12
$\begingroup$

Let $f:X \rightarrow (Y, \mathcal{Y})$ be an abstract function, with $\mathcal{Y}$ a $\sigma$-algebra on $Y$. Endow $X$ with $f^{-1}(\mathcal{Y})$. Is then $f(X)$ a measurable set in $Y$? If not, are there simple conditions on $f$ making $f(X)$ measurable? If $\mathcal{Y}$ were a $\sigma$-ring, would this modify anything?

More concisely (and generally): when is the image of a measurable set under a measurable function a measurable set?

Improve this question
$\endgroup$
1
  • 6
    $\begingroup$ Take $f$ to be the inclusion map of any non-measurable set to get a negative answer for the first question. $\endgroup$
    – user1688
    Commented Jun 8, 2013 at 20:17

4 Answers 4

Reset to default
13
$\begingroup$

There are actually positive results if you change the context a little bit. Suppose that $X$ is a separable complete metric space, i.e., a Polish space, and assume that $Y$ is something like $\mathbb R^n$, a Polish space that carries a measure that interacts nicely with the topology like the Lebesgue measure.

Now, if $f:X\to Y$ is Borel measurable, then for every Borel set $B\subseteq X$ the image $f[B]$ is not necessarily Borel in $Y$, but it is Lebesgue measurable in $Y$.

Improve this answer
$\endgroup$
3
  • 2
    $\begingroup$ Can you provide a reference? $\endgroup$
    – Dan Glasscock
    Commented Sep 25, 2013 at 13:59
  • 2
    $\begingroup$ Kechris' Classical Descriptive Set Theory should contain all the relevant information. The key word here is "analytic set". A set is analytic if it is a continuous image of a Borel set in a Polish space. Since graphs of Borel measurable functions are Borel, the image of a Borel set under a Borel measurable function is analytic, being the projection of a Borel subset of the graph of the function. And analytic sets are Lebesgue measurable (see the wikipedia entry on analytic sets.) $\endgroup$
    – Stefan Geschke
    Commented Dec 3, 2013 at 7:51
  • 1
    $\begingroup$ I believe the above comment considers only Borel measurable functions $f:V_1\rightarrow V_2$ where $V_1, V_2$ are Polish spaces, so $V_1 \times V_2$ is a Polish space, and the "continuous function" is the projection map $h:V_1\times V_2\rightarrow V_2$ given by $h(v_1,v_2)=v_2$, which can project the graph set $\{(v,f(v)): v \in V_1\}\subseteq V_1 \times V_2$ to the set $f(V_1)$. $\endgroup$
    – Michael
    Commented Apr 1, 2022 at 20:47
3
$\begingroup$

Consider $\mathcal{Y}=\left\{\emptyset,Y\right\}$. Every non-surjective function from $f$ on a non-empty set $X$ is measurable, but the image of any non-empty subset is not measurable. Using similar constructions you can get many counterexamples. I would not expect any nice, simple and non-trivial conditions in the case of general $\sigma$-algebras or $\sigma$-rings.

Improve this answer
$\endgroup$
1
$\begingroup$

Let $X$ and $Y$ be standard Borel spaces, and $f : X \to Y$ be Borel.

If $A \subseteq X$ is Borel, and $f | _A : A \to Y$ is injective, then $f(A)$ is Borel in $Y$.

(This is corollary 15.2 in "Classical descriptive set theory" by Kechris.)

Improve this answer
$\endgroup$
1
$\begingroup$

It is well known that for any continuous function $f$, $f$ sends all measurable sets to measurable ones if and only if $f$ has Luzin's-(N)-property. I.e. $f$ sends all null sets to null ones. By [1], Luzin's-(N)-property is a $\Pi^1_1$-complete property. So there is no simple description.

[1].P. Holick ́y, S. P. Ponomarev, L. Zaj ́ıˇcek & M. Zelen ́y (1998/99): Structure of the set of continuous functions with Luzin’s property (N). Real Anal. Exchange 24(2), pp. 635–656.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .