# 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 W c P(X) be a s-ring.
Let y : W -> [-oo,oo] be countably additive and y(O)=0.
Let A[n] be a decreasing sequence of sets from W.
Suppose that y(A[k]) :- |R, for some k:-|N.
What can we conclude about the value of y( //\\(n:-|N) A[n] ) ?
y( //\\(n:-|N) A[n] ) = lim(n->oo) y(A[n])
the limit is shown to exist in the course of the proof
In the proof, we use the previous item.
page 158 in 1st measure
page 55 in the 2nd measure notebook
Let W c P(X) be a ring.
Let y : W -> [-oo,oo] be additive.
Let A,B :- W.
Suppose that A c B and y(A) = oo.
What can we conclude?
y(B) = oo
A c B, hence B = A u (B\A) disjointly.
Since y is additive, y(B) = y(A) + y(B\A).
Since y(A) = oo, so is y(B).
Let W c P(X) be a ring.
Let y : W -> [-oo,oo] be additive.
Let A,B :- W.
Suppose that A c B and y(A) = -oo.
What can we conclude?
y(B) = -oo
A c B, hence B = A u (B\A) disjointly.
Since y is additive, y(B) = y(A) + Y(B\A).
Since y(A) = -oo, so is y(B).
Let W c P(X) be a ring.
Let y : W -> [0,oo] be additive.
Let A,B :- W.
Suppose that A c B.
What can we conclude?
y(A) <= y(B)
Since A c B, B = A u (B\A) disjointly.
Since y is additive, y(B) = y(A) + y(B\A).
Since y(B\A)>=0, we get y(A) <= y(A) + y(B\A) = y(B).
Hence y(A) <= y(B).
Let W be a ring of sets.
Let y : W -> [0,oo] be countably additive.
Can we conclude that y is countably subadditive on W?
Yes.
hint: U(n:-|N) A[n] = \\*//(n:-|N) [ A[n] \ U(k=1 to n-1)_A[k] ]
page 84 in 1st measure
Let W be a ring of sets.
Let y : W -> [0,oo[ be additive.
Prove that
/\B,A:-W |y(A) - y(B)| <= y(A+B)
We will use [ E c F => y(E) <= y(F). ]
y(A) <= y(A u B) = y( (A+B) u (AnB) ) = y(A+B) + y(AnB) <= y(A+B) + y(B)
Hence y(A) <= y(A+B) + y(B).
Since the values are real numbers, we get y(A) - y(B) <= y(A+B).
Let W be a semi-ring of sets. Let y : W -> [-oo,oo] be finitely additive.
Let E be written, in two different ways, as a finite disjoint union of sets in W.
What can we conclude?
Let E = U(i=1 to n) C[i] = U(j=1 to m) D[j]. And C[i],D[j] :- W.
We can conclude:
+(i=1 to n) y(C[i]) = +(j=1 to m) y(D[j])
page 86 in 1st measure
Notice here that finite additivity is required.
Let J be a semi-ring of sets.
Let y : J -> [-oo,oo] be finitely additive.
Extend function y so that y : R(J) -> [-oo,oo] is additive.
page 87, 88 in 1st measure
Notice that finite additivity is required.
Let a,b,c belong to [0,oo].
Prove that (a+b)*c = (a*c) + (b*c).
___
Let C and D be sets.
Suppose that
1) (a,b) :- C x D
2) (x,y) :- C x D
What can we conclude about (a,y) and (x,b) ?
(a,y) :- C x D
(x,b) :- C x D