nyuuzyou/wikis huggingface dataset:
Riemann Zeta Function of 6/Proof 1

== Theorem ==
== Proof == 
By Fourier Series: $x^6$ over $-\pi$ to $\pi$, for $x \in \closedint {-\pi} \pi$:
:$\ds x^6 = \frac {\pi^6} 7 + \sum_{n \mathop = 1}^\infty \frac {12 n^4 \pi^4 - 240 n^2 \pi^2 +1440} {n^6} \, \map \cos {n \pi} \, \map \cos {n x}$
Setting $x = \pi$: 
Category:Riemann Zeta Function of 6
Riemann Zeta Function of 6/Proof 2

== Theorem ==
== Proof == 
Dividing out the x factor on both sides and equating the product with the sum, we have:
Equating the $x^2$ term on both sides of the equation, we obtain the value of the sum of the individual terms in the Basel Problem:
Equating the $x^4$ term on both sides of the equation, we obtain the value of the sum of the product of every unique combination of $2$ terms from the Basel Problem:
Equating the $x^6$ term on both sides of the equation, we obtain the value of the sum of the product of every unique combination of $3$ terms from the Basel Problem:
When we take the cube of a sum, we have:
Let $A = \dfrac 1 {1^2}, B = \dfrac 1 {2^2}, C = \dfrac 1 {3^2}, \cdots $
Then the (Cube of Sum) becomes:
and the first term on the (Sum of Cubes) becomes:
To obtain the remaining two terms on the , we have:
Finally, we have:
Category:Riemann Zeta Function of 6
Riemann Zeta Has No Zeroes With Real Part One

== Theorem ==
Let $\zeta$ be the Riemann zeta function.
Then for all $t \in \R$:
:$\map \zeta {1 + i t} \ne 0$
== Proof ==
Throughout, the complex variable $s$ is $s = \sigma + i t$.
We have, for $\sigma > 1$,
Therefore,
:$(1): \quad \ds -\map \Re {\dfrac {\map {\zeta'} s} {\map \zeta s} } = \sum_{n \mathop \ge 1} \map \Lambda n n^{-\sigma} \map \cos {t \log n}$
Now observe that:
:$3 + 4 \cos \theta + \map \cos {2 \theta} = 2 \paren {1 + \cos \theta}^2 \ge 0$
Because for all $n \ge 1$ we have $\map \Lambda n n^{-\sigma} \ge 0$, we have:
Now let:
:$\map \eta s = \map \zeta s^3 \cdot \map \zeta {s + i t}^4\cdot \map \zeta {s + 2 i t}$
Then the above computation has shown that
:$\map \Re {\dfrac {\map {\eta'} s} {\map \eta s} } \le 0$
By Poles of Riemann Zeta Function we know that $\zeta$ has a simple pole at $s = 1$ with residue $1$.
 that $1 + i t$ is a zero of $\zeta$ of order $d \ge 1$.
Therefore, at $s = 1$, $\eta$ has a zero of order $4 d - 3 \ge 0$, that is:
:$\map \eta s \sim (s - 1)^{4 d - 3}$
as $s \to 1^+$, where:
:$\sim$ indicates asymptotic equality
:superscript $+$ denotes a limit from the right along the real line.
Therefore:
:$\dfrac {\map {\eta'} s} {\map \eta s} \sim \dfrac {4 d - 3} {s - 1}$
as $s \to 1^+$.
Since $\map \Re {\dfrac {4 d -3} {s - 1} } \to + \infty$ as $s \to 1^+$, it follows that:
:$\map \Re {\dfrac {\map {\eta'} s} {\map \eta s} } \to \infty$
as $s \to 1^+$.
But we have already shown that:
:$\map \Re {\dfrac {\map {\eta'} s} {\map \eta s} } \le 0$
which is a contradiction.
Category:Riemann Zeta Function
Riemannian Geodesic is Locally Minimizing

