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JohnyBoy
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Post Number: 1
Posted on Tuesday, 29 May, 2012 - 01:12 pm:   

Does anyone have an idea of how to do the following measure theory question:

Let epsilon = {empty set, X} and let mu be defined by mu(X) = 1,
mu(empty set) = 0. Find all the measurable functions for the sigma algebra epsilon.
lebesgue
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Post Number: 2797
Posted on Tuesday, 29 May, 2012 - 01:29 pm:   

Welcome to Nrich! Can you find at least one function f:X->R that is measurable and to show that it is measurable? Can you find at least one function g:X->R that is not measurable and to show that it is not measurable?
JohnyBoy
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Post Number: 2
Posted on Tuesday, 29 May, 2012 - 01:43 pm:   

Thanks. From what I understand the idea in proving that a function f:X->R is measurable is to show that f^-1((t, infinity)) is measurable for any t element of R.

So for a measurable function I would say that the constant function is an example of a measurable function since if f(x) = c, where c is a constant, then f^-1((t,infinity)) = X or f^-1((t,infinity)) = empty set, depending on the value of t. Am I on the right track? Thanks.
lebesgue
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Post Number: 2798
Posted on Tuesday, 29 May, 2012 - 01:55 pm:   

You are right, constant functions are measurable. (This bit would be true for EACH sigma algebra, not only for this trivial one.) Can a nonconstant function g:X->R that attains at least two values a and b, a<b, be measurable given that your sigma algebra is rather trivial?
JohnyBoy
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Post Number: 3
Posted on Tuesday, 29 May, 2012 - 02:08 pm:   

No, I don't think so, since if we choose t element of R such that a < t < b then g^-1((t, infinity)) would not give X or the empty set, so g would not be measurable.
lebesgue
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Post Number: 2799
Posted on Tuesday, 29 May, 2012 - 02:14 pm:   

Well done; so you discovered that only constant functions are measurable with this sigma algebra.
JohnyBoy
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Post Number: 4
Posted on Tuesday, 29 May, 2012 - 02:19 pm:   

Great, thanks!

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