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I was looking at proofs of Jensen's inequality and noticed that they usually assume that for a convex function $g$ and a random variable $X$, the expression $\mathbb{E}(g(X))$ is well defined, which implies that $g(X)$ is a random variable.

I was thinking whether it was possible that all real-valued functions of random variables are random variables themselves, which is why the proofs skip the step of proving that $g(X)$ is a random variable.

But then I thought if $A$ is a Borel set, then $(g \circ X)^{-1}(A) = X^{-1}(g^{-1}(A)) $ may not itself be $\mathcal{F}$-measurable if $g^{-1}(A)$ is not Borel. But it is true if convex functions $g:\mathbb{R} \to \mathbb{R}$ are $\mathcal{B}$-measurable.

Is it true that all convex functions are $\mathcal{B}$-measurable, then? I know the sub-level sets of a convex function are convex, so if we can show that a convex set is Borel, that would work too. And is there a counterexample of a real-valued function of a random variable that isn't a random variable?

Ilham
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  • Are you worried about ending up with a non-measurable function, or something like that? – Gregory Grant May 13 '15 at 20:55
  • @GregoryGrant Yes, I am worried that the function $g \circ X$ is not $\mathcal{F}$-measurable (when $X$ is $\mathcal{F}$-measurable). I don't know whether left compositions of $X$ with any real-valued function $g$ will make $g \circ X$ measurable, or if we can only say that if we have $\mathcal{B}$-measurable $g$ - I would also like to know whether all convex, real functions $g$ are $\mathcal{B}$-measurable. – Ilham May 13 '15 at 23:18
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    Yes, all convex functions are measurable. In fact it is true that any convex function can be written as the sup of a countable collection of linear functions, i.e, there exist $a_n,b_n \in \mathbb{R}$ with $g(x) = \sup_x (a_nx+b_n)$. Since the sup of a countable collection of measurable functions is measurable, it follows that $g$ is measurable. – shalin May 13 '15 at 23:37
  • @shalop thanks for the hint! So can a composition of a function with a random variable not be measurable? – Ilham May 13 '15 at 23:39
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    What do you mean? Yes, a composition of a non-measurable function with a random variable can be non-measurable. But when you have a convex function, it is measurable (by my previous comment), so its composition with a random variable will be measurable... – shalin May 14 '15 at 03:55
  • Random variable $\equiv$ measurable function. – Did May 14 '15 at 06:49

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