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Why not: Given a measurable space $(\Omega, \mathcal{A})$, a function $X:\Omega \to \mathbb{R}$ with Borel $\sigma$-algebra $\mathfrak{B}(\mathbb{R})$, is measurable if $\forall A \in \mathcal{A}, X(A) = B \in \mathfrak{B}(\mathbb{R})$ ?

I'm confident that my version is wrong somehow---very smart people don't write definitions casually---but for the life of me I can't see why my (simpler?) version would not work. Assuming that I'm right about being wrong, a simple example where it doesn't work would be better than a technical explanation as I'm very much a novice.

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    $\begingroup$ Are you familiar with the idea of Lebesgue integration? You need the inverse image of "nice" (measurable) sets to be "nice" in order to carry it out. See e.g., here. Also compare with the notion of "continuous", which requires inverse images of open set to be open, rather than direct images. $\endgroup$
    – Arturo Magidin
    Jul 2 at 23:25
  • $\begingroup$ @ArturoMagidin. I've heard of it but that's about it. Thanks for the link, it looks interesting. If the reason for the $X^{-1}(B)$ etc., is an aspect of this subject I haven't seen yet then I'm okay to wait until I get there. Maybe you could post something like that as an answer and I'll close this out, $\endgroup$
    – TonyK
    Jul 2 at 23:40
  • $\begingroup$ I had a follow up thought: if it weren't for the Lebesgue integral, would my version be a reasonable definition? In other words, if all we ultimately want to do is ``push'' a measure from one measurable space to another, would my version work? $\endgroup$
    – TonyK
    Jul 2 at 23:51
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    $\begingroup$ The whole point of measure theory is integration. Saying "if it weren't for the Lebesgue integral, could we do it this way" is kind of like asking "were it not for wanting them to fly, could we design airplanes so they are giant cubes instead?" $\endgroup$
    – Arturo Magidin
    Jul 2 at 23:56
  • $\begingroup$ What about probability -- if the measurable space has a probability measure on it then the function is a r.v. Doesn't that count? $\endgroup$
    – TonyK
    Jul 3 at 2:28

1 Answer 1

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With your definition the characteristic function of any set would be measurable, since the image $\{1\}$ is Borel.

Now, working with the usual Vitali non-measurable set $E$ , you have a set $E\subset[0,1] $ such that $$ [0,1]=\bigcup_r (E+r), $$ where $r$ runs over the rationals in $(0,1)$.

Now, the whole point of measure theory is to define integration. An integral should be linear and, when trying to generalize Riemannian integration, translation invariant. Also, the spirit of an integral is that for a positive function on the real line it should be "the area under the curve" or, more generally, of the form "value of the function times size of the region". So you would want $$\tag1 \int_{\mathbb R}1_{[0,1]}=1=\sum_r\int_{\mathbb R}1_{E+r}. $$ The problem is that if the integral is going to be translation invariant, then all integrals on the right should be equal; if they are nonzero then the sum is infinite; and if they are zero then the sum is zero. So there is no way to make the equality $(1)$ to work.

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  • $\begingroup$ "Characteristic function". Edited. $\endgroup$
    – Martin Argerami
    Jul 3 at 2:40
  • $\begingroup$ Thanks for weighting in. Two comments/questions. You, like Arturo say the whole point of measure theory is integration. Granted that is how it started but random variables play a role in probability (my interest). That was where I was coming from. Second, I thought I'd excluded non-measurable sets by specifying a measurable space as the domain and the real line with Borel as the co-domain. Or does that not get me out of jail? $\endgroup$
    – TonyK
    Jul 3 at 2:54
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    $\begingroup$ It doesn't because you will always want to consider subsets. Any definition of integral will require some kind of subdividing, as you want to obtain a number out of what is happening locally. As for random variables, I guess you will want to do things with them. Like averaging them (which requires integration), analyzing them in terms of a distribution (which requires integration), analyzing its moments (which requires integration). I cannot answer authoratively because my knowledge of probability is minimal, but there is integration everywhere. $\endgroup$
    – Martin Argerami
    Jul 3 at 2:59
  • $\begingroup$ Mmm, more work to be done. Thanks. $\endgroup$
    – TonyK
    Jul 3 at 3:13
  • $\begingroup$ Probability is measure theory restricted to the case where the measure of the entire space is 1. $\endgroup$
    – Teepeemm
    Jul 3 at 11:27

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