nyuuzyou/wikis huggingface dataset:
Schanuel's Conjecture Implies Algebraic Independence of Pi and Euler's Number over the Rationals

== Theorem ==
Let Schanuel's Conjecture be true.
Then $\pi$ (pi) and $e$ (Euler's number) are algebraically independent over the rational numbers $\Q$.
== Proof ==
Assume the truth of Schanuel's Conjecture.
Let $z_1 = 1$ and $z_2 = i \pi$.
Note that $z_1$ is wholly real and $z_2$ is wholly imaginary.
Hence, by Wholly Real Number and Wholly Imaginary Number are Linearly Independent over the Rationals, they are linearly independent over $\Q$.
By Schanuel's Conjecture, the extension field $\map \Q {z_1, z_2, e^{z_1}, e^{z_2} }$ has transcendence degree at least $2$ over $\Q$.
That is, the extension field $\map \Q {1, i \pi, e, -1}$ has transcendence degree at least $2$ over $\Q$.
However, $1$ and $-1$ are algebraic.
Therefore $i \pi$ and $e$ must be algebraically independent over $\Q$.
 $\pi$ and $e$ are not algebraically independent over $\Q$.
Then, then there would be a non-trivial polynomial $\map g {x, y}$ with rational coefficients satisfying:
:$\map g {\pi, e} = 0$
Then, one can construct a non-trivial polynomial $\map f {x, y} = \map g {i x, y} \map g {-i x, y}$ with rational coefficients satisfying:
:$\map f {i \pi, e} = 0$
which is contradictory to the previous statement that no such polynomials exist.
Therefore, if Schanuel's Conjecture is true, then $\pi$ and $e$ are algebraically independent over $\Q$.
Category:Transcendental Numbers
Category:Pi
Category:Euler's Number
Category:Schanuel's Conjecture
Schanuel's Conjecture Implies Algebraic Independence of Pi and Log of Pi over the Rationals

== Theorem ==
Let Schanuel's Conjecture be true.
Then $\pi$ (pi) and the logarithm of $\pi$ (pi):
:$\ln \pi$
are algebraically independent over the rational numbers $\Q$.
== Proof ==
Assume the truth of Schanuel's Conjecture.
Let $z_1 = \ln \pi$, $z_2 = i \pi$.
Note that $z_1$ is wholly real and $z_2$ is wholly imaginary.
Hence, by Wholly Real Number and Wholly Imaginary Number are Linearly Independent over the Rationals, they are linearly independent over $\Q$.
By Schanuel's Conjecture, the extension field $\Q \left({z_1, z_2, e^{z_1}, e^{z_2}}\right)$ has transcendence degree at least $2$ over $\Q$.
That is, the extension field $\Q \left({\ln \pi, i \pi, \pi, e^{i \pi}}\right)$ has transcendence degree at least $2$ over $\Q$.
However, by Euler's Identity, $e^{i \pi} = -1$ is algebraic.
Also, $i \pi$ and $\pi$ are not algebraically independent, as they satisfy $x^2 + y^2 = 0$, where $x = i \pi$ and $y = \pi$.
Therefore, if Schanuel's Conjecture holds, then $\ln \pi$ and $\pi$ are algebraically independent.
Category:Transcendental Numbers
Category:Pi
Category:Logarithms
Category:Schanuel's Conjecture
Schanuel's Conjecture Implies Transcendence of 2 to the power of Euler's Number

== Theorem ==
Let Schanuel's Conjecture be true.
Then $2$ to the power of Euler's number $e$:
:$2^e$
is transcendental, where $e$ is Euler's number.
== Proof ==
Assume the truth of Schanuel's Conjecture.
Let $z_1 = \ln \ln 2$, $z_2 = 1 + \ln \ln 2$, $z_3 = \ln 2$, and $z_4 = e \ln 2$.
By Lemma, they are linearly independent over the rational numbers $\Q$.
Observe that $z_3 = e^{z_1}$ and $z_4 = e^{z_2}$.
By Schanuel's Conjecture, the extension field $\map \Q {z_1, z_2, z_3, z_4, e^{z_1}, e^{z_2}, e^{z_3}, e^{z_4} }$ has transcendence degree at least $4$ over the rational numbers $\Q$.
That is, the extension field $\map \Q {\ln \ln 2, 1 + \ln \ln 2, \ln 2, e \ln 2, \ln 2, e \ln 2, 2, 2^e}$ has transcendence degree at least $4$ over $\Q$.
However, $2$ is algebraic.
Also, $\ln \ln 2$ and $1 + \ln \ln 2$ are not algebraically independent over $\Q$.
Therefore, $\ln \ln 2$, $\ln 2$, $e \ln 2$, and $2^e$ must all be transcendental.
Therefore, if Schanuel's Conjecture holds, then $2^e$ is transcendental.
Category:Transcendental Numbers
Category:2
Category:Euler's Number
Category:Schanuel's Conjecture
Schanuel's Conjecture Implies Transcendence of 2 to the power of Euler's Number/Lemma

== Lemma ==
Let:
:$z_1 = \ln \ln 2$
:$z_2 = 1 + \ln \ln 2$
:$z_3 = \ln 2$
:$z_4 = e \ln 2$
Let Schanuel's Conjecture be true.
Then $z_1$, $z_2$, $z_3$ and $z_4$ are linearly independent over the rational numbers $\Q$.
== Proof ==
Assume the truth of Schanuel's Conjecture.
$2$ is algebraic.
Hence, by the Corollary of the weaker Hermite-Lindemann-Weierstrass theorem, $\ln 2$ is transcendental.
Now, we will prove that $z_1$ and $z_3$ are linearly independent over the rational numbers $\Q$.
Equivalently, by Linearly Independent over the Rational Numbers iff Linearly Independent over the Integers, they are linearly independent over the integers $\Z$.
Let $a, b \in \Z$ such that:
:$a z_1 + b z_3 = 0$
Substituting:
:$a \ln \ln 2 + b \ln 2 = 0$
Applying the exponential function to both sides:
:$\left({\ln 2}\right)^a 2^b = 1$
Since $\ln 2$ is transcendental, the above equation only has solution when $a = b = 0$.
Thus, $z_1$ and $z_3$ are linearly independent over the rational numbers $\Q$.
By Schanuel's Conjecture, the extension field $\Q \left({z_1, z_3, e^{z_1}, e^{z_3}}\right)$ has transcendence degree at least $2$ over the rational numbers $\Q$.
That is, the extension field $\Q \left({\ln \ln 2, \ln 2, \ln 2, 2}\right)$ has transcendence degree at least $2$ over $\Q$.
However, $2$ is algebraic.
Therefore, $\ln \ln 2$ and $\ln 2$ must be algebraically independent over $\Q$.
That is, $z_1$, $z_3$ must be algebraically independent over rational numbers $\Q$.
It follows that $z_1$, $z_3$, and $1$ are linearly independent over the rational numbers $\Q$.
By Schanuel's Conjecture, the extension field $\Q \left({z_1, z_3, 1, e^{z_1}, e^{z_3}, e}\right)$ has transcendence degree at least $3$ over the rational numbers $\Q$.
That is, the extension field $\Q \left({\ln \ln 2, \ln 2, 1, \ln 2, 2, e}\right)$ has transcendence degree at least $3$ over $\Q$.
However, $1$ and $2$ are algebraic.
Therefore, $\ln \ln 2$, $\ln 2$, $e$ must be algebraically independent over $\Q$.
That is, $z_1$, $z_3$, and $e$ must be algebraically independent over rational numbers $\Q$.
It follows that $z_1$, $z_3$, $e z_3$ must be algebraically independent over rational numbers $\Q$.
That is, $z_1$, $z_3$, $z_4$ must be algebraically independent over $\Q$.
Therefore, $z_1$, $1$, $z_3$, $z_4$ must be linearly independent over $\Q$.
Hence, $z_1$, $1 + z_1$, $z_3$, $z_4$ must be linearly independent over $\Q$.
That is, if Schanuel's Conjecture holds, then $z_1$, $z_2$, $z_3$, $z_4$ are linearly independent over the rational numbers $\Q$.
Category:Transcendental Numbers
Category:2
Category:Euler's Number
Category:Schanuel's Conjecture
Schanuel's Conjecture Implies Transcendence of Euler's Number to the power of Euler's Number

