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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.

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Erm, isn't d continuous, and hence automatically measurable? –  Matthew Daws Mar 26 '10 at 10:40
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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 '10 at 10:50
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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 '10 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 '10 at 11:36
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@Robin: Please make your comments into an answer. –  François G. Dorais Mar 26 '10 at 12:00

2 Answers 2

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!).

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Why was this answer accepted? Of course this answer was very helpful; but it doesn't actually answer the question. –  Regenbogen Mar 26 '10 at 22:39
    
I've thought it was immediate to show that linear bijection are Leb-Leb 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? –  Nicolò Mar 26 '10 at 22:59
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@Nicolò: Cover $N$ with balls of very small total measure and think about what $T$ does to each ball... –  François G. Dorais Mar 27 '10 at 1:11
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Yes, but there are several things to show.. –  Nicolò Mar 27 '10 at 17:43
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T is Lipschitz, hence zero-sets go to zero-sets –  Carlo Mantegazza Jan 21 '13 at 15:20

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.

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