Math ASCII Notation Demo

Mathematical content on Apronus.com is presented in Math ASCII Notation which can be properly displayed by all Web browsers because it uses only the basic set of characters found on all keyboards and in all fonts.

The purpose of these pages is to demonstrate the power of the Math ASCII Notation. In principle, it can be used to write mathematical content of any complexity. In practice, its limits can be seen when trying to write complicated formulas (containing, for example, variables with many indexes or multiple integrals).

Despite its limitations the Math ASCII Notation has much expressive power, as can be seen from browsing through these pages.

Let F be a s-algebra in X.
(1) f : X -> |R*
(2) n:-|N
(3) a[1], ..., a[n] :- |R
(4) A[1], ..., A[n] :- F
(5) X = \\*//(k=1 to n) A[k]
(6) /\x:-X f(x) = +(k=1 to n) a[k]*1(A[k])(x)
f is simple and F-measurable
page 106 in 2nd measure
page 141 in 1st measure
Let F be a s-algebra in X.
Suppose that f : X -> |R is simple and F-measurable.
How can function f be conveniently expressed?
There exist:
(1) n:-|N
(2) a[1], ... , a[n] :- |R ; all different ; n numbers altogether
(3) A[1], ... , A[n] :- F ; disjoint and non-empty
such that
(4) X = \\*//(k=1 to n) A[k]
(5) /\x:-X f(x) = +(k=1 to n) a[k]*1(A[k])(x)
Let A[1], A[2], A[3], ..., A[n], ... be a sequence of sets such that
/\n:-|N A[n] = U(k=1 to n) A[k]\A[k-1], where A[0] = O.
Can we conclude that A[n] is increasing?
YES
page 99 in 2nd measure
Let A[1], A[2], A[3], ..., A[n], ... be a sequence of sets such that
/\n:-|N A[n] = U(k=1 to n) A[k]\A[k-1], where A[0] = O.
What can we conclude from this?
(1) /\n:-|N A[n] c A[n+1]
(2) {A[k]\A[k-1]}(k:-|N) is a disjoint sequence
(3) /\n:-|N A[n] = \\*//(k=1 to n) A[k]\A[k-1], where A[0] = O
(4) \\//(n:-|N) A[n] = \\*//(n:-|N) A[n]\A[n-1], where A[0] = O.
page 99 in 2nd measure
Let A[1], A[2], A[3], ..., A[n], ... be a sequence of sets such that
/\n:-|N A[n] = U(k=1 to n) A[k]\A[k-1], where A[0] = O.
Can we conclude that { A[k]\A[k-1] }(k:-|N) is a disjoint sequence?
YES. page 99 in 2nd measure
We can even conclude much more:
(1) /\n:-|N A[n] c A[n+1]
(2) {A[k]\A[k-1]}(k:-|N) is a disjoint sequence
(3) /\n:-|N A[n] = \\*//(k=1 to n) A[k]\A[k-1], where A[0] = O
(4) \\//(n:-|N) A[n] = \\*//(n:-|N) A[n]\A[n-1], where A[0] = O.
Let A[1], A[2], A[3], ..., A[n], ... be a sequence of sets such that
{A[k]\A[k-1]}(k:-|N) is a disjoint sequence.
Can we conclude that A[n] is increasing?
NO
A[n] = {n}
Let A[1], A[2], A[3], ..., A[n], ... be a sequence of sets such that
\\//(n:-|N) A[n] = \\*//(n:-|N) A[n]\A[n-1], where A[0] = O.
Can we conclude that A[n] is increasing?
NO
A[n] = {n}
If a :- [-oo,oo], what does (a+) mean?
(a+) = max(0 , a)
page 103 in 2nd measure
If a :- [-oo,oo], what does (a-) mean?
(a-) = max(0 , -a)
page 103
In the context of measure theory,
what does it mean: f~g ?
In this context:
(1) F is a s-algebra in X
(2) y : F -> [0,oo] is countably additive
(3) f,g : X -> |R* are F-measurable
f~g <=> y( {x:-X : f(x)!=g(x)} ) = 0