Existence of a measure-preserving bijection

Question

Let $f, g \, \colon \mathbb{R}^n \rightarrow \mathbb{R}$ be two Borel-measurable functions such that $f$ is non negative and

• g is radially symmetric,
• the function $ (0, \infty )\ni t \mapsto g (tx)$ is increasing and
• their distribution functions coincide, i.e., for all values $\lambda >0$ : $\vert \{ f(y) < \lambda \} \vert = \vert \{ g(y) < \lambda \} \vert < \infty$, where $\vert A \vert$ denotes the Lebesgue measure of $A \subset \mathbb{R}^n$.

I would like to show that there is a measure-preserving bijection $\phi$ such that $ g = f \circ \phi$.

So far, I have no idea how to tackle this problem. I would be really grateful for any suggestions. Thanks!

Answer 1

I presume that you are looking for a bijection in the measure category. Then your first two conditions are not really necessary, and existence of such a bijection follows from very general measure theory considerations.

First, by appropriate partitioning of $\mathbb R^n$ the problem can be reduced to the situation when both functions are bounded, and are non-zero on subsