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,g : X -> |R be simple functions.

Can we conclude that f+g is also a simple function?

Can we conclude that f+g is also a simple function?

YES

Use the theorem on page 149 in 1st measure.

Use the theorem on page 149 in 1st measure.

Let W be a s-algebra in X. Let y : W -> [0,oo] be additive.

Let f : X -> [0,oo) be simple and W-measurable.

Let 0 <= a < oo.

Prove that LebS(X,af,dy) = a*LebS(X,f,dy).

Let f : X -> [0,oo) be simple and W-measurable.

Let 0 <= a < oo.

Prove that LebS(X,af,dy) = a*LebS(X,f,dy).

page 151 in 1st measure

Notice that y,f,a are all assumed to be non-negative

in order to avoid the possibility of adding: oo + (-oo).

Notice that y,f,a are all assumed to be non-negative

in order to avoid the possibility of adding: oo + (-oo).

Let W be a s-algebra in X. Let y : W -> [0,oo] be additive.

Let E :- W and let f : X -> [0,oo) be simple and measurable.

Define the Lebesgue integral of function f over set E with respect to y.

LebS(E,f,dy) = ???

Let E :- W and let f : X -> [0,oo) be simple and measurable.

Define the Lebesgue integral of function f over set E with respect to y.

LebS(E,f,dy) = ???

LebS(E,f,dy) = LebS(X,1(E)*f,dy)

Notice that here we have to check that 1(E)*f is simple and measurable.

page 152 in 1st measure

Notice that here we have to check that 1(E)*f is simple and measurable.

page 152 in 1st measure

Let W be a s-algebra in X. Let y : W -> [0,oo] be additive.

Let f,g : X -> [0,oo) be simple and W-measurable.

Prove that LebS(X,f,dy) + LebS(X,g,dy) = LebS(X,f+g,dy).

Let f,g : X -> [0,oo) be simple and W-measurable.

Prove that LebS(X,f,dy) + LebS(X,g,dy) = LebS(X,f+g,dy).

page 151 in 1st measure

Notice that y,f,g are all assumed to be non-negative

in order to avoid the possibility of adding: oo + (-oo).

We also want the formula /\a,b,c:-[0,oo] a*(b+c) = (a*b)+(a*c).

Also notice that here we have to check if f+g is simple.

Notice that y,f,g are all assumed to be non-negative

in order to avoid the possibility of adding: oo + (-oo).

We also want the formula /\a,b,c:-[0,oo] a*(b+c) = (a*b)+(a*c).

Also notice that here we have to check if f+g is simple.

Let W be a s-algebra in X. Let y : W -> [0,oo] be additive.

Let f,g : X -> [0,oo) be simple and W-measurable.

Suppose that f <= g.

Prove that LebS(X,f,dy) <= LebS(X,g,dy).

Let f,g : X -> [0,oo) be simple and W-measurable.

Suppose that f <= g.

Prove that LebS(X,f,dy) <= LebS(X,g,dy).

page 151 in 1st measure

Let W be a s-algebra in X. Let y : W -> [0,oo] be additive.

Let f,g : X -> [0,oo) be simple and W-measurable.

Let E :- W.

Suppose that /\x:-E f(x)<=g(x).

Prove that LebS(E,f,dy) <= LebS(E,g,dy).

Let f,g : X -> [0,oo) be simple and W-measurable.

Let E :- W.

Suppose that /\x:-E f(x)<=g(x).

Prove that LebS(E,f,dy) <= LebS(E,g,dy).

page 152 in 1st measure

Let W be a s-algebra in X. Let y : W -> [0,oo] be additive.

Let f,g : X -> [0,oo) be simple and W-measurable.

Let E :- W.

Prove that LebS(E,f,dy) + LebS(E,g,dy) = LebS(E,f+g,dy).

Let f,g : X -> [0,oo) be simple and W-measurable.

Let E :- W.

Prove that LebS(E,f,dy) + LebS(E,g,dy) = LebS(E,f+g,dy).

page 152 in 1st measure

Let W be a s-algebra in X. Let y : W -> [0,oo] be additive.

Let f : X -> [0,oo) be simple and W-measurable.

Let 0 <= a < oo. Let E :- W.

Prove that LebS(E,af,dy) = a*LebS(E,f,dy).

Let f : X -> [0,oo) be simple and W-measurable.

Let 0 <= a < oo. Let E :- W.

Prove that LebS(E,af,dy) = a*LebS(E,f,dy).

page 152 in 1st measure

Let W be a s-algebra in X. Let y : W -> [0,oo] be additive.

Let f : X -> [0,oo) be simple and measurable.

Let Y : W -> [0,oo] be defined Y(E) = LebS(E,f,dy).

Can we conclude that function Y is additive?

Let f : X -> [0,oo) be simple and measurable.

Let Y : W -> [0,oo] be defined Y(E) = LebS(E,f,dy).

Can we conclude that function Y is additive?

Y is additive, Y(O)=0

Notice that when proving Y(O)=0 we used /\x:-|R* 0*x = 0.

In measure theory, it is essential that 0*oo = 0.

page 119 in 1st measure

Notice that when proving Y(O)=0 we used /\x:-|R* 0*x = 0.

In measure theory, it is essential that 0*oo = 0.

page 119 in 1st measure

Let W be a s-algebra in X. Let y : W -> [0,oo] be countably
additive.

Let f : X -> [0,oo) be simple and measurable.

Let Y : W -> [0,oo] be defined Y(E) = LebS(E,f,dy).

Can we conclude that function Y is countably additive?

Let f : X -> [0,oo) be simple and measurable.

Let Y : W -> [0,oo] be defined Y(E) = LebS(E,f,dy).

Can we conclude that function Y is countably additive?

Y is countably additive, Y(O)=0

Notice that when proving Y(O)=0 we used /\x:-|R* 0*x = 0.

In measure theory, it is essential that 0*oo = 0.

page 155 in 1st measure

Notice that when proving Y(O)=0 we used /\x:-|R* 0*x = 0.

In measure theory, it is essential that 0*oo = 0.

page 155 in 1st measure