumption that $\forall i \in \closedint 1 n: s_i > 0$.
Let $\ds u = \sum_{i \mathop = 1}^n \size {x_i - y_i}^r = \sum_{i \mathop = 1}^n s_i^r$, and $v = \dfrac 1 r$.
From Derivative of Function to Power of Function‎, $\map {D_r} {u^v} = v u^{v - 1} \map {D_r} u + u^v \ln u \map {D_r} v$.
Here:
:$\ds \map {D_r} u = \sum_{i \mathop = 1}^n s_i^r \ln s_i$ from Derivative of Exponential Function and Sum Rule for Derivatives
:$\map {D_r} v = - \dfrac 1 {r^2}$ from Power Rule for Derivatives
In the case where $r = 1$, we have:
:$\map {D_r} {u^v} = 0$
When $r > 1$, we have:
where $K > 0$ because all of $s_i, r > 0$.
For the same reason, $\ds \dfrac{s_j^r} {\sum_{i \mathop = 1}^n s_i^r} < 1$ for all $j \in \set {1, \ldots, n}$.
From Logarithm of 1 is 0 and Logarithm is Strictly Increasing, their logarithms are therefore negative.
So for $r > 1$:
:$\ds \map {D_r} {\paren {\sum_{i \mathop = 1}^n s_i^r}^{1 / r} } < 0$
So, from Derivative of Monotone Function, it follows that (given the conditions on $r$ and $s_i$) $\ds \paren {\sum_{i \mathop = 1}^n s_i^r}^{1 / r}$ is decreasing.
As we assumed $r \le t$, we have $\map {d_r} {x, y} \ge \map {d_t} {x, y}$.
Category:P-Product Metrics on Real Vector Space are Topologically Equivalent
P-Seminorm is Seminorm

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $p \in \hointr 1 \infty$.
Let $\map {\LL^p} {X, \Sigma, \mu}$ be the Lebesgue $p$-space on $\struct {X, \Sigma, \mu}$. 
Let $\norm {\, \cdot \,}_p$ be the $p$-seminorm on $\map {\LL^p} {X, \Sigma, \mu}$.
Then $\norm {\, \cdot \,}_p$ is a seminorm on $\map {\LL^p} {X, \Sigma, \mu}$.
== Proof ==
Let $f \in \map {\LL^p} {X, \Sigma, \mu}$.
From the construction of the integral of a positive $\Sigma$-measurable function, we have: 
:$\ds \paren {\int \size f^p \rd \mu}^{1/p} \ge 0$
so:
:$\norm f_p \ge 0$ 
We also have: 
:$\ds \int \size f^p \rd \mu < \infty$
So $\norm {\, \cdot \,}$ maps from $\map {\LL^p} {X, \Sigma, \mu}$ to the non-negative reals.
We now verify and .
=== Proof of ===
Let $f \in \map {\LL^p} {X, \Sigma, \mu}$ and $\lambda \in \R$, we have: 
=== Proof of ===
This is Minkowski's Inequality on Lebesgue Space.
Category:Lebesgue Spaces
Category:Seminorms
P-Seminorm of Function Zero iff A.E. Zero

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space and let $p \in \closedint 1 \infty$.
Let $\map {\LL^p} {X, \Sigma, \mu}$ be the Lebesgue $p$-space. 
Let $f \in \map {\LL^p} {X, \Sigma, \mu}$.
Then:
:$\norm f_p = 0$ $f = 0$ $\mu$-almost everywhere
where $\norm \cdot_p$ is the $p$-seminorm. 
== Proof ==
=== Case 1: $1 \le p < \infty$ ===
We have that: 
:$\norm f_p = 0$ $\norm f_p^p = 0$. 
That is, $\norm f_p = 0$ :
:$\ds \int \size f^p \rd \mu = 0$ 
From Measurable Function Zero A.E. iff Absolute Value has Zero Integral, this is equivalent to:
:$\size f^p = 0$ almost everywhere.
From Pointwise Exponentiation preserves A.E. Equality, we have: 
:$\size f = 0$ almost everywhere.
So:
:$f = 0$ almost everywhere.
=== Case 2: $p = \infty$ ===
We have that: 
:$\norm f_\infty = 0$ 
:
:$\inf \set {c > 0 : \map \mu {\set {x \in X : \size {\map f x} \ge c} } = 0} = 0$
From the definition of infimum, for each $\epsilon$ there exists $c < \epsilon$ such that: 
:$\map \mu {\set {x \in X : \size {\map f x} \ge c} } = 0$
We show that it follows that: 
:$\map \mu {\set {x \in X : \size {\map f x} \ge c} } = 0$
for all $c > 0$.
 not, and there exists $M > 0$ such that: 
:$\map \mu {\set {x \in X : \size {\map f x} \ge M} } > 0$
Then, from Survival Function is Decreasing, we have: 
:$\map \mu {\set {x \in X : \size {\map f x} \ge c} } > 0$
for all $c \le M$.
But we have shown that there must exist $c \le M$ such that: 
:$\map \mu {\set {x \in X : \size {\map f x} \ge c} } = 0$
which is a contradiction. 
So: 
:$\map \mu {\set {x \in X : \size {\map f x} \ge c} } = 0$
for all $c > 0$.
Then, we have: 
Then, if $\map f x \ne 0$ we have: 
:$x \in \set {x \in X : \size {\map f x} > 0}$
which is a $\mu$-null set.
So: 
:$f = 0$ $\mu$-almost everywhere.
Category:Lebesgue Spaces
P-Sequence Space of Real Sequences is Metric Space