== Theorem ==
Let Schanuel's Conjecture be true.
Then Euler's number $e$ to the power of itself:
:$e^e$
is transcendental. 
== Proof ==
Assume the truth of Schanuel's Conjecture.
Let $z_1 = 1$, $z_2 = e$.
By Euler's Number is Irrational, $z_1$ and $z_2$ are linearly independent over $\Q$.
By Schanuel's Conjecture, the extension field $\Q \left({z_1, z_2, e^{z_1}, e^{z_2}}\right)$ has transcendence degree at least $2$ over $\Q$.
That is, the extension field $\Q \left({1, e, e, e^e}\right)$ has transcendence degree at least $2$ over $\Q$.
However, $1$ is algebraic.
Therefore, if Schanuel's Conjecture holds, $e^e$ must be transcendental.
Category:Transcendental Numbers
Category:Euler's Number
Category:Schanuel's Conjecture
Schanuel's Conjecture Implies Transcendence of Log Pi

== Theorem ==
Let Schanuel's Conjecture be true.
Then the logarithm of $\pi$ (pi):
:$\ln \pi$
is transcendental. 
== Proof ==
Assume the truth of Schanuel's Conjecture.
From Schanuel's Conjecture Implies Algebraic Independence of Pi and Log of Pi over the Rationals, $\ln \pi$ and $\pi$ are algebraically independent over the rational numbers $\Q$.
Therefore, if Schanuel's Conjecture holds, $\ln \pi$ must be transcendental.
Category:Transcendental Numbers
Category:Pi
Category:Logarithms
Category:Schanuel's Conjecture
Schanuel's Conjecture Implies Transcendence of Pi by Euler's Number

== Theorem ==
Let Schanuel's Conjecture be true.
Then $\pi \times e$ is transcendental. 
== Proof ==
Assume the truth of Schanuel's Conjecture.
By Schanuel's Conjecture Implies Algebraic Independence of Pi and Euler's Number over the Rationals, $\pi$ and $e$ are algebraically independent over the rational numbers $\Q$.
That is, no non-trivial polynomials $\map f {x, y}$ with rational coefficients satisfy:
:$\map f {\pi, e} = 0$
 $\pi \times e$ is algebraic.
Then there would be a non-trivial polynomial $\map g z$ with rational coefficients satisfying:
:$\map g {\pi \times e} = 0$
However, $\map f {x, y} := \map g {x \times y}$ would be a non-trivial polynomial with rational coefficients satisfying:
:$\map f {\pi, e} = 0$
which contradicts the earlier statement that no such polynomials exist.
Therefore, if Schanuel's Conjecture holds, $\pi \times e$ is transcendental.
Category:Transcendental Numbers
Category:Pi
Category:Euler's Number
Category:Schanuel's Conjecture
Schanuel's Conjecture Implies Transcendence of Pi plus Euler's Number

== Theorem ==
Let Schanuel's Conjecture be true.
Then $\pi + e$ is transcendental. 
== Proof ==
Assume the truth of Schanuel's Conjecture.
By Schanuel's Conjecture Implies Algebraic Independence of Pi and Euler's Number over the Rationals, $\pi$ and $e$ are algebraically independent over the rational numbers $\Q$.
That is, no non-trivial polynomials $\map f {x, y}$ with rational coefficients satisfy:
:$\map f {\pi, e} = 0$
 $\pi + e$ is algebraic.
Then there would be a non-trivial polynomial $\map g z$ with rational coefficients satisfying:
:$\map g {\pi + e} = 0$
However, $\map f {x, y} := \map g {x + y}$ would be a non-trivial polynomial with rational coefficients satisfying:
:$\map f {\pi, e} = 0$
which contradicts the earlier statement that no such polynomials exist.
Therefore, if Schanuel's Conjecture holds, $\pi + e$ is transcendental.
Category:Transcendental Numbers
Category:Pi
Category:Euler's Number
Category:Schanuel's Conjecture
Schanuel's Conjecture Implies Transcendence of Pi to the power of Euler's Number