== Theorem ==
Let $\struct {M, g}$ be a Riemannian manifold.
Let $\gamma$ be a geodesic.
Then $\gamma$ is locally minimizing.
== Proof ==
== Sources ==
* : $\S 6$: Geodesics and Distance. Uniformly Normal Neighborhoods
Category:Riemannian Manifolds
Category:Geodesics
Riemannian Manifold as Metric Space

== Theorem ==
Let $\struct {M, g}$ be a connected Riemannian manifold with or without boundary.
Let $d_g$ be the Riemannian distance.
Then $\struct {M, d_g}$ is a metric space whose metric topology is the same as the given manifold topology.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Lengths and Distances
Category:Riemannian Geometry
Riemannian Manifold has Zero Gaussian Curvature iff Euclidean

== Theorem ==
Let $M$ be a Riemannian manifold of dimension $2$.
The Gaussian curvature on $M$ is zero the Riemannian metric on $M$ is the same as the Euclidean metric.
== Proof ==
== Historical Note ==
== Sources ==
* : Chapter $\text {A}.32$: Riemann ()
Category:Riemannian Geometry
Riemannian Manifold has Zero Gaussian Curvature iff Euclidean/Historical Note

== Historical Note on Riemannian Manifold has Zero Gaussian Curvature iff Euclidean ==
 demonstrated that a Riemannian Manifold has Zero Gaussian Curvature iff Euclidean in a posthumous paper on heat conduction.
Thus the Riemann-Christoffel tensor measures how much a Riemannian manifold deviates from a Euclidean space.
== Sources ==
* : Chapter $\text {A}.32$: Riemann ()
Category:Historical Notes
Riemannian Metric gives rise to Connection

== Theorem ==
Let $M$ be a smooth manifold.
Let $g$ be a Riemannian metric on $M$.
Then $g$ gives rise to a connection.
It is not necessarily the case that a connection gives rise to a Riemannian metric.
== Proof ==
== Sources ==
* 
* 
Category:Riemannian Metrics
Category:Connections
Riemannian Volume Form of Orientable Hypersurface

== Theorem ==
Let $\struct {\tilde M, \tilde g}$ be an oriented Riemannian manifold.
Let $\struct {M, g}$ be an orientable hypersurface with an induced metric.
Let $N$ be a unit global normal vector for $M$.
Let $\rd V_g$ be the Riemannian volume form.
Then Riemannian volume form is given by:
:$\rd V_g = \valueat {\paren {N \lrcorner \rd V_{\tilde g} } } M$
where $\lrcorner$ denotes the interior multiplicaiton.
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Basic Constructions on Riemannian Manifolds
Category:Riemannian Geometry
Riemannian Volume Form under Orientation-Preserving Isometry

== Theorem ==
Let $\struct {M, g}$ and $\struct {\tilde M, \tilde g}$ be oriented Riemannian manifolds.
Let $\phi : M \to \tilde M$ be an orientation-preserving isometry.
Let $\rd V_g$ be the Riemannian volume form.
Then:
:$\phi^* \rd V_{\tilde g} = \rd V_g$
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Basic Constructions on Riemannian Manifolds
Category:Isometries (Riemannian Manifold)
Category:Riemannian Geometry
Riesel Problem/Historical Note

== Historical Note on Riesel Problem ==
As of $24$ February $2021$, there remain $48$ values of $k$ below $509 \, 203$ for which no primes have been found of the form:
:$k 2^n - 1$
for any positive integer $n$.
They are as follows:
:$2293, 9221, 23669, 31859, 38473,$
:$46663, 67117, 74699, 81041, 93839,$
:$97139, 107347, 121889, 129007, 143047,$
:$161669, 192971, 206039, 206231,$
:$215443, 226153, 234343, 245561, 250027,$
:$315929, 319511, 324011, 325123,$
:$327671, 336839, 342847, 344759, 362609,$
:$363343, 364903, 365159, 368411, 371893,$
:$384539, 386801, 397027, 409753, 444637,$
:$470173, 474491, 477583, 485557, 494743$
Any of these may therefore be a Riesel number smaller than $509 \, 203$.
Research is ongoing.
Category:Historical Notes
Riesz's Convergence Theorem

