# Random Variables that aren't measurable

I've been reading through a math. stats. book, and I'm a little confused with the concept of measurable random variables. The book states:

Let $(E, \mathcal{E})$ and $(F,\mathcal{F})$ be two measurable spaces. A random variable $X : E → F$ is called measurable (relative to $\mathcal{E}$ and $\mathcal{F}$) if $X^{−1}(\Lambda) \in \mathcal{E}$ for all $\Lambda \in \mathcal{F}$.

Now, here's my question: If some says "$X:(\Omega,\mathcal{A}) → (\mathbb{R}, \mathcal{B})$" without defining $\mathcal{B}$, then I can assume $\mathcal{B}$ is just the image of $X$, right? But, with this assumption, wouldn't all random variables be measurable? It seems like all random variables would be measurable, then, unless $\mathcal{B}$ was constructed in such a way that $X$ would not map onto $\mathcal{B}$. Is that true? Can anyone give me an example of a meaningful scenario where we'd have non-measurable random variables?

The reason I ask is because I'm working on this homework problem:

Given $(\Omega,\mathcal{A}, P)$, let $\mathcal{A}'=\{A ∪ N:A ∈ \mathcal{A},N ∈ \mathcal{N}\}$, where $\mathcal{N}$ are the null sets (as in Theorem 6.4). Suppose $X = Y$ almost surely where $X$ and $Y$ are two real-valued functions on $Ω$. Show that $X: (Ω,\mathcal{A}') → (\mathbb{R}, \mathcal{B})$ is measurable if and only if $Y : (Ω,\mathcal{A}') → (\mathbb{R}, \mathcal{B})$ is measurable.

But, if these two $\mathcal{B}$'s are just defined as the images of the functions, isn't this trivial?

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I guess that $\mathcal{B}$ is meant to be the Borel $\sigma$-algebra on $\mathbb{R}$. –  usul Apr 11 at 21:51
I thought of that too... But, then this seems like a counterexample (and maybe here's where I'm wrong): Let $\Omega=\mathbb{R}$, and let $X$ be the identity mapping. Suppose P([0,1])=1. Then $Y$ could map every number not in [0,1] to 2. Doesn't this satisfy the conditions but violate the statement to be proved? –  Josh Browning Apr 11 at 21:58
Random variables are always measurable. This part of the definition is badly phrased. –  Did Apr 11 at 23:07
Sorry, how is that a counterexample? Explaining it may help. –  Sharkos Apr 11 at 23:09
In my "counterexample", $X$ is an onto function (any element in $\mathbb{R}$ is mapped to). Also, $X=Y$ almost surely because the two functions agree on all values with non-zero probability. But, $Y$ isn't measurable because no value is mapped to 1.5, for example. –  Josh Browning Apr 11 at 23:18

First of all, if you have the same $\cal B$ in two places within the same context then they represent the same object. So it is probably safe to assume it is not something defined by $X$ itself (then $Y$ would have to define something slightly different, which would be weird).

Secondly in the context of probability and measure theory, it is usually the case that $\cal B$ denotes the Borel $\sigma$-algebra of the space, in this case $\Bbb R$.

Lastly it seems that you have taken this from some text somewhere. Look in the text preceding your quote and find out what does $\cal B$ stands for. If this was taken from a homework assignment, send an email to your TA or professor and ask for clarification.

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Thanks for the response! Here's my responses: 1. Sure, I'd agree that the two $\mathcal{B}$'s seem like they should be the same. 2. Yeah, that makes sense. But, I'm having a hard time believing this statement then, because I don't see a reason why $X$ or $Y$ would have to map onto all of $\mathcal{B}$, or why one mapping onto all of $\mathcal{B}$ implies the other does. 3. Yes, this is from a text. But, the problem doesn't clarify what $\mathcal{B}$ is, and I'm working through this textbook for my own education, not a course. So, I'm outta luck there :) –  Josh Browning Apr 11 at 23:23
Josh, let me work backwards. (3) if you are working through a textbook then you can definitely find the definition of $\cal B$ somewhere in there. Look for a symbol index at the end, or perhaps for "Borel" in the general index and look for a notational convention. (2) There is no need for $X$ to be surjective or anything. We just require that whenever we take a Borel set, its preimage is from the $\sigma$-algebra associated with the domain of $X$. –  Asaf Karagila Apr 11 at 23:26
So, if I'm understanding correctly, $X$ being measurable basically means this: If $B \in \mathcal{B}$ and $X(A)=B$ for some $A \in 2^\Omega$, then $A \in \mathcal{A'}$. My confusion is that I had thought measurable meant that: For all $B \in \mathcal{B}$, there exists some $A \in \mathcal{A}$ such that $X(A)=B$. –  Josh Browning Apr 12 at 11:45
@Josh: No. This is not what this means. It is possible that there is some $A\notin\cal A'$ such that $X(A)\in\cal B$. What we do require that whenever $B\in\cal B$ then $\{a\mid X(a)\in B\}\in\cal A'$. If $X$ is not injective then there is a possible difference between the two notions. –  Asaf Karagila Apr 12 at 11:46
Got it! Thanks so much! –  Josh Browning Apr 12 at 11:56