== Theorem ==
Let Schanuel's Conjecture be true.
Then $\pi$(pi) to the power of Euler's number $e$:
:$\pi^e$
is transcendental.
== Proof ==
Assume the truth of Schanuel's Conjecture.
Let $z_1 = \ln \ln \pi$, $z_2 = 1 + \ln \ln \pi$, $z_3 = \ln \pi$, $z_4 = e \ln \pi$, and $z_5 = i \pi$.
By Lemma, they are linearly independent over the rational numbers $\Q$.
Observe that $z_3 = e^{z_1}$ and $z_4 = e^{z_2}$.
By Schanuel's Conjecture, the extension field $\Q \left({z_1, z_2, z_3, z_4, z_5, e^{z_1}, e^{z_2}, e^{z_3}, e^{z_4}, e^{z_5}}\right)$ has transcendence degree at least $5$ over the rational numbers $\Q$.
That is, the extension field $\Q \left({\ln \ln \pi, 1 + \ln \ln \pi, \ln \pi, e \ln \pi, i \pi, \ln \pi, e \ln \pi, \pi, \pi^e, e^{i \pi}}\right)$ has transcendence degree at least $5$ over $\Q$.
However, by Euler's Identity, $e^{i \pi} = -1$ is algebraic.
Also, $\ln \ln \pi$ and $1 + \ln \ln \pi$ are not algebraically independent over $\Q$.
Also, $\pi$ and $i \pi$ are not algebraically independent over $\Q$.
Therefore, $\ln \ln \pi$, $\ln \pi$, $e \ln \pi$, $\pi$, and $\pi^e$ must all be transcendental.
Therefore, if Schanuel's Conjecture holds, then $\pi^e$ is transcendental.
Category:Transcendental Numbers
Category:Pi
Category:Euler's Number
Category:Schanuel's Conjecture
Schanuel's Conjecture Implies Transcendence of Pi to the power of Euler's Number/Lemma

== Lemma ==
Let Schanuel's Conjecture be true.
Let $z_1 = \ln \ln \pi$, $z_2 = 1 + \ln \ln \pi$, $z_3 = \ln \pi$, $z_4 = e \ln \pi$, and $z_5 = i \pi$.
Then, $z_1$, $z_2$, $z_3$, $z_4$, and $z_5$ are linearly independent over the rational numbers $\Q$.
== Proof ==
Assume the truth of Schanuel's Conjecture.
Now, we will prove that $z_1$ and $z_3$ are linearly independent over the rational numbers $\Q$.
Equivalently, they are linearly independent over the integers $\Z$.
Let $a, b \in \Z$ such that:
:$a z_1 + b z_3 = 0$
Substituting:
:$a \ln \ln \pi + b \ln \pi = 0$
Applying the exponential function to both sides:
:$\paren {\ln \pi}^a \pi^b = 1$
By Schanuel's Conjecture Implies Algebraic Independence of Pi and Log of Pi over the Rationals, the above equation only has solution when $a = b = 0$.
Thus, $z_1$ and $z_3$ are linearly independent over the rational numbers $\Q$.
Since $z_5$ is wholly imaginary, $z_1$, $z_3$, and $z_5$ are linearly independent over the rational numbers $\Q$.
By Schanuel's Conjecture, the extension field $\map \Q {z_1, z_3, z_5, e^{z_1}, e^{z_3}, e^{z_5} }$ has transcendence degree at least $3$ over the rational numbers $\Q$.
That is, the extension field $\map \Q {\ln \ln \pi, \ln \pi, i \pi, \ln \pi, \pi, e^{i \pi} }$ has transcendence degree at least $3$ over $\Q$.
However, by Euler's Identity, $e^{i \pi} = -1$ is algebraic.
Also, $\pi$ and $i \pi$ are not algebraically independent over $\Q$.
Therefore, $\ln \ln \pi$, $\ln \pi$, and $i \pi$ must be algebraically independent over $\Q$.
That is, $z_1$, $z_3$, and $z_5$ must be algebraically independent over rational numbers $\Q$.
It follows that $z_1$, $z_3$, $z_5$, and $1$ are linearly independent over the rational numbers $\Q$.
By Schanuel's Conjecture, the extension field $\map \Q {z_1, z_3, z_5, 1, e^{z_1}, e^{z_3}, e^{z_5}, e}$ has transcendence degree at least $4$ over the rational numbers $\Q$.
That is, the extension field $\map \Q {\ln \ln \pi, \ln \pi, i \pi, 1, \ln \pi, \pi, e^{i \pi}, e}$ has transcendence degree at least $4$ over $\Q$.
However, $1$ is algebraic.
Moreover, by Euler's Identity, $e^{i \pi} = -1$ is algebraic.
Also, $\pi$ and $i \pi$ are not algebraically independent over $\Q$.
Therefore, $\ln \ln \pi$, $\ln \pi$, $i \pi$, and $e$ must be algebraically independent over $\Q$.
That is, $z_1$, $z_3$, $z_5$, and $e$ must be algebraically independent over rational numbers $\Q$.
It follows that $z_1$, $z_3$, $e z_3$, and $z_5$ must be algebraically independent over rational numbers $\Q$.
That is, $z_1$, $z_3$, $z_4$, and $z_5$ must be algebraically independent over $\Q$.
Therefore, $z_1$, $1$, $z_3$, $z_4$, and $z_5$ must be linearly independent over $\Q$.
Hence, $z_1$, $1 + z_1$, $z_3$, $z_4$, and $z_5$ must be linearly independent over $\Q$.
That is, if Schanuel's Conjecture holds, then $z_1$, $z_2$, $z_3$, $z_4$, and $z_5$ are linearly independent over the rational numbers $\Q$.
Category:Transcendental Numbers
Category:Logarithms
Category:Pi
Category:Euler's Number
Category:Schanuel's Conjecture
Schatunowsky's Theorem

== Theorem ==
Let $n \in \Z_{>0}$ be a strictly positive integer.
Let $\map w n$ denote the number of primes strictly less than $n$ which are not divisors of $n$.
Let $\map \phi n$ denote the Euler $\phi$ function of $n$.
Then $30$ is the largest integer $n$ such that:
:$\map w n = \map \phi n - 1$
== Proof ==
The above equation is equivalent to the property that all numbers greater than $1$ that are coprime to it but less are prime.
For an integer to have this property:
If it is greater than $p^2$ for some prime $p$, then it must be divisible by $p$.
If not, it will be coprime to $p^2$, a composite number.
Let $p_n$ denote the $n$th prime.
Suppose $N$ has this property.
By the argument above, if $p_{n + 1}^2 \ge N > p_n^2$, we must have $p_1 p_2 \cdots p_n \divides N$.
By Absolute Value of Integer is not less than Divisors, we have $p_1 p_2 \cdots p_n \le N$.
Bertrand-Chebyshev Theorem asserts that there is a prime between $p_n$ and $2 p_n$.
Thus we have $2 p_n > p_{n + 1}$.
Hence for $n \ge 5$:
This is a contradiction. 
Hence we must have $N \le p_5^2 = 121$.
From the argument above we also have:
:$2 \divides N$ for $4 < N \le 9$
:$2, 3 \divides N$ for $9 < N \le 25$
:$2, 3, 5 \divides N$ for $25 < N \le 49$
:$2, 3, 5, 7 \divides N$ for $49 < N \le 121$
So we end up with the list $N = 1, 2, 3, 4, 6, 8, 12, 18, 24, 30$.
This list is verified in Integers such that all Coprime and Less are Prime.
== Also see ==
* Integers such that all Coprime and Less are Prime
* Odd Integers whose Smaller Odd Coprimes are Prime
== Sources ==
* : $\text {4-2}$ Residue Systems: Exercise $4$
* : $30$
* : $30$
* 
Category:Euler Phi Function
Category:Prime Numbers
Schatunowsky's Theorem/Mistake

== Source Work ==
:
:The Dictionary
::$30$
:
:The Dictionary
::$30$
== Mistake ==
:$30$ is the greatest number such that all the numbers less than it and prime to it are themselves primes. The other numbers with this property are $2$, $3$, $4$, $6$, $8$, $12$, $18$ and $24$.
== Correction ==
While all integers are coprime to $1$ (and $30$ in particular), remember that $1$ is not classified as a prime number.
The canonical exposition of this result is for $1$ to be specifically mentioned.
== Sources ==
* : $30$
* : $30$
Category:Mistakes/Curious and Interesting Numbers
Schauder Basis is Linearly Independent

