Lebesgue integral once again

  • Thread starter r4nd0m
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  • #1
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Main Question or Discussion Point

I have one more question about the Lebesgue integral:

What if we defined the Lebesgue integral like this:

Let X be a measurable space and f any nonnegative function from X to R.

Then the Lebesgue integral of f as [tex]\int_X f d\mu = sup(I_X)[/tex] where [tex]I_X[/tex] is the integral of a simple function and the sup is taken over all simple measurable functions on X, such that 0<=s<=f.

As you see this definition is the same as the original, except, that the assumption that f is measurable is missing.

My question is: What would be wrong with this definition?
 
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Answers and Replies

  • #2
HallsofIvy
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If f is not measurable, then sup(Ix) will not exist.
 
  • #3
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Why shouldn't the supremum exist?
 
  • #4
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Why shouldn't the supremum exist?
What would be the meaning of a simple integral over a non-measurable set?

The set is non-measurable, so we cannot apply the definition, and [tex]I_{X}[/tex] is empty and thus has no supremum!
 
  • #5
StatusX
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You could define it like that for any functions, but you'd lose nice properties, like linearity and the convergence theorems.
 
  • #6
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What would be the meaning of a simple integral over a non-measurable set?

The set is non-measurable, so we cannot apply the definition, and [tex]I_{X}[/tex] is empty and thus has no supremum!
The set IS measurable, the FUNCTION is not measurable.

You could define it like that for any functions, but you'd lose nice properties, like linearity and the convergence theorems.
Thanks, this seems to be reasonable.
 
  • #7
Hurkyl
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More concretely, I would expect your definition, for nonmeasurable functions, to have bad behavior akin to Lebesgue inner measure for nonmeasurable sets.
 
  • #8
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More concretely, I would expect your definition, for nonmeasurable functions, to have bad behavior akin to Lebesgue inner measure for nonmeasurable sets.
I don't really understand what you mean, can you give an example?

To StatusX:
Can it be proven, that if f is not measurable, then the integral is not linear?

Basically, what I want to know is - I try to imagine that I'm in the position of Henri Lebesgue and I have to define a new kind of integral as general as possible. The definition of the measure seems to be very natural.

But I don't understand how did he come to the definition of a measurable function.
I see, that there are no problems with that definition and that we get many nice properties from it, but I think that the natural question is, can we make it more general and still keep the nice properties?
Or can the opposite be PROVEN, that if we changed the definition we would lose the properties?
 
  • #9
Hurkyl
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The definition of measurable function is "obvious". There is an interesting class of sets (measurable sets), so one naturally wants to know what sorts of functions play nice with them.

open sets : continuous functions :: measurable sets : measurable functions

From this, one (maybe) can intuit why measurable functions are precisely the functions that behave nicely w.r.t. integration.


Incidentally, here's a simle class of nonmeasurable functions that might help you build counterexamples: for any nonmeasurable set, its characteristic function is nonmeasurable.
 
  • #10
jle
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need measurable set

Lebesgue integral of a function is based on measure of sets defined by inverse images. the inverse image need to belong to the sigma algebra on which the measure is defined. A function is not measurable with respect to a measure if the inverse image does not belong to the sigma algebra. In that case you cannot give a measure of this inverse image and therefore the integral cannot be calculated.
 

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