Ashley P.

asked • 06/01/23

Measurable Functions

Question:


Let (X,@) be a measurable space.


Prove that if for each a in R, {x in X | f(x) <= a } is measurable, then f is measurable.


My approach


Let a belong to R.


Then,

{x in X | f(x) <= a } = {x in X | f(x) > a }c

since the right hand side is measurable if f is measurable, left hand side implies that f is measurable.


Is this a correct approach?

Roger R.

tutor
The idea is right, but the wording sounds like a circular argument: "If f is a measurable function, the RHS is a measurable set. Therefore, the LHS proves that f is a measurable function." Depending on the definition of "measurable function" you use, you could say something along the lines "Let S be the measurable set {x in X | f(x) <= a }. The complement S' is also a measurable set (because measurable sets form an algebra of sets). But {x in X | f(x) > a} = S', therefore, {x in X | f(x) > a } is a measurable set. Since a is arbitrary, f is a measurable function."
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06/02/23

1 Expert Answer

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Jeremie F. answered • 06/02/23

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Yes this is the correct way to approach this problem since the complement of a measurable set is measurable.

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Ashley P.

I've noticed somewhere that the converse does not hold in general. Since I've done the proof as the converse, is this true?
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06/02/23

Jeremie F.

tutor
A couple of remarks: 1) It is true that in general the converse of a statement does not hold (though in some cases it does think if and only if statements). 2) In this case the two sets you are using are equal, which means they must both have the same properties. Next you note that the complement of the set on the right hand side has a particular property. Finally that property holds in this scenario for complements of sets (this is NOT always true but in this case it is true!). So the logic is sound here. Now the only place where the logic is not "complete" is that you have not proven that {x in X | f(x)>a} is measurable and you also have not proven that the complement of a measurable set is measurable, these are known results however. I am not sure if you have proven those results previously or not, but if you have not, you would need to do so to complete the above proof.
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06/02/23

Ashley P.

Thank you for the comments!
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06/02/23

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