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $p \in \R$, $p \ge 1$.
Let $\sequence {f_n}_{n \mathop \in \N}, f_n: X \to \R$ be a sequence in Lebesgue $p$-space $\map {\LL^p} \mu$.
Suppose that the pointwise limit $f := \ds \lim_{n \mathop \to \infty} f_n$ exists $\mu$-almost everywhere, and that $f \in \map {\LL^p} \mu$.
Then the following are equivalent:
:$(1): \quad \ds \lim_{n \mathop \to \infty} \norm {f - f_n}_p = 0$
:$(2): \quad \ds \lim_{n \mathop \to \infty} \norm {f_n}_p = \norm f_p$
where $\norm {\, \cdot \,}_p$ denotes the $p$-seminorm.
== Proof ==
=== From $(1)$ to $(2)$ ===
This follows from the reverse triangle inequality:
:$\size {\norm f_p - \norm {f_n}_p} \le \norm {f - f_n}_p$
=== From $(2)$ to $(1)$ ===
By Power of Absolute Value is Convex Real Function, we have:
:$\forall a, b \in \R : \size {\dfrac {a - b} 2}^p \le \dfrac { {\size a}^p + {\size {-b} }^p} 2 = \dfrac { {\size a}^p + {\size b}^p} 2$
In particular:
:$\map {h_n} x := 2^{p - 1} \paren {\size {\map f x}^p + \size {\map {f_n} x}^p} - \size {\map f x - \map {f_n} x}^p$
is a positive measurable function.
By Fatou's Lemma:
:$\ds \int \liminf_{n \mathop \to \infty} h_n \rd \mu \le \liminf_{n \mathop \to \infty} \int h_n \rd \mu$
Observe:
and:
Thus we have:
:$\ds 2^p \norm f_p ^p \le 2^p \norm f_p ^p - \limsup_{n \mathop \to \infty} \int \size {f - f_n}^p \rd \mu$
By Real Number Ordering is Compatible with Addition, adding $- 2^p \norm f_p ^p$ to the both sides:
:$\ds 0 \le - \limsup_{n \mathop \to \infty} \int \size {f - f_n}^p \rd \mu$
By Order of Real Numbers is Dual of Order of their Negatives:
:$\ds \limsup_{n \mathop \to \infty} \int \size {f - f_n}^p \rd \mu \le 0$
== Sources ==
* : $12.10$
Category:Lebesgue Spaces
Riesz's Lemma

== Theorem ==
Let $X$ be a normed vector space.
Let $Y$ be a proper closed linear subspace of $X$. 
Let $\alpha \in \openint 0 1$.
Then there exists $x_\alpha \in X$ such that: 
:$\norm {x_\alpha} = 1$
with:
:$\norm {x_\alpha - y} > \alpha$
for all $y \in Y$.
== Proof 1 ==
 
== Proof 2 ==
Category:Riesz's Lemma
Category:Functional Analysis
Riesz's Lemma/Proof 1

== Theorem ==
== Proof ==
Since $Y < X$:
:$X \setminus Y$ is non-empty. 
Since $Y$ is closed:
:$X \setminus Y$ is open.
Let $x \in X \setminus Y$.
Then there exists $\epsilon > 0$ such that: 
:$\map {B_\epsilon} x \subset X \setminus Y$
So, for all $y \in Y$, we must have: 
:$\norm {x - y} \ge \epsilon$
That is: 
:$\inf \set {\norm {x - y} \colon y \in Y} \ge \epsilon$
For brevity, let:
:$d = \inf \set {\norm {x - y} \colon y \in Y}$
Since $\alpha^{-1} > 1$, there exists $z \in Y$ with: 
:$\norm {x - z} < d \alpha^{-1}$
otherwise the infimum would be at least $d \alpha^{-1} > d$, a contradiction.
Since $x \in X \setminus Y$ and $z \in Y$, we clearly have $x \ne z$.
So, we can set: 
:$ x_\alpha = \dfrac {x - z} {\norm {x - z} }$
Clearly:
:$\norm {x_\alpha} = \dfrac {\norm {x - z} } {\norm {x - z} } = 1$
Now, for any $y \in Y$ we have: 
We've already seen that: 
:$\norm {x - z} < d \alpha^{-1}$
Since $Y$ is closed under linear combination, we have:
:$z + \norm {x - z} y \in Y$
and so: 
:$\norm {x - \paren {z + \norm {x - z} y} } \ge d$
We conclude that: 
Category:Riesz's Lemma
Riesz's Lemma/Proof 2

