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Definition of Random Variable: Suppose $(\Omega,\Sigma,\mathbb P)$ is a probability space. If $\mathbf Y : \Omega \mapsto \mathbb R$ is measurable w.r.t. the Borel $\sigma$-algebra on $\mathbb R$, then $\mathbf Y$ is called a Random Variable. The $\sigma$-algebra generated by $\mathbf Y$ is: $\sigma(\mathbf Y)=\{ \mathbf Y^{-1}(A):A\in\mathscr B(\mathbb R)\}\subset\Sigma.$

Definition of Independent Random Variables: Two random variables $\mathbf X$, $\mathbf Y$ are independent if events A and B are independent (i.e. $\mathbb P(A\cap B)=\mathbb P(A)\mathbb P(B)$), whenever $A\in\sigma(\mathbf X)$ and $B\in\sigma(\mathbf Y)$.

Question 1: Why do we only consider $\mathscr B(\mathbb R)$ here? The measurable collection of set on $\mathbb R$ could be extended beyond the Borel sets if we give a measure (e.g. Lebesgue measure). Although the extension is to include null-sets, we get a lot more sets (by intersection/union of null-sets with Borel sets), why do we do not require $\mathbf Y$ to be measurable over them as well? Plus, even if N is a null-set on $(\mathbb R,\mathscr A,\mu)$, it does not mean $\mathbf Y^{-1}(N)$ is a null-set on $(\Omega,\Sigma,\mathbb P)$ - it could has a positive probability, and thus is not boring/trivial.

Question 2: When it comes to independence of $\mathbf X$ and $\mathbf Y$, why do we only consider events in $\sigma(\mathbf X)$ and $\sigma(\mathbf Y)$? Why do we not care about events such as $L\in(\Sigma\setminus\sigma(\mathbf X))$ and $M\in(\Sigma\setminus\sigma(\mathbf Y))$? (i.e. $\mathbb P(L\cap M)\neq\mathbb P(L)\mathbb P(M)$ does not affect the independence of $\mathbf X$ and $\mathbf Y$). Note that for $L\in(\Sigma\setminus\sigma(\mathbf X))$ and $M\in(\Sigma\setminus\sigma(\mathbf Y))$ with $\mathbb P(L\cap M)\neq\mathbb P(L)\mathbb P(M)$, although $L\notin\sigma(\mathbf X)$, it is still somehow relevant to $\mathbf X$, because $\mathbf X$ is defined on every point of the set L, and maps those points to $\mathbb R$. For example, there could be set $P\in\sigma(\mathbf X)$ and $P\cap L\neq\emptyset$, so X could be defined on some points of $L$ despite of the fact that $L\notin\sigma(\mathbf{X})$.

UPDATE: summarize answers + my thoughts

Big thanks for both answers, very helpful!

For Question 1: To summarize the answers: examples were given about a mapping between probability space (which is e.g. $(\mathbb R,\mathscr M,\mathcal{L})$) to $\mathbb R$. And that mapping will map a $\mathcal{L}$-measurable set inversely back to a non-$\mathcal{L}$-measurable set in probability space. For example (from Halmos's Measure theory §19. Problem 3.) let f(x) = 0.5*(x + c(x)), where c(.) is a Cantor function. And then let $C$ be the Cantor set, $\exists A \subset C$ s.t. $A$ is $\mathcal L$-measurable but not a Borel set, and $F^{-1}(A)$ is not $\mathcal L$-measurable (Refer to this post). Extended discussion: the above issue could be solved by two ways: 1) we give up Lebesgue sets but go with Borel sets (which is exactly our definition); but we could also do 2) Maintain Lebesgue sets, but give up those mappings as a valid random variable. In terms of why is it important to keep functions like above (i.e. maps Lebesgue measurable sets inversely back to a non-measurable set) to be a random variable, and thus we eventually choose 1) over 2)? - Please see @zoli's comments below