== Theorem ==
Let $A$ be the set of all real sequences $\sequence {x_i}$ such that the series $\ds \sum_{i \mathop \ge 0} x_i^2$ is convergent.
Let $d_p$ be the $p$-sequence metric on $\R$.
Then $\ell^p := \struct {A, d_p}$ is a metric space.
== Proof ==
By definition of the $p$-sequence metric on $\R$:
Let $A$ be the set of all real sequences $\sequence {x_i}$ such that the series $\ds \sum_{i \mathop \ge 0} x_i^2$ is convergent.
Then $\ell^p := \struct {A, d_2}$ where $d_p: A \times A: \to \R$ is the real-valued function defined as:
:$\ds \forall x = \sequence {x_i}, y = \sequence {y_i} \in A: \map {d_p} {x, y} := \paren {\sum_{k \mathop \ge 0} \size {x_k - y_k}^p}^{\frac 1 p}$
=== Proof of ===
So holds for $d_p$.
=== Proof of ===
Let $z = \sequence {z_i}\in A$.
So holds for $d_p$.
=== Proof of ===
So holds for $d_p$.
=== Proof of ===
So holds for $d_p$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.18$
Category:P-Sequence Metrics
P-Sequence Space with P-Norm forms Normed Vector Space

== Theorem ==
A $p$-Sequence Space with a $p$-norm forms a normed vector space.
== Proof ==
We have that:
:a $p$-sequence space is a vector space
:the $p$-norm on the $p$-sequence space is a norm
By definition, $\struct {\ell^p, \norm {\, \cdot \,}_p}$ is a normed vector space.
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:P-Norms
Category:Examples of Normed Vector Spaces
P-Sequence Space with Pointwise Addition and Pointwise Scalar Multiplication on Ring of Sequences forms Vector Space

== Theorem ==
Let $\ell^p$ be the p-sequence space.
Let $\struct {\R, +_\R, \times_\R}$ be the field of real numbers.
Let $\paren +$ be the pointwise addition on the ring of sequences.
Let $\paren {\, \cdot \,}$ be the pointwise multiplication on the ring of sequences.
Then $\struct {\ell^p, +, \, \cdot \,}_\R$ is a vector space.
== Proof ==
Let $\sequence {a_n}_{n \mathop \in \N}, \sequence {b_n}_{n \mathop \in \N}, \sequence {c_n}_{n \mathop \in \N} \in \ell^p$.
Let $\lambda, \mu \in \R$.
Let $\sequence 0 := \tuple {0, 0, 0, \dots}$ be a real-valued function.
Let us use real number addition and multiplication.
Define pointwise addition as:
:$\sequence {a_n + b_n}_{n \mathop \in \N} := \sequence {a_n}_{n \mathop \in \N} +_\R \sequence {b_n}_{n \mathop \in \N}$.
Define pointwise scalar multiplication as:
:$\sequence {\lambda \cdot a_n}_{n \mathop \in \N} := \lambda \times_\R \sequence {a_n}_{n \mathop \in \N}$
Let the additive inverse be $\sequence {-a_n} := - \sequence {a_n}$.
=== Closure Axiom ===
By assumption, $\sequence {a_n}_{n \mathop \in \N}, \sequence {b_n}_{n \mathop \in \N} \in \ell^p$.
By definition:
:$\ds \sum_{n \mathop = 1}^\infty \size {a_n}^p < \infty$
:$\ds \sum_{n \mathop = 1}^\infty \size {b_n}^p < \infty$
Consider the sequence $\sequence {a_n + b_n}$.
Then: 
Hence:
:$\sequence {a_n + b_n} \in \ell^p$
=== Commutativity Axiom ===
By Pointwise Addition on Ring of Sequences is Commutative, $\sequence {a_n} + \sequence {b_n} = \sequence {b_n} + \sequence {a_n}$
=== Associativity Axiom ===
By Pointwise Addition on Ring of Sequences is Associative, $\paren {\sequence {a_n} + \sequence {b_n} } + \sequence {c_n} = \sequence {a_n} + \paren {\sequence {b_n} + \sequence {c_n} }$.
=== Identity Axiom ===
=== Inverse Axiom ===
=== Distributivity over Scalar Addition ===
=== Distributivity over Vector Addition ===
=== Associativity with Scalar Multiplication ===
=== Identity for Scalar Multiplication ===
== Sources ==
* : $\S 1.1$: Vector Space
Category:Examples of Vector Spaces
Category:Functional Analysis
P-Value/Examples/Arbitrary Example 1

== Example of $p$-Value ==
Consider a $t$-test performed on a sample of $10$ observations from a normal distribution with unknown expectation $\mu$.
To test the null hypothesus that $\mu = 0$ against the alternative hypothesis that $\mu \ne 0$, the statistic $t$ has $9$ degrees of freedom.
Suppose $t = 2.41$.
Then:
:$\map \Pr {\size t \ge 2.41} = 0.0392$
which indicates a significance at the exact $0.0392$ level.
A result is significant at a pre-chosen level $\alpha$ $p \le a$.
== Sources ==
* 
* 
Category:Examples of P-Values