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A function $f : \mathbb{R} \to \mathbb{R}$ is called Lebesgue-measurable if preimages of Borel-measurable sets are Lebesgue-measurable.

I don't understand why we would pick this definition, rather than saying that a function is measurable if preimages of Lebesgue-measurable sets are Lebesgue-measurable.

In fact, Wikipedia says that

A measurable function is a function between the underlying sets of two measurable spaces that preserves the structure of the spaces: the preimage of any measurable set is measurable

So this would mean that the Lebesgue-measurable functions are morphisms of measure spaces $(\mathbb{R}, \mathrm{Lebesgue}) \to (\mathbb{R}, \mathrm{Borel})$, rather than $(\mathbb{R}, \mathrm{Lebesgue}) \to (\mathbb{R}, \mathrm{Lebesgue})$.

So why do we care more about "Lebesgue-Borel measurable functions" than "Lebesgue-Lebesgue measurable functions", and why do we use the term "Lebesgue measurable" to refer to those rather than Lebesgue-Lebesgue measurable functions?

user735382
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    One is the "analyst" and the other is the "probabilist" definition of measurability. In many cases of practical use, they coincide. – Giuseppe Negro Dec 16 '19 at 15:58
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    The main reason is that with the other definition, nice functions, like continuous functions, can be non-measurable. For example. – egorovik Dec 16 '19 at 16:13
  • in the more general framework of the Bochner integral a $\mu$-measurable function is defined as the almost everywhere pointwise limit of a sequence of simple functions. It can be shown that if $f$ is $\mu$-measurable in this sense then it is also measurable in the sense of $f^{-1}(A)$ is $\mu$-measurable for every Borel set $A$ of it codomain – Masacroso Dec 16 '19 at 17:13
  • See also https://mathoverflow.net/q/31603/4832 – Nate Eldredge Jun 18 '20 at 19:21

2 Answers2

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As egorovik said in the comments, the problem is that there aren't enough Lebesgue-Lebesgue measurable functions to actually do analysis, because not all continuous functions are Lebesgue-Lebesgue measurable. Namely if you define the functions

  • $f : [0,1] \to [0,1]$ is the Cantor function

  • $g : [0,1] \to [0,2],g(x)=f(x)+x$

  • $h : [0,2] \to [0,1],h=g^{-1}$

then $h$ is a continuous function with the property that there is a measurable subset of $[0,1]$ such that $h^{-1}(A)$ is not measurable. This $A$ can be given as $g^{-1}(B)$ where $B$ is any nonmeasurable subset of $g(C)$, where $C$ is the Cantor set.

The defect in the Lebesgue-Borel definition is that the composition of measurable functions isn't measurable...but it is surprisingly rare for this to be a problem.

Ian
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One reason is that the condition that preimages of Borel-measurable sets are Lebesgue-measurable is a weaker condition than the condition that preimages of Lebesgue-measurable sets are Lebesgue-measurable. One can find functions that are "Lebesgue-Borel measurable" but not "Lebesgue-Lebesgue measurable" (see the comment by egorovik), but not vice versa.

This also means that it is easier to check if a function is "Lebesgue-Borel measurable" than "Lebesgue-Lebesgue measurable" (In this case one only needs to check preimages of sets of the form $(-\infty,c)$ for $c\in\mathbb R$, because these sets generate the Borel measurable sets).

Generally, we consider measurable functions because we want to do measure theory (integration theory, etc.). It turns out, that this theory works for both "Lebesgue-Lebesgue measurable functions" and "Lebesgue-Borel measurable functions".

Thus, if we establish a theory for "Lebesgue-Borel measurable functions", we have established a theory for a larger class of functions. Mathematicians like it if their theory works for a larger class of objects.

supinf
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