Needless to say, I know that the answer is trivially "because that's part of the definition of a random variable", but what I'm really looking for is why that's part of the definition. Why do we want our random variables to be $\mathcal{F}$-measurable? What useful properties would we lose if this was not the case? What would it mean in practical terms for an r.v. to not be $\mathcal{F}$-measurable?


Because formally, the distribution of a random variable is the image measure of the probability measure in the underlying probability space. And the image measure is only defined for measurable functions.

That is, let $(\Omega, \mathbb F, P)$ be a probability space and $(\mathcal X, \mathbb E)$ be a measurable space and let $X: \Omega \to \mathcal X$ be a $\mathbb E - \mathbb F$ measurable map. Then the distribution of $X$ is the image measure $X(P)$.

I am not sure whether this avoids being a simple appeal to definition in your eyes?

More "practically" the problem is simply one of assigning probabilities to events. If the event of interest cannot be assigned a probability then the distribution of $X$ cannot be described.

  • $\begingroup$ I'm tempted to say yes, although in doing so I'd probably have to admit my lack of familiarity with image measures, however if the answer really does boil down to "because if that wasn't the case then we wouldn't even be able to define a random variable", then I'll be forced to accept what you've said as a correct answer. $\endgroup$ – J. Min Oct 28 '17 at 19:08
  • $\begingroup$ @J.Min : I have tried to address the "practical" part. I could construct a trivial example if it helps? It would simply be something to do with $\mathbb F$ not having, say, the event of "two heads" in it, even though the (candidate) random variable would be able to take that value... $\endgroup$ – Ketil B T Oct 28 '17 at 19:12
  • $\begingroup$ A trivial example would be nice, although I seem to already be happy with what you've said. $\endgroup$ – J. Min Oct 28 '17 at 19:14
  • $\begingroup$ Ah, I've just thought through the description of the proposed trivial example. I'm now satisfied, thank you. $\endgroup$ – J. Min Oct 28 '17 at 19:28
  • $\begingroup$ Ketil, perhaps an example would be useful for other users who will read this thread in the future. $\endgroup$ – Richard Hardy Oct 28 '17 at 19:41

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