== Theorem ==
== Proof ==
Consider the normed quotient vector space $X / Y$ with quotient mapping $\pi$.
From Operator Norm of Quotient Mapping in Quotient Normed Vector Space is 1, we have: 
:$\norm \pi_{\map B {X, X/Y} } = 1$ 
Since $\alpha \in \openint 0 1$, there exists $x_\alpha \in X$ with $\norm {x_\alpha} = 1$ and: 
:$\norm {\map \pi {x_\alpha} }_{X/Y} > \alpha$
by the definition of the norm on the space of bounded linear transformations.
That is, by the definition of the quotient norm:
:$\ds \inf_{z \mathop \in Y} \norm {x_\alpha - z} > \alpha$
So there exists $y \in Y$ such that: 
:$\norm {x_\alpha - y} > \alpha$
Category:Riesz's Lemma
Riesz-Fischer Theorem

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $p \in \R$, $p \ge 1$.
The Lebesgue $p$-space $\map {\LL^p} \mu$, endowed with the $p$-norm $\norm {\cdot}_p$, is a Banach space.
=== Corollary ===
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $p \in \R$, $p \ge 1$.
If a sequence $\sequence {f_k}$ in $\map {\LL^p} \mu$ converges to $f$,
then there is a subsequence $\sequence {f_{k_j}}$ that converges pointwise a.e. to $f$.
== Proof ==
From Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach, to prove $\map {\LL^p} \mu$ is complete, it suffices to prove that every absolutely summable sequence in $\map {\LL^p} \mu$ is summable.
Let $\sequence {f_n}$ be an absolutely summable sequence in $\map {\LL^p} \mu$
Define:
:$\ds \sum_{k \mathop = 1}^\infty \norm {f_k}_p =: B < \infty$
Also define:
:$\ds G_n := \sum_{k \mathop = 1}^n \size {f_k}$
and:
:$\ds G = \sum_{k \mathop = 1}^\infty \size {f_k}$
It is clear that the conditions of the Monotone Convergence Theorem (Measure Theory) hold, so that:
:$\ds \int_X G^p = \lim_{n \mathop \to \infty} \int_X G_n^p$
By observing that:
we can also say that:
:$\ds \int_X \size {G_n}^p \le B^p$
and therefore:
:$\ds \lim_{n \mathop \to \infty} \int_X \size {G_n}^p \le B^p$
Therefore we have that:
:$\ds \int_X G^p \le B^p < \infty$
This confirms:
:$G \in \map {\LL^p} \mu$
In particular:
:$G \in \map{\LL^p} \mu$
entails that:
:$G < \infty$ a.e.
So $\sequence {f_k}$ is absolutely summable a.e..
By Absolutely Convergent Series is Convergent/Real Numbers:
:$\ds F = \sum_{k \mathop = 1}^\infty f_k$
converges a.e.
Because $\size F \le G$:
:$F \in \map {\LL^p} \mu$
It only remains to show that:
:$\ds \sum_{k \mathop = 1}^n f_k \to F$ in $\norm {\cdot}_p$
which we can accomplish by Lebesgue's Dominated Convergence Theorem.
Because $\ds \size {F - \sum_{k \mathop = 1}^n f_k}^p \le (2G)^p \in \map{\LL^1}\mu$, the theorem applies.
We infer:
:$\ds \norm {F - \sum_{k \mathop = 1}^n f_k}_p^p = \int_X \size {F - \sum_{k \mathop = 1}^n f_k}^p \to 0$
Therefore by in $\map{\LL^p}\mu$ we have that $\ds \sum_{k \mathop = 1}^\infty f_k$ converges in $\map{\LL^p}\mu$.
This shows that $\sequence {f_k}$ is summable, as we were to prove. 
== Historical Note ==
== Sources ==
* : $6.1$ Theorem $6.6$
* : $12.7$
Category:Lebesgue Spaces
Category:Banach Spaces
Riesz-Kakutani Representation Theorem

