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I write $\mathcal{M}_{Bor}, \mathcal{M}_{Leb}$ for the Borel/ Lebesgue $\sigma$- algebras on $\mathbb{R}$.

Let $A \in \mathcal{M}_{Leb}$. Let $f: A\to \mathbb{R}$. Then $f$ is Lebesgue- measurable iff $ \forall B \in \mathcal{M}_{Bor}: f^{-1}(B) \in \mathcal{M}_{Leb}$.

What goes wrong if we instead require $ \forall E \in \mathcal{M}_{Leb}: f^{-1}(E) \in \mathcal{M}_{Leb}$, for $f$ to be measurable? Superficially this definition seems more natural, so why isn't it used?

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    It can become the case that with such a definition there exist continuous functions that are not measurable. – rubikscube09 Jun 18 '20 at 17:30
  • Devils staircase perhaps? –  Jun 18 '20 at 17:31
  • close - consider $g(x) = f(x) + x$ where here $f$ is the devil's staircase. – rubikscube09 Jun 18 '20 at 17:32
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    And $g^{-1}$ will take a zero-measure set into an unmeasurable set. – Zerox Jun 18 '20 at 17:34
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    Is there any intuition for this? Why might one expect it to be the case that the $\forall E \in \mathcal{M}{Leb}: f^{-1}(E) \in \mathcal{M}{Leb}$ definition causes this problem? –  Jun 18 '20 at 18:39
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    The problem is that we have a lot of measurable subsets (it's in bijection with all the subsets of $\mathbb{R}$), and they are not really that well-behaved. Therefore, it's a somewhat arbitrary to ask for a function to have the property that the preimage of measurable subset to be measurable. Then, asking the preimage of a borel subset to be measurable to be measurable is equivalent to ask that the preimage of an open subset is measurable, which is reasonable. – QuinnLesquimau Jun 18 '20 at 19:04
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    One reason we should worry about the $(\mathcal{M}{Leb},\mathcal{M}{Leb})$-approach is that with respect to $\mathcal{M}{Leb}$, null sets are "too good." Specifically, no subset of a null-set is non-measurable. This means that no bijection $b$ between a positive-measure set $S$ and a null set $N$ can be $(\mathcal{M}{Leb},\mathcal{M}{Leb})$-measurable: consider $b[A]$ for $A\subseteq S$ non-measurable. Since there are continuous bijections between some positive-measure sets and some null sets, continuous functions won't in general be $(\mathcal{M}{Leb},\mathcal{M}_{Leb})$-measurable. – Noah Schweber Jun 18 '20 at 19:05
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    (By contrast, there is no such "too good"-ness phenomenon in the class of Borel sets: you can't be "Borel by accident.") More abstractly, note that Borel-ness is a purely topological property whereas Lebesgue measurability isn't - so there's no reason to expect $(\mathcal{M}{Leb},\mathcal{M}{Leb})$-measurability to "play well" with even relatively simple topological notions like continuity. – Noah Schweber Jun 18 '20 at 19:07
  • I see! Does $\mathcal{M}_{Leb}$ form a topology on $\mathbb{R}$? –  Jun 18 '20 at 19:09
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    @user No - every singleton is measurable, but an arbitrary union of singletons need not be measurable. Fundamentally measure is about more than just topology. – Noah Schweber Jun 18 '20 at 19:13
  • OK, so is that why we run into the issues mentioned? –  Jun 18 '20 at 19:14
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    See https://mathoverflow.net/q/31603/4832 and also https://math.stackexchange.com/questions/3478537/why-this-definition-for-lebesgue-measurable-functions?rq=1 – Nate Eldredge Jun 18 '20 at 19:20

2 Answers2

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Elaborating on my comments above:

One reason we should worry about the $(\mathcal{M}_{Leb}, \mathcal{M}_{Leb})$-approach is that with respect to $\mathcal{M}_{Leb}$, null sets are "too good." Specifically, no subset of a null-set is non-measurable. This means that no bijection $b$ between a positive-measure set $S$ and a null set $N$ can be $(\mathcal{M}_{Leb},\mathcal{M}_{Leb})$-measurable: consider $b[A]$ for $A\subseteq S$ non-measurable. Since there are continuous bijections between some positive-measure sets and some null sets, continuous functions won't in general be $(\mathcal{M}_{Leb},\mathcal{M}_{Leb})$-measurable.

More abstractly, we're seeing here that $\mathcal{M}_{Leb}$ and $\mathcal{M}_{Bor}$ are fundamentally different types of object:

  • $\mathcal{M}_{Leb}$ involves more than just the topology of $\mathbb{R}$. Let $C$ be the usual Cantor set and $F$ the fat Cantor set. Then there is an autohomeomorphism $h$ of $\mathbb{R}$ with $h[C]=F$. Consequently, by the above reasoning membership in $\mathcal{M}_{Leb}$ is not "ambient-isomorphism-invariant."

  • By contrast, $\mathcal{M}_{Bor}$ is purely topological: if $B\in\mathcal{M}_{Bor}$ and $h$ is an autohomeomorphism of $\mathbb{R}$ then $h[B]$ is also Borel.

    • Note that we have to be very careful here: the continuous image of a Borel set is not in general Borel! (The continuous preimage of a Borel set is Borel, however, and that's what's at work here.)

    • Also note that I'm not saying that $\mathcal{M}_{Bor}$ constitutes a topology on $\mathbb{R}$ - it doesn't, since all singletons are Borel but not all unions of singletons are Borel. I'm just saying that it is "reducible to" the topology in some sense. Indeed, every topological space comes with a notion of "Borel-ness" (and variations!) while nothing of the sort is true as far as measurability goes.

So in general we should expect Borel-ness to play well with topological concepts, but be suspicious of Lebesgue measurability's topological behavior. And the situation with continuous functions is a good example of one such discrepancy.

Finally, there's another vague theme here besides "topological nature:" structural rigidity. In a sense, some null sets are "measurable by accident" (namely, those which are very topologically different from any positive-measure measurable set). By contrast, nothing is "Borel by accident." This suggests that the class of Borel sets has much better structure overall than the class of Lebesgue measurable sets. This is true, and thinking along these lines takes us into the realm of descriptive set theory, but that's a far ways off. Still, it's worth pointing out here since the "flavor" at least is detectable already.

Noah Schweber
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It has do with the construction of the specific underlying measure (in your case Lebesgue measure).

In the beginning, you start with intervals (step functions) or continuous functions. Then, through some machinery (Daniell's approach or Lebesgue-Charatheodoty approach) you build integration theory (you extend you notion of integral or measure) Then you realize there is much more stuff that what you started with (Lebesgue sets and Lebesgue measurable functions) Step functions and continuous functions are akin to Borel sets (countable opertions with closed/open intervals, preimages of intervals by continuous functions, etc).

If you are to do the same with a different measure $\mu$, that is build integration with respect another measure staring with step functions or continuous functions, then you may get a different extension (different measurable and measurable functions). Which extension is the most natural? In any case, both extensions contain the Borel sets and any other additional set (or function) we obtained can be approximated by Borel sets.

You can also ask about the intersection of all possible extensions one gets starting with step functions (or continuous function). That intersection contains the Borel sets properly, but is much smaller that the collection of Lebesgue sets (of functions). That intersection, the universal $\sigma$-algebra, is the natural extension and it is important in the study of stochastic processes and in Measure theory.

Things become more complicated, but the same principles apply, when you start doing integration in topological spaces.

Mittens
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