4

Let $T\colon\mathbb{R}^n \to \mathbb{R}^m$ ($n>m$) be a linear map. Is it true that $T(A)$ is a Lebesgue measurable set for the Borel set $A$?

$T(A)$ for the compact set $A$ is compact set, so is measurable. If $A$ is an open set, then $A$ is countable sum of compact sets and $T(A)$ is measurable. For closed set is the same situation. I don't know what else for other type of Borel sets?

Of course, it may by used the open mapping theorem if we assume that $T$ is surjective.

Mick
  • 41
  • https://math.stackexchange.com/questions/1533077/is-there-a-borel-set-a-and-a-linear-map-f-such-that-fa-is-not-borel –  Jul 23 '19 at 10:03
  • There is another question than mine. I know that $T(A)$ usually is not borel set. – Mick Jul 23 '19 at 10:13

2 Answers2

5

If by measurable set you mean a Borel set then the answer is NO. There exist Borel sets in $\mathbb R^{2}$ whose images under the projection $(x,y) \to x$ are not Borel.

See A Borel set whose projection onto the first coordinate is not a Borel set

Answer for Lebesgue mesaurability: YES, $T(A)$ is Lebesgue measurable. This can be proved using the theory of analytic sets. $A$ is analytic and its image under any Borel measurable map is analytic. So $T(A)$ is analytic. This implies that $T(A)$ is universally measurable, in particular Lebesgue measurable. [Reference: Measure Theory by Cohn].

PS There is easier proof using the fact that $T$ is Lipschitz. I fact we can write $A$ is the union of a null set and a sigma compact set. Since $T$ maps null sets to null sets it follows that $T(A)$ is also of the same type.

Kavi Rama Murthy
  • 359,332
  • I mean the Lebesgue measure not Borel measure. – Mick Jul 23 '19 at 10:15
  • @Mick I have revised my answer based on your comment. – Kavi Rama Murthy Jul 23 '19 at 10:25
  • I think that referring to analytic sets is a bit an overkill here. Measurability of $T(A)$ can be proved directly. – Giuseppe Negro Jul 23 '19 at 10:31
  • I try by Lipschitz but image null set may be not measurable. Let $X$ be Vitali set embedded in $\mathbb{R}^2$. It is measurable in $\mathbb{R}^2$, but not in $\mathbb{R}$. – Mick Jul 23 '19 at 10:51
  • @Mick: That's for Borel sets. The Lebesgue measure is complete, meaning that all subsets of a null set are measurable. – Giuseppe Negro Jul 23 '19 at 10:56
  • @ Giuseppe Negro: I know, but in proof for invertible linear map you prove by Lipschitz that image of null set is null set. Here it is imposible, because my counterexample. – Mick Jul 23 '19 at 11:04
  • @Kavi Rama Murthy: But how do you prove that $T$ maps borel null sets to null sets? This was my problem. – Mick Jul 23 '19 at 11:52
  • 2
    @Mick By regularity of Lebesgue measure we can find compact sets $K_n \subset A$ such that $m(A\setminus \cup_n K_n)=0$. Since $T(A)=\cup_n T(K_n) \cup T(A\setminus \cup_n K_n)$ it follows that $T(A)$ is the union of a countable number of compact sets and a set of measure $0$. – Kavi Rama Murthy Jul 23 '19 at 11:56
  • 1
    @Kavi Rama Murthy: How do you know that $T(A\setminus \cup_n K_n)$ has zero measure? – Mick Jul 23 '19 at 12:05
2

If $T$ is not of full rank, then its image is a null set, thus $T(A)$ is contained in a null set and, in particular, it is measurable. If $T$ is full rank, then it is surjective, and you already know how to answer in this case.

Giuseppe Negro
  • 34,508