Let $X:\mathbb{R}\rightarrow\mathbb{R}$ be a non-constant function, measurable with respect to the Borel-Algebra $\mathcal{B}$ and $\sigma(X)$ the sigma-algebra generated by $X$. Let $\mathcal{A}\subsetneqq\sigma(X)$, $\mathcal{A}\neq\{\emptyset, \mathbb{R}\}$ be a sub-sigma algebra of $\sigma(X)$.


Does there always exist a $\mathcal{B}$-measurable function $Y:\mathbb{R}\rightarrow\mathbb{R}$ such that

  • $\sigma(X)\cap\sigma(Y)=\mathcal{A}$
  • But neither is $\sigma(X)\subset\sigma(Y)$ nor $\sigma(Y)\subset\sigma(X)$.

It was easy for me to find a few specific examples where $\sigma(X)$ has only a finite number of elements, but I am interested in the general case.

EDIT: The example given by Eric answers the question as stated. But due to his comment and answer I noticed that I actually would like to understand something slightly(?) different. I am really interested in the case where $\mathcal{A}$ is also generated by a function, i.e. $\mathcal{A}=\sigma(f\circ X)$. The countable-uncountable sigma-algebra is not generated by a function since it is not countably generated (see here) but a sigma-algebra generated by a function is.

  • 1
    $\begingroup$ Your "equivalent formulation" seems quite different from the original question. How does $f(X)=g(Y)$ say that $\sigma(X)\cap\sigma(Y)=\mathcal{A}$? $\endgroup$ Commented Jul 11, 2016 at 20:50
  • $\begingroup$ Yes, Thanks for pointing this out! I naively assumed that every $\mathcal{A}$ being a sub-algebra were also generated by a function. Now that I start to understand your answer, I see this is not correct. I need to think about this. Probably I need to edit/revise the original question as well. $\endgroup$
    – g g
    Commented Jul 12, 2016 at 8:40

1 Answer 1


This is certainly not always possible. For instance, if $X$ is the identity function, then $\sigma(X)=\mathcal{B}$, so for any Borel-measurable $Y$, $\sigma(Y)\subseteq\sigma(X)$.

For a slightly less trivial example, let $X$ be given by $X(t)=t$ unless $t=0$ and $X(0)=1$. Then $\sigma(X)$ is the algebra of all Borel sets that can't distinguish $0$ and $1$. Let $\mathcal{A}$ consist of all elements of $\sigma(X)$ that are either countable or cocountable. If $Y$ is Borel-measurable and $\mathcal{A}\subset\sigma (Y)$, then $Y$ must be injective on $\mathbb{R}\setminus\{0\}$; in particular, all the fibers of $Y$ must be countable.

Now let $$r=\inf\{r\in\mathbb{R}:Y^{-1}((r,\infty))\text{ is countable}\}$$ and $$s=\inf\{s\in\mathbb{R}:Y^{-1}((-\infty,s))\text{ is countable}\}.$$ Then clearly $s\leq r$, and in fact $s<r$ since if $r=s$ then the fact that $Y^{-1}(\{r\})$ is countable would imply $Y^{-1}(\mathbb{R})$ is countable. Choose $t\in(s,r)$ and let $S=(-\infty,t)\setminus\{Y(0),Y(1)\}$. Then $Y^{-1}(S)$ is Borel, is neither countable nor cocountable, and does not distinguish $0$ and $1$. Thus $Y^{-1}(S)\in\sigma(X)\setminus\mathcal{A}$. In particular, this shows that $\sigma(Y)\cap\sigma(X)$ cannot be equal to $\mathcal{A}$ for any Borel-measurable $Y$.


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