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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 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).
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).
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) = ???
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 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).
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.
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).
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).
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).
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).
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?
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 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?
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

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