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Re: Measure theory help (extended real line)
Posted:
Jul 17, 2005 8:14 AM
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Carl R. wrote:
> Hi, can someone please explain the following proof:
>
> Write R for the set of all real numbers.
> Let (X,A) be a measurable space and let f be an extended real-valued
> function (i.e f: X-> R U {-infinity,infinity} ).
>
> Theorem:
>
> f is measurable iff the sets A= { x in X : f(x) = +oo},
> B= { x in X : f(x) = -oo} belong to A and the real
> valued function f_1 defined by
>
> f_1(x) = f(x) if x is not in A U B,
> 0 if x is in A U B
>
> is measurable.
>
> Proof: I can see why A and B are measurable, if f is measurable.
> The part I don't see clear is that if f is measurable then
> f_1(x) is a measurable function. The book proves this as follows:
>
> Let f be measurable and let a be a real number. If a>=0
> then {x in X | f_1(x) > a } = { x in X : f(x) > a } \ A.
>
> My question is why do you have to substract the set A ?
Because f(x) > a includes the possibility f(x) = infinity. If f(x) =
infinity then f_1(x) = 0 so if a >= 0 and f_1(x) > a then, in
particular, f_1(x) is nonzero so f(x) cannot be infinity.
> Now if a<0 then { x in X : f_1(x) > a} = { x in X : f(x) > a} U B
>
> Again, why do you have to take the union with the set B ?
If a < 0 then f_1(x) > a includes the possibility that f_1(x) =
infinity or -infinity. In the second set's definition, the condition
f(x) > a takes care of the first possibility but you need to throw in B
to take care of the second possibility, namely that f(x) = -infinity.
Per
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