== Theorem ==
Let $X$ be a Hausdorff compact space.
Let $\map \BB X$ be the Borel $\sigma$-algebra on $X$.
Let $\map C {X, \R}$ be the space of real-valued continuous functions.
Let $\norm {\, \cdot \,}_\infty$ be the supremum norm on $X$.
Let $\struct {C', \norm {\, \cdot \,}_{C'} }$ be the normed dual space of $\struct {\map C {X, \R}, \norm {\, \cdot \,}_\infty}$.
Then $\ell \in C'$ there is a unique signed measure $\mu$ on $\struct {X, \map \BB X}$ such that:
:the variation $\size \mu$ of $\mu$ is finite
:$\ds \forall f \in \map C {X, \R} : \map \ell f = \int_X f \rd \mu$
In addition:
:$\norm \ell_{C'} = \map {\size \mu} X$
== Proof ==
== Sources ==
* : $8.3$: Reflexive Spaces
Category:Functional Analysis
Category:Riesz-Markov-Kakutani Representation Theorem
Riesz-Markov-Kakutani Representation Theorem/Construction of mu and M

== Theorem ==
=== Notation ===
== Construction of $\mu$ and $\MM$ ==
For every $V \in \tau$, define:
:$\map {\mu_1} V = \sup \set {\Lambda f: f \prec V}$
Note that $\mu_1$ is monotonically increasing.
That is, for all $V, W \in \tau$ such that $V \subset W$, we have:
For every other subset $E \subset X$, define:
:$\map \mu E = \inf \set {\map {\mu_1} V: V \supset E \land V \in \tau}$
Since $\mu_1$ is monotonically increasing:
:$\mu_1 = \mu {\restriction_\tau}$
Define:
:$\MM_F = \set {E \subset X : \map \mu E < \infty \land \map \mu E = \sup \set {\map \mu K: K \subset E \land K \text { compact} } }$
Define:
:$\MM = \set {E \subset X : \forall K \subset X \text { compact}: E \cap K \in \MM_F}$
Category:Riesz-Markov-Kakutani Representation Theorem
Riesz-Markov-Kakutani Representation Theorem/Lemma 1

== Lemma for Riesz-Markov-Kakutani Representation Theorem ==
=== Notation ===
=== Construction of $\mu$ and $\MM$ ===
== Lemma ==
$\mu$ is countably subadditive.
== Proof ==
Let $V_1$ and $V_2$ be open subsets of $X$.
Let $g \in \map {C_c} X: g \prec V_1 \cup V_2$.
Let $\set {h_1, h_2}$ be a partition of unity subordinate to $\set {V_1, V_2}$.
By linearity of $\Lambda$ and the definition of $\mu$:
Since $g$ was arbitrary, by the definition of $\mu$, $\mu$ is subadditive over $\tau$.
Applying Principle of Mathematical Induction yields the countable subadditivity of $\mu$ over $\tau$.
Let $\sequence{E_i}\in X^\N$.
By definition of $\mu$, for all $E_i$, for all $\epsilon \in \R_{>0}$, there exists an open $V_i \supset E_i$ such that:
:$\map \mu {V_i} \le \map \mu {E_i} + 2^{-i} \epsilon$
Therefore, by $\ds \bigcup_{i \mathop = 1}^\infty V_i \supset \bigcup_{i \mathop = 1}^\infty E_i$, monotonicity of $\mu$, and countable subadditivity of $\mu$ over $\tau$:
:$\ds \map \mu {\bigcup_{i \mathop = 1}^\infty E_i} \le \map \mu {\bigcup_{i \mathop = 1}^\infty V_i} \le \epsilon + \sum_{i \mathop = 1}^\infty \map \mu {E_i}$
Since $\epsilon$ was arbitrary, $\mu$ is countably subadditive over $X$.
Category:Riesz-Markov-Kakutani Representation Theorem
Riesz-Markov-Kakutani Representation Theorem/Lemma 2