For Question 2: To summarize the answers: The parity example by @zhoraster is a good one, for illustrating "independence meaning probability able to multiply" v.s. "normal meaning of generally not dependent"), but this is NOT the issue confusing me. What really bothers is that the property (independence) of v.r. is defined based on $\mathit subsets$ of $\Omega$; yet v.r. itself based on $\mathit elements$ of $\Omega$. @zoli mentions wiki does give definition of independence with intervals and cdf, but per @Did's comment, and I also personally feel that, independence itself does not necessarily need to rely on those concepts - instead, could be just by the independence of $\sigma$-algebra generated by the r.v. Extended discussion: I think @zoli's comment could be close, that "Events not belonging to the $\sigma$−algebra generated by a random variable cannot be described using only statements about the random variable." - My own thoughts on this: although we define $\mathbb P$ over $\Omega$ for all elements in $\Sigma$, what could really be carry through the r.v. $\mathbf X$ is $\mathbb P_X(A)=\mathbb P(\{\omega|\mathbf X(\omega)\in A\})$, where $A\in\mathscr{B}(\mathbb R)$, thus set $L$ or $M$ above might have a measure under $\mathbb P$, we cannot reflect that via $\mathbb{P}_X$ or $\mathbb{P}_Y$. Further, if we do care about those events, we'll pick a r.v. $\mathbf Z$ s.t. $L\in\sigma(\mathbf Z)$. BTW, from wiki: "the underlying probability space $\Omega$ is a technical device...In practice, one often...just puts a measure on ${\mathbb {R} }$...".

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  • $\begingroup$ I could not find any better explanation: (1) In the case of real to real functions, defining measurability with sets like $(-\infty,x)$ or $(-\infty,]$ coincides with the definition based on the inverse images of Borel sets. (2) Defining measurability with the inverse images of Lebesgue sets leads to problematic cases. As far as strange functions. Those exist we cannot exclude them. $\endgroup$
    – zoli
    Feb 27, 2017 at 8:56
  • $\begingroup$ One more thing. The class of Baire functions is the smallest set of functions that contains all the continuous functions and is closed wrt limits. That is, if $\{f_n\}$ is a convergent sequence of Baire functions then the limit is also a Baire function. Theorem: The class of Borel measurable functions equals the class of Baire functions. (Gikhman- Skorokhod, The Theory of Stochastic Processes 2.2. Theorem 4.) $\endgroup$
    – zoli
    Feb 27, 2017 at 9:54
  • $\begingroup$ Thanks @zoli, and I think I'm convinced! $\endgroup$
    – Jay Zha
    Feb 27, 2017 at 15:27
  • $\begingroup$ I've updated the question to summarize the answers and also plus my own thoughts. Thanks all! $\endgroup$
    – Jay Zha
    Feb 27, 2017 at 15:35

2 Answers 2

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EDITED

On the definition of measurability of real valued functions

The serious question is related to the definition of measurability: Let's condider the measurable space $[\Omega,\mathcal A]$ and the real valued function $f$. Why do we demand only that the inverse images of Borel sets be in $\mathcal A$?

Answer:

Let $[\Omega,\mathcal A]=[\mathbb R, \mathcal L]$ and let $$f:[\mathbb R, \mathcal L]\to \mathbb R.$$

And let's define the Lebesgue measurability based on the inverse images of Lebesgue sets.

There is a Caveat. Furthermore, there exists at least one function and there exists at least one Lebesgue measurable set whose inverse image is not Lebesgue measurable. (See also: Halmos's Measure theory §19. Problem 3. and this Post.) With the inverse images of Borel sets such problems do not arise.

Independence of real random variables

The definition of independence of real valued random variables is independent of either the Borel measurable sets or the Lebesgue measurable sets. It is Based on the inverse images of intervals like

$$ \ (-\infty,x]\ \text{ or }\ (-\infty,x).$$

This is because the independence of random variables depends only on the behavior of their cdf's (common and individual) and $f$.

(The theorem behind the statements above claims the equivalence of the two definitions of measurability in the case of real valued functions (1): with inverse images of Borel sets and (2) with intervals like above. See Halmos, §18, Theorem A.

Events not belonging to the $\sigma-$ algebra generated by a random variable cannot be described using only statements about the random variable. So the probability of those events will not arise when talking about independence.

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    $\begingroup$ The reduction of independence to intervals that you suggest is strange. Actually, the notion of independence is relevant in a wider setting, for random variables that are not real valued hence, stressing that one can translate it in terms of CDFs in the real valued case is (unnecessary and) misleading. $\endgroup$
    – Did
    Feb 24, 2017 at 8:31
  • $\begingroup$ @Did: I am just using the statement in wiki (en.wikipedia.org/wiki/…) claiming that for real valued random variables the two definitions (with Borel sets and with cdf's) are equivalent. $\endgroup$
    – zoli
    Feb 24, 2017 at 9:19
  • $\begingroup$ Thanks @zoli! I'll check for your answer. $\endgroup$
    – Jay Zha
    Feb 25, 2017 at 4:56
  • $\begingroup$ I've updated my question by summarizing your answers, adding my new thoughts and understanding, and also ask two follow-up questions. Could you help to take a further look, I think I'm close. thx. @zoli $\endgroup$
    – Jay Zha
    Feb 26, 2017 at 6:03
  • $\begingroup$ @YujieZha: I will read your edit an try to answer. $\endgroup$
    – zoli
    Feb 26, 2017 at 9:17
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Answer to 1: First, the Borel $\sigma$-algebra(s) is perhaps the most natural when it comes to probabilistic questions. Speaking of random variables/vectors/functions/elements in $\mathcal X$, one usually does not care if there is some "natural" measure on $\mathcal X$ like the Lebesgue measure.

Moreover, trying to define probabilities of events $\{X\in A\}$, where $A$ is Lebesgue measurable, is potentially dangerous. Take, for example, some Lebesgue measurable non-Borel set $B$ and set $A = C(B)$, where $C$ is the Cantor staircase. Then $A$ is Lebesgue measurable, but it is not possible to define the probability of $\{X\in A\}$, where $X$ has distribution corresponding to $C$ (e.g., $F_X(x) = C(x), x\in[0,1]$).

Answer to 2: We don't care of those events since we want to define independence of $X$ and $Y$; we don't want anything not related to $X$ and $Y$ to be involved in the definition.

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  • $\begingroup$ For Answer 1 @zhoraster, (correct me if I am wrong) I feel the existence of probability of {X∈A} does not ultimately rely on X or A, but it relies on the probability space (Ω,Σ,P). So if $X^{-1}$ maps A back to a subset of Ω that belongs to Σ, then the probability exists, otherwise it does not exist. For the example you gave, all non-zero probability concentrates in the cantor set. But I still did not get it why not possible to define P(X∈A) - could you explain? To me I feel if X is defined in a way that {w: X(w)∈A} ∈ Σ, then it should have a probability.. $\endgroup$
    – Jay Zha
    Feb 25, 2017 at 4:53
  • $\begingroup$ For Answer 2 @zhoraster, actually I cannot see why set L is not related to r.v. X. I understand that set L is not related to σ(X), but it is still related to X (and even defining X, to some extent): X is defined as a map that relates all points w∈Ω with some points in the real line. Thus although L is not in σ(X), X is defined on every point of the set L. And so is Y. $\endgroup$
    – Jay Zha
    Feb 25, 2017 at 4:54
  • $\begingroup$ @YujieZha, the Cantor staircase on $[0,1]$ is in fact a Borel isomorphism between $[0,1]$ and the Cantor set; moreover, the Lebesgue measure is mapped to Cantor measure. Therefore, the set $A$ isn't (Caratheodory) measurable w.r.t. the Cantor measure, as it is an image of a non-Lebesgue measurable set. $\endgroup$
    – zhoraster
    Feb 25, 2017 at 15:49
  • $\begingroup$ Concerning 2, I fail to see your point, sorry. Suppose you flip a coin and you want to test the independence of result from something else, say, presidential election in the US. You flip, say, tails and look whether this affects probability of outcomes of the election. You don't need to look at other things like temperature in Alaska. $\endgroup$
    – zhoraster
    Feb 25, 2017 at 16:24
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    $\begingroup$ I suspect that you are misinterpreting the statistical notion of independence (which is different from common understanding). For example, if we flip the coin twice, then the parity of the number of heads is independent of the first outcome, in the probabilistic sense. It also independent of the second outcome. But it clearly depends on both outcomes. $\endgroup$
    – zhoraster
    Feb 25, 2017 at 16:25

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