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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? 
YES Use the theorem on page 149 in 1st measure. 
Let W be a salgebra in X. Let y : W > [0,oo] be additive. Let f : X > [0,oo) be simple and Wmeasurable. 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 nonnegative in order to avoid the possibility of adding: oo + (oo). 
Let W be a salgebra 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) = ??? 
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 
Let W be a salgebra in X. Let y : W > [0,oo] be additive. Let f,g : X > [0,oo) be simple and Wmeasurable. 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 nonnegative 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 salgebra in X. Let y : W > [0,oo] be additive. Let f,g : X > [0,oo) be simple and Wmeasurable. Suppose that f <= g. Prove that LebS(X,f,dy) <= LebS(X,g,dy). 
page 151 in 1st measure 
Let W be a salgebra in X. Let y : W > [0,oo] be additive. Let f,g : X > [0,oo) be simple and Wmeasurable. 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 salgebra in X. Let y : W > [0,oo] be additive. Let f,g : X > [0,oo) be simple and Wmeasurable. 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 salgebra in X. Let y : W > [0,oo] be additive. Let f : X > [0,oo) be simple and Wmeasurable. 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 salgebra 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? 
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 
Let W be a salgebra 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? 
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 