== Lemma for Riesz-Markov-Kakutani Representation Theorem ==
=== Notation ===
=== Construction of $\mu$ and $\MM$ ===
== Lemma ==
Let $K \subset X$ be compact.
Then:
:$K \in \MM_F$
and:
:$\map \mu K = \inf \set {\Lambda f : K \prec f}$
== Proof ==
Let $f \in \map {C_c} X: K \prec f$.
Let $\alpha \in \R \cap \openint 0 1$.
Define:
:$V_\alpha = \map {f^{-1} } {\openint \alpha \to}$
Then $K \in V_\alpha$ and:
:$\forall g: \map {C_c} X: g \prec V_\alpha: \alpha g \le f$
Thus:
Since $\alpha$ can be arbitrarily close to $1$:
:$\map \mu K \le \inf \set {\Lambda f: K \prec f}$
By definition of $\mu$, for all $\epsilon \in \R_{>0}$, there exists a $V \in \tau: V \supset K$ such that:
:$\map \mu V < \map \mu K + \epsilon$
By Urysohn's Lemma, there exists an Urysohn function $f$ for $K$ and $V$.
By definition of $\mu$:
:$\Lambda f \le \map \mu V < \map \mu K + \epsilon$
Since $\epsilon$ was arbitrary:
:$\map \mu K \ge \inf \set {\Lambda f: K \prec f}$
Therefore:
:$\map \mu K = \inf \set {\Lambda f: K \prec f}$
By definition of positive linear functional, $\Lambda$ is finite.
Thus:
:$\map \mu K < \infty$
So:
:$K \in \MM_F$
Category:Riesz-Markov-Kakutani Representation Theorem
Riesz-Markov-Kakutani Representation Theorem/Lemma 3

== Lemma for Riesz-Markov-Kakutani Representation Theorem ==
=== Notation ===
=== Construction of $\mu$ and $\MM$ ===
== Lemma ==
$\mu$ is countably additive over pairwise disjoint collections of compact sets.
== Proof ==
Let $K_1$ and $K_2$ be disjoint compact subsets of $X$.
By Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods, there exists an open $V\supset K_1$ such that $V$ and $K_2$ are disjoint.
By Urysohn's Lemma, there exists a mapping $f \in \map{C_c} X$ such that:
:$\map f {K_1} = \set 1$
and:
:$\map f {K_2} = \set 0$
By Lemma 2 and union of compact sets is compact:
:$\forall \epsilon \in \R_{>0}: \exists g \in \map {C_c} X: K_1 \cup K_2 \prec g \text{ and } \Lambda g < \map \mu {K_1 \cup K_2} + \epsilon$
Now:
:$K_1 \prec f g$
and:
:$K_2 = \prec \paren {1 - f} g$
By linearity of $\Lambda$ and Lemma 2:
Thus, by Lemma 1, $\mu$ is additive for disjoint compact sets.
Applying the Principle of Mathematical Induction yields countable additivity.
Category:Riesz-Markov-Kakutani Representation Theorem
Riesz-Markov-Kakutani Representation Theorem/Lemma 4

