Which is the cleanest way to show that the difference, $d:R^n\times R^n\rightarrow R^n$, $d(x,y)= xy$, is LebesgueLebesgue measurable? (i.e. foreach A lebesgue measurable in $R^n$, $d^{1}(A)$ is Lebesgue measurable in $R^n\times R^n$). Thanks in advance.

1$\begingroup$ Erm, isn't d continuous, and hence automatically measurable? $\endgroup$– Matthew DawsMar 26, 2010 at 10:40

16$\begingroup$ Continuity does not imply measurability in Nicolo's strong sense. There are continuous bijections mapping sets of positive measure to sets of zero measure (e.g. Cantor sets). Each subset of a zeromeasure Cantor set has Lebesgue measure zero but its inverse image need not be measurable. Decomposing the subtraction map as a composite of $(x,y)\mapsto(xy,x)$ and $(x,y)\mapsto x$ does it cleanly enough for me. The usual definition of a Lebesgue measurable function requires the inverse image of a Borel set to be Lebesgue integrable: this is weaker than Nicolo's condition. $\endgroup$– Robin ChapmanMar 26, 2010 at 10:50

5$\begingroup$ Another example: the embedding $\mathbb{R}\to\mathbb{R}^2$ taking $x$ to $(x,0)$ isn't LebesgueLebesgue measurable in Nicolo's sense. $\endgroup$– Robin ChapmanMar 26, 2010 at 11:28

2$\begingroup$ @Robin: Please make your comments into an answer. $\endgroup$– François G. DoraisMar 26, 2010 at 12:00

1$\begingroup$ By coincidence (?) we just came to this exact exercise in: G. Folland, Real Analysis. This is Exercise 5 on page 245. I won't post more until after the class discusses it... $\endgroup$– Gerald EdgarMar 31, 2010 at 17:59
2 Answers
Nicolo is asking about functions where the inverse image of a Lebesgue measurable set is Lebesgue measurable. This is stronger than the usual definition of measurability where it is required only the inverse image of each Borel set must be Lebesgue measurable. Continuous functions need not be measurable by this stronger criterion. If $B$ has zero Lebesgue measure and $A=f^{1}(B)$ has nonzero measure then each subset of $B$ is Lebesgue measurable but its inverse image may be nonmeasurable. A simple example is given by $f:x\mapsto (x,0)$ from $\mathbb{R}$ to $\mathbb{R}^2$. Taking $A$ to be a nonmeasurable subset of $\mathbb{R}$ and $B=f(A)$ we see this $f$ is not LebesgueLebesgue measurable. More interesting examples occur on the real line when there are continuous homeomorphisms from $\mathbb{R}$ to itself taking Cantor sets of positive measure to Cantor sets of zero measure.
To return to Nicolo's example. Each surjective linear map from $\mathbb{R}^m\to\mathbb{R}^n$ is LebesgueLebesgue measurable as it can be decomposed as a composition of linear bijections and the projection map $\mathbb{R}^m\to\mathbb{R}^n$ mapping onto the first $n$ coordinates (both these types of maps can be seen to be LebesgueLebesgue measurable). By definition, the class LebesgueLebesgue measurable maps is closed under composition (unlike the class of Lebesguemeasurable maps!).

$\begingroup$ Why was this answer accepted? Of course this answer was very helpful; but it doesn't actually answer the question. $\endgroup$ Mar 26, 2010 at 22:39

$\begingroup$ I've thought it was immediate to show that linear bijection are LebLeb measurable, but thinking a little more on it there is a thing to show: if $T$ is a linear bijection and $N$ is a Borelian of null measure, is $T(N)$ of null measure? $\endgroup$– NicolòMar 26, 2010 at 22:59

2$\begingroup$ @Nicolò: Cover $N$ with balls of very small total measure and think about what $T$ does to each ball... $\endgroup$ Mar 27, 2010 at 1:11

1$\begingroup$ Yes, but there are several things to show.. $\endgroup$– NicolòMar 27, 2010 at 17:43

2$\begingroup$ T is Lipschitz, hence zerosets go to zerosets $\endgroup$ Jan 21, 2013 at 15:20
Unitary matrices preserve measure. A diagonal matrix of full rank is a LesbesgueLesbesgue measurable transformation. Linear maps over the reals have a singular value decomposition.