== Theorem ==
Let $\Bbb F \in \set {\R, \C}$. 
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\Bbb F$.
Let $\set {e_n : n \in \N}$ be a Schauder basis for $X$. 
Then $\set {e_n : n \in \N}$ is linearly independent.
== Proof ==
Suppose that: 
:$\ds \sum_{k \mathop = 1}^n \alpha_{i_k} e_{i_k} = 0$
for some $n \in \N$, $i_1, \ldots, i_n \in \N$ and $\alpha_{i_1}, \ldots, \alpha_{i_n} \in \Bbb F$.
Define a sequence $\sequence {\alpha_j}_{j \mathop \in \N}$ in $\Bbb F$ by: 
:$\ds \alpha_j = \begin{cases}\alpha_{i_k} & \text { if there exists } k \text { such that } j = i_k \\ 0 & \text { otherwise}\end{cases}$
Then, we have: 
:$\ds \sum_{j \mathop = 1}^\infty \alpha_j e_j = \sum_{k \mathop = 1}^n \alpha_{i_k} e_{i_k} = 0$
From the definition of Schauder basis, we then have: 
:$\alpha_j = 0$ for each $j \in \N$
and in particular: 
:$\alpha_{i_k} = 0$ for each $k$. 
Since the coefficients $\alpha_{i_1}, \ldots, \alpha_{i_n}$ and $n \in \N$ were arbitrary, we have that:
:$\set {e_n : n \in \N}$ is linearly independent.
== Sources ==
* $9.1$: Schauder Bases in Normed Spaces
Category:Schauder Bases
Category:Linear Independence
Scheffé's Lemma

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f_n$ be a sequence of $\mu$-integrable functions that converge almost everywhere to another $\mu$-integrable function $f$.
Then $f_n$ converges to $f$ in $L^1$ $\ds \int_X \size {f_n} \rd \mu$ converges to $\ds \int_X \size f \rd \mu$.
=== Corollary ===
== Proof ==
=== Sufficient Condition ===
Let $f_n \to f$ in $L^1$.
Then:
and so $f_n \to f$ in $L^1$ implies the of this inequality goes to $0$ as $n$ grows to infinity.
Since the of the inequality is non-negative, it also goes to $0$ as $n$ grows to infinity.
=== Necessary Condition ===
Let $\ds \int_X \size {f_n} \rd \mu \to \int_X \size f \rd \mu$.
We wish to show that:
:$\ds \int_X \size {f - f_n} \rd \mu \to 0$
First, by Triangle Inequality:
:$\forall a, b \in \R: \size a + \size b - \size {a - b} \ge 0$
Therefore, for each $x \in X$:
:$\size {\map f x} + \size {\map {f_n} x} - \size {\map f x - \map {f_n} x} \ge 0$
Thus, we may employ Fatou's Lemma for Integrals on the expression $\ds \int_X \liminf_n \size f + \size {f_n} - \size {f - f_n} \rd \mu$ to yield:
:$(1): \ds \int_X \liminf_n \size f + \size {f_n} - \size {f - f_n} \rd \mu \le \liminf_n \int_X \size f + \size {f_n} - \size {f - f_n} \rd \mu$
Now, the integrand on the of $(1)$ equals $2 \size f$ almost everywhere since $f_n \to f$ pointwise almost everywhere.
So the integral on the of $(1)$ is:
:$\ds 2 \int_X \size f \rd \mu$
We may thus rewrite $(1)$ as:
Rearranging the and of this inequality, we get:
:$\ds \limsup_n \int_X \size {f - f_n} \rd \mu \le 0$
This implies that:
:$\ds \int_X \size {f - f_n} \rd \mu \to 0$
Category:Measure Theory
Category:Scheffé's Lemma
Scheffé's Lemma/Corollary

== Corollary to Scheffé's Lemma ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $f_n$ be a sequence of $\mu$-integrable functions that convergence in measure to another $\mu$-integrable function $f$.
Then $f_n$ converges to $f$ in $L^1$ $\ds \int_X \size {f_n} \rd \mu$ converges to $\ds \int_X \size f \rd \mu$.
== Proof ==
Let $f_n$ converges to $f$ in measure instead.
The proof of the first direction remains unchanged from the above.
In the other direction, suppose:
:$\ds \int_X \size {f_n} \rd \mu \to \int_X \size f \rd \mu$
We wish to show again that:
:$\ds \int_X \size {f - f_n} \rd \mu \to 0$.
 this is false.
Since the integral is non-negative, we can find some $\epsilon > 0$ and an infinite subsequence $g_n$ of $f_n$ such that:
:$\ds \int_X \size {f - g_n} \rd \mu \ge \epsilon$
Since $g_n$ still converges in measure to $f$, by Convergence in Measure Implies Convergence a.e. of Subsequence $g_n$ has a further subsequence $h_n$ that converges almost everywhere to $f$.
But we also have:
:$\ds \int_X \size {h_n} \rd \mu \to \int_X \size f \rd \mu$
Hence from Scheffé's Lemma:
:$\ds \int_X \size {f - h_n} \rd \mu \to 0$
This is a contradiction, as $h_n$ is a subsequence of $g_n$, hence $\ds \int_X \size {f - h_n} \rd \mu$ must remain larger than $\epsilon$.
Category:Scheffé's Lemma
Schreier-Zassenhaus Theorem

== Theorem ==
Let $G$ be a finite group.
Let $\HH_1$ and $\HH_2$ be two normal series for $G$.
Then $\HH_1$ and $\HH_2$ have refinements of equal length whose factors are isomorphic.
== Proof ==
Suppose that:
:$(1): \quad \set e = G_0 \lhd G_1 \lhd \cdots \lhd G_{n - 1} \lhd G_n = G$
and:
:$(2): \quad \set e = H_0 \lhd H_1 \lhd \cdots \lhd H_{m - 1} \lhd H_m = G$
are two normal series for $G$.
Let a new series be formed:
:$(3): \quad \set e = \hat G_0 \subseteq \hat G_1 \subseteq \cdots \subseteq \hat G_{n m - 1} \subseteq \hat G_{n m} = G$
such that:
:$\hat G_k = G_q \paren {G_{q + 1} \cap H_r}$
for $k = q m + r$, where $0 \le q < n$ and $0 \le r \le m$.
For completeness, define $\hat G_{n m} = G_n$.
Note that, by the above construction:
:$\hat G_{q m} = G_q \paren {G_{q + 1} \cap H_0} = G_q \paren {\set e} = G_q$
This implies that each group in $(1)$ also appears in $(3)$.
It needs to be demonstrated that $\hat G_k$ is well-defined.
Consider the ambiguity of $k = q m + r$ in the following cases:
:When $m \nmid k$, there is none.
:When $m \divides k$, we have:
:::$k = q m + m = \paren {q + 1} m$
For $(3)$ to be well-defined, we require:
:$G_q \paren {G_{q + 1} \cap H_m} = G_{q + 1} \paren {G_{q + 2} \cap H_0}$
Note that the is just:
:$G_{q + 1} \paren {G_{q + 2} \cap \set e} = G_{q + 1} \paren {\set e} = G_{q + 1}$
The is:
:$G_q \paren {G_{q + 1} \cap H_m} = G_q \paren {G_{q + 1} \cap G} = G_q G_{q + 1} = G_{q + 1}$
Thus the construction is verified.
It next needs to be demonstrated that:
:$\hat G_k \lhd \hat G_{k + 1}$
Choose the representation of $k$ in which $r < m$.
It follows that:
:$\hat G_{k + 1} = G_q \paren {G_{q + 1} \cap H_{r + 1} }$
Since $H_{r + 1} \rhd H_r$:
:$\hat G_{k + 1} = G_q \paren {G_{q + 1} \cap H_{r + 1} } \rhd G_q \paren {G_{q + 1} \cap H_r} = \hat G_k$
That is, $\hat G_k$ is a normal subgroup of $\hat G_{k + 1}$.
$(A): \quad$ But note that $(3)$ may not actually be a normal series as it is possible that $\hat G_k = \hat G_{k + 1}$ for some $k$.
Having created series $(3)$, we use the same procedure to form:
:$(4): \quad \set e = \hat H_0 \subseteq \hat H_1 \subseteq \cdots \subseteq \hat H_{n m - 1} \subseteq \hat H_{n m} = G$
such that:
:$\hat H_k = H_q \paren {H_{q + 1} \cap G_r}$
for $k = q n + r$, where $0 \le q < m$ and $0 \le r \le n$.
The same statements that were made about $(3)$ also hold about $(4)$.
Now let $k = u m + v$ and $l = v n + u$.
It follows that:
So $(3)$ and $(4)$ have isomorphic factors.
In $(A)$ it was remarked that some of the factors in these isomorphic series may have redundant elements where $\hat G_k = \hat G_{k + 1}$ and similarly for $\hat H_l = \hat H_{l + 1}$.
As the series consist of isomorphic elements, all we now have to do is remove these redundant elements from both (they will occur at the same place).
Hence we end up with two refinements of equal length whose factors are isomorphic.
== Also known as ==
The Schreier-Zassenhaus Theorem is also known as the Schreier Refinement Theorem.
== Sources ==
* : Chapter $2$: Normal and Composition Series: $\S 72$. The Schreier-Zassenhous Theorem
Category:Normal Series