== Lemma for Riesz-Markov-Kakutani Representation Theorem ==
=== Notation ===
=== Construction of $\mu$ and $\MM$ ===
== Lemma ==
$\mu$ is countably additive over pairwise disjoint collections of subsets of $\MM_F$.
== Proof ==
Let $\sequence {E_i} \in \paren {\MM_F}^\N$ be pairwise disjoint with union $E$.
Let $\map \mu E = \infty$.
Then, by countable subadditivity:
:$\ds \infty = \map \mu E \le \sum_{i \mathop = 1}^\infty \map \mu {E_i}$
So:
:$\ds \map \mu E = \sum_{i \mathop = 1}^\infty \map \mu E$
Suppose $\map \mu E < \infty$.
By definition of $\MM_F$, for all $\epsilon \in \R_{>0}$, for each $i$, there exists a compact $H_i \subset E_i$ such that:
:$\map \mu {H_i} > \map \mu {E_i} - 2^{-i} \epsilon$
So:
This holds for all $n \in \N$.
So $\mu$ is countably superadditive over pairwise disjoint collections of subsets of $\MM_F$.
Therefore, by Lemma 1, $\mu$ is countably additive over pairwise disjoint collections of subsets of $\MM_F$.
Category:Riesz-Markov-Kakutani Representation Theorem
Riesz-Markov-Kakutani Representation Theorem/Lemma 5

== Lemma for Riesz-Markov-Kakutani Representation Theorem ==
=== Notation ===
=== Construction of $\mu$ and $\MM$ ===
== Lemma ==
$\set {V \in \tau: \map \mu V < \infty} \subset \MM_F$
== Proof ==
Since $\mu$ and $\mu_1$ coincide on $\tau$, by definition of $\mu_1$, for all $\alpha < \map \mu V$, there exists some $f \in \map {C_c} X: f \prec V$ such that $\Lambda f > \alpha$.
Let $K = \supp f$.
For all $W \in \tau$ such that $K \subset W$:
:$f \prec W$
So, by definition of $\mu_1$:
:$\Lambda f \le \map \mu W$
By definition of $\mu$:
:$\Lambda f \le \map \mu K$
So:
:$\alpha < \map \mu K$
That is, for all $\alpha < \map \mu V$, there exists some compact $K \subset V$ such that:
:$\alpha < \map \mu K \le \map \mu V$
Since $V$ was arbitrary:
:$\set {V \in \tau: \map \mu V < \infty} \subset \MM_F$
Category:Riesz-Markov-Kakutani Representation Theorem
Riesz-Markov-Kakutani Representation Theorem/Lemma 6

== Lemma for Riesz-Markov-Kakutani Representation Theorem ==
=== Notation ===
=== Construction of $\mu$ and $\MM$ ===
== Lemma ==
For all $E \in \MM_F$ and $\epsilon \in \R_{>0}$, there exist some compact $K \in X$ and some $V \in \tau$ such that:
:$K \subset E \subset V$
and:
:$\map \mu {V \setminus K} < \epsilon$
== Proof ==
Let $E \in \MM_F$.
Then, by definition of $\mu$ and $\MM_F$, there exist compact $K \subset E$ and open $V \supset E$ such that:
:$\map \mu V - \dfrac \epsilon 2 < \map \mu E < \map \mu K + \dfrac \epsilon 2$
By Compact Subspace of Hausdorff Space is Closed:
:$V \setminus K \in \tau$
So, by Lemma 5:
:$V \setminus K \in \MM_F$
By Lemma 4:
:$\map \mu {V \setminus - K} = \map \mu V - \map \mu K < \epsilon$
Category:Riesz-Markov-Kakutani Representation Theorem
Riesz-Markov-Kakutani Representation Theorem/Lemma 7

