# Class of Lebesgue-Lebesgue measurable functions?

A function $f:\mathbb{R}^n\to\mathbb{R}^m$ is Lebesgue-Lebesgue measurable iff inverse images of Lebesgue measurable sets are Lebesgue measurable.

Seems to me that since projections* and arithmetic operations are Lebesgue-Lebesgue measurable surjective linear transformations have to be Lebesgue-Lebesgue. Is it possible to generalize and say that all surjective multivariable polynomials (rational/analytic functions) are Lebesgue-Lebesgue?

For surjective rational functions in one variable I had the following idea: Let $f:\mathbb{R}\to\mathbb{R}$ be a rational function, dissect $\mathbb{R}$ to a finite number of intervals s.t. at any interval $I_k$, $f:I_k \to f(I_k)$ is a diffeomorphism. Then glue them together to get a function that preserves measurability in both directions. (maybe i can extend that to the case where $f$ had countable number of zeros and singularities?)

*By which i don't mean to include "projections" like $f:x \mapsto (x,0)$

• A rational function may not be defined on all of $\mathbb{R}^{n}$, but if you restrict it to its natural domain, then the answer is affirmative. Commented Dec 2, 2014 at 0:34
• @BerrickFillmore Yeah, that's what i meant. My progress so far is something along the lines of: By Sard's theorem it's almost everywhere locally lipschits continuous with lipschits continuous local inverse. Am i on the right track?
– user116457
Commented Dec 2, 2014 at 0:38
• What do you mean by projections? The linear map $f:\mathbb{R} \to \mathbb{R}^2$ given by $f(x) = (x,0)$ is not Lebesgue-Lebesgue measurable. See mathoverflow.net/a/19421/31729. Commented Dec 2, 2014 at 0:39
• That map is an inclusion, the opposite of a projection — $\pi(x,y)=x$. Commented Dec 2, 2014 at 2:16
• Given that $x\mapsto(x,0)$ is not Lebesgue-Lebesgue measurable, why do you think arbitrary polynomial functions are? Commented Dec 2, 2014 at 7:57

This seems to have to do with the so-called coarea formula; see e.g. the book by Krantz and Parks "Geometric integration theory".

Assume for simplicity that your function $f:\mathbb R^n\to \mathbb R^m$ is $\mathcal C^1$.

To say that $f$ is Lebesgue-Lebesgue is the same as to say that $f^{-1}(E)$ is a Lebesgue-null set in $\mathbb R^n$ for every Lebesgue-null set $E\subset\mathbb R^m$. And since every Lebesgue null set is contained in a Borel set with Lebesgue measure $0$, it is enough to consider only Borel sets $E$.

Now, the coarea formula reads as follow (assuming that $n\geq m$): for any measurable set $A\subset\mathbb R^n$, we have $$\int_A J_mf(x)\, d\mathcal L^n(x)=\int_{\mathbb R^m} \mathcal H^{n-m}\bigl( A\cap f^{-1}(y)\bigr)\,d\mathcal L^m(y)\, .$$ Here, $\mathcal L^n$ is Lebesgue measure and $\mathcal H^{n-m}$ denotes the $(n-m)$-dimensional Hausdorff measure on $\mathbb R^n$. Finally, $J_mf(x)$ is the $m$-dimensional Jacobian of $f$ at $x$, i.e. $$J_mf(x)=\sqrt{{\rm det}(Df(x)Df(x)^*)}\, .$$

Now, let $E\subset \mathbb R^m$ be a Borel set with Lebesgue measure $0$, and take $A:= f^{-1}(E)$. This is a Borel set, so we can apply the coarea formula to it; and we want to show that (under some assumption on $f$), the set $A$ has Lebesgue measure $0$.

Observe that if $y\in \mathbb R^m$, then $A\cap f^{-1}(y)=f^{-1}(E)\cap f^{-1}(y)$ is empty if $y\not\in E$ and equal to $f^{-1}(y)$ is $y\in E$. So we get $$\int_{f^{-1}(E)} J_mf(x)\, d\mathcal L^n(x)=\int_E \mathcal H^{n-m}\bigl( f^{-1}(y)\bigr)\,d\mathcal L^m(y)=0\, ,$$ because $\mathcal L^m(E)=0$.

It follows that $f^{-1}(E)$ does have Lebesgue measure $0$ provided $J_mf(x)> 0$ almost everywhere, which means that $Df(x)Df(x)^*$ is invertible almost everywhere. Presumably, this holds in the cases you are considering; but I didn't check that.