2
$\begingroup$

Suppose there is a function $f:A\to B$ where $A,\,B\subseteq\mathbb{R}$, is there any example for this function being NOT Borel function?

Well the question came up to be when I was reading the probability theory and it states that the two random variable is independent if and only if $\mathbb{E}[f(X)g(Y)] = \mathbb{E}[f(X)]\,\mathbb{E}[g(Y)]$ for any $f,\,g$ being Borel function. This is not hard to proof but I cannot think of any function that is not Borel when the function takes real value and also provides real value. Any comment is extremely welcomed :)

$\endgroup$
  • 1
    $\begingroup$ This may be of interest. $\endgroup$ – David Mitra Apr 30 '13 at 20:35
1
$\begingroup$

A couple of silly examples, using a well-ordering of the reals:

There are only $\mathfrak c$ Borel functions whose domain and range are contained in $\mathbb R$ (because Borel functions have Borel graphs, and there are only $\mathfrak c$ Borel subsets of $\mathbb R^2$). List them as $(f_\alpha\mid \alpha<\mathfrak c)$, list $\mathbb R$ as $(x_\alpha\mid\alpha<\mathfrak c)$, and, for each $\alpha<\mathfrak c$, define $f(x_\alpha)=0$ or $1$, whichever is different from $f_\alpha(x_\alpha)$ (including the possibility that $x_\alpha$ is not in the domain of $f_\alpha$). In this example, $f$ only takes the values $0$ and $1$, so it is the characteristic function of a (necessarily, non-Borel) set.

We can also produce an injective example, by letting $f(x_\alpha)=x_\beta$, where $\beta$ is least such that $x_\beta\notin\{f_\alpha(x_\alpha)\}\cup\{f(x_\tau)\mid \tau<\alpha\}$. Since at each stage only fewer than $\mathfrak c$ values have been excluded, $x_\beta$ is defined.

$\endgroup$
  • $\begingroup$ Thanks, great example for someone like me who has yet taken serious measure-theory $\endgroup$ – chooingbobo Apr 30 '13 at 21:08
  • $\begingroup$ For the fact that there are only $\mathfrak c$ Borel sets: math.stackexchange.com/q/70880/462. The fact that Borel functions have Borel graphs is due to Sierpiński, see for example page 124 of A second course on real functions, by A.C.M. van Rooij and W.H. Schikhof, books.google.com/… $\endgroup$ – Andrés E. Caicedo Apr 30 '13 at 22:58
1
$\begingroup$

Royden's book on real analysis discusses a non-measurable subset of the line. The indicator (or characteristic function in analysis parlance) of this subset is not Borel measurable.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.