Welcome to MathOverflow

Question

9

Which is the cleanest way to show that the difference, $d:R^n\times R^n\rightarrow R^n$, $d(x,y)= x-y$, is Lebesgue-Lebesgue 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.

flag	asked Mar 26 2010 at 10:29 Nicolò 512●3●17

1

Erm, isn't d continuous, and hence automatically measurable? – Matthew Daws Mar 26 2010 at 10:40

14

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 zero-measure Cantor set has Lebesgue measure zero but its inverse image need not be measurable. Decomposing the subtraction map as a composite of $(x,y)\mapsto(x-y,x)$ 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. – Robin Chapman Mar 26 2010 at 10:50

3

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

Robin is right, every continuous function is Lebesgue-Borel measurable, but it is not said to be Lebesgue-Lebesgue measurable. My problem arise from showing that if f is measurable, then is f(x-y). Unlucky if two function $f, g$ are measurable (i.e. Lebesgue-Borel measurable), their composition $f\circ g$ not needs to be. It is if the $g$ is Lebesgue-Lebesgue measurable – Nicolò Mar 26 2010 at 11:36

1

@Robin: Please make your comments into an answer. – François G. Dorais♦ Mar 26 2010 at 12:00

show 2 more comments

Colin McQuillan · Answer 1 · 2010-03-26 13:25:41Z UTC

4

Unitary matrices preserve measure. A diagonal matrix of full rank is a Lesbesgue-Lesbesgue measurable transformation. Linear maps over the reals have a singular value decomposition.

edited Mar 26 2010 at 13:25

answered Mar 26 2010 at 12:57

Colin McQuillan
541●1●7

Robin Chapman · Answer 2 · 2010-03-26 13:31:34Z UTC

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 non-measurable. A simple example is given by $f:x\mapsto (x,0)$ from $\mathbb{R}$ to $\mathbb{R}^2$. Taking $A$ to be a non-measurable subset of $\mathbb{R}$ and $B=f(A)$ we see this $f$ is not Lebesgue-Lebesgue 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 Lebesgue-Lebesgue 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 Lebesgue-Lebesgue measurable). By definition, the class Lebesgue-Lebesgue measurable maps is closed under composition (unlike the class of Lebesgue-measurable maps!).

How to show that x-y is Lebesgue-Lebesgue measurable

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

2 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.