== Lemma for Riesz-Markov-Kakutani Representation Theorem ==
=== Notation ===
=== Construction of $\mu$ and $\MM$ ===
== Lemma ==
The union, if of finite measure, of countable pairwise disjoint subsets of $\MM_F$ is in $\MM_F$.
== Proof ==
Let $\sequence {E_i} \in \paren {\MM_F}^\N$ be pairwise disjoint.
Suppose $\ds E = \bigcup_{i \mathop = 1}^\infty E_i$ has finite measure.
By Lemma 4, for all $\epsilon > 0$, there exists some $N \in \N$ such that:
:$\ds \map \mu E \le \epsilon \sum_{i \mathop = 1}^\infty \map \mu {E_i}$
By definition of $\MM_F$, for all $i$, there exists some compact $H_i \subset E_i$ such that:
:$\map \mu {H_i} > \map \mu {E_i} - 2^{-i} \epsilon$
Then:
:$\ds \map \mu E \le \map \mu {\bigcup_{i \mathop = 1}^N H_i} + 2 \epsilon$
By Finite Union of Compact Sets is Compact:
:$\ds \bigcup_{i \mathop = 1}^N \subset E$
Therefore:
:$E \in \MM_F$
Category:Riesz-Markov-Kakutani Representation Theorem
Riesz-Markov-Kakutani Representation Theorem/Lemma 8

== Lemma for Riesz-Markov-Kakutani Representation Theorem ==
=== Notation ===
=== Construction of $\mu$ and $\MM$ ===
== Lemma ==
$\MM_F$ is closed under set difference, union and intersection.
== Proof ==
Let $\tuple {A, B} \in \paren {\MM_F}^2$.
By Lemma 6, there exist compact sets $K_1, K_2$ and open sets $V_1, V_2$ such that:
:$K_1 \subset A \subset V_1$
:$K_2 \subset B \subset V_2$
and:
:$\forall i \in \set {1, 2}: \map \mu {V_i \setminus K_i} < \dfrac \epsilon 2$
Now:
So, by Lemma 1:
:$\map \mu {A_B} \le \map \mu {K_1 \setminus V_2} + \epsilon$
By Closed Subspace of Compact Space is Compact:
:$K_1 \setminus V_2$ is compact.
Since $K_1 \setminus V_2 \subset A \setminus B$, there exist compact subsets of $A \setminus B$ arbitrarily close in measure to $A \setminus B$.
So:
:$A \setminus B \in \MM_F$. 
Now, by Lemma 7:
:$A \cup B = \paren {A \setminus B} \cup B \in \MM_F$
and:
:$A \cap B = A \setminus \paren {A \setminus B} \in \MM_F$.
Category:Riesz-Markov-Kakutani Representation Theorem
Riesz-Markov-Kakutani Representation Theorem/Lemma 9

== Lemma for Riesz-Markov-Kakutani Representation Theorem ==
=== Notation ===
=== Construction of $\mu$ and $\MM$ ===
== Lemma ==
$\MM_F = \set {E \in \MM: \map \mu E < \infty}$
== Proof ==
By Lemma 2 and Lemma 8:
:$E \in \MM_F \implies \paren {\forall K \text{ compact}: E \cap K \in \MM_F \implies E \in \MM}$
That is:
:$\MM_F \subset \MM$
Conversely, suppose that:
:$E \in \MM$
:$\map \mu E < \infty$
:$\epsilon \in \R{>0}$
By definition of $\mu$, there exists an open $V \supset E$ such that:
:$\map \mu V < \infty$
By Lemma 5 and Lemma 6, there exists a compact $K \subset V$ such that:
:$\map \mu {V \setminus K} < \epsilon$
By definition of $\MM$:
:$E \cap K \in \MM_F$.
So by definition of $\MM_F$, there exists a compact $H \subset E \cap K$ such that:
:$\map \mu {E \cap K} < \map \mu H + \epsilon$
Since $E \subset \paren {E \cap K} \cup \paren {V \setminus K}$, by Lemma 1:
:$\map \mu E \le \map \mu {E \cap K} + \map \mu {V \setminus K} + 2 \epsilon$
Thus:
:$E \in \MM_F$.
Therefore:
:$\MM_F = \set {E \in \MM: \map \mu E < \infty}$
Category:Riesz-Markov-Kakutani Representation Theorem