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.

(1) W is a s-algebra in X
(2) f,g : X -> |R \ {-oo,oo} are W-measurable
(3) F : |R^2 -> |R is continuous
(4) h : X -> |R , h(x) = F(f(x),g(x))
What can we conclude from this?
h is W-measurable
page 137 in 1st measure
(1) D is an open connected subset of |C
(2) g,h : D -> |C are diffable
(3) /\z:-D g'(z) = h'(z)
What can we conclude from this?
\/A:-|C /\z:-D g(z) = h(z) + A
hint:
(g-h)' = g' - h' = 0
g-h is constant
Let F be a s-algebra in X.
A:-F <=> ???
A:-F <=> 1(A) is F-measurable
1(A) is the characteristic function of A
1(A) : X -> |R*
Let F be a s-algebra in X. Let A c X.
1(A) is the characteristic function of A.
1(A) is F-measurable <=> ???
1(A) is F-measurable <=> A:-F
f : |R -> (-1;1) , f(x) = x / ( |x|+1 )
What is interesting about this function?
It shows that |R is homeomorphic with (-1;1).
It is even continuously differentiable.
It can be used to define a metric on [-oo,oo].
Then [-oo,oo] is homeorphic with [-1,1].
Let (X,F,y) be a measure space.
Let f:X->[0,oo] be measurable.
Let A,B:-F and AnB=O.
Prove that Leb*(AuB,f,dy) = ???.
Leb*(AuB,f,dy) = Leb*(A,f,dy) + Leb*(B,f,dy)
page 170 in 1st measure
Let F be a s-algebra in X.
Let f,g : X -> |R* be F-measurable.
What can we conclude about the set {x:-X : f(x)>g(x)} ?
{x:-X : f(x)>g(x)} :- F
hint:
{x: f(x)>g(x)} = \\//a:-|Q {x: f(x)>a} n {x: a>g(x)}
page 139 in 1st measure
Let F be a s-algebra in X.
Let f,g : X -> |R* be F-measurable.
What can we conclude about the set {x:-X : f(x)>=g(x)} ?
{x:-X : f(x)>=g(x)} :- F
hint:
{x: f(x)<g(x)} = \\//a:-|Q {x: f(x)<a} n {x: a<g(x)}
page 139 in 1st measure
Let F be a s-algebra in X.
Let f,g : X -> |R* be F-measurable.
What can we conclude about the set {x:-X : f(x)=g(x)} ?
{x:-X : f(x)=g(x)} :- F
hint:
{x: f(x)<g(x)} = \\//a:-|Q {x: f(x)<a} n {x: a<g(x)}
page 139 in 1st measure
Let F be a s-algebra in X.
Let f,g : X -> |R \ {-oo,oo} be F-measurable.
What are two two ways of proving that f+g is F-measurable?
WAY 1:
(1) F : |R^2 -> |R is continuous
(2) h : X -> |R , h(x) = F(f(x),g(x))
thesis: h is F-measurable
Let F(x,y)=x+y (page 138 in 1st measure)
WAY 2:
(-1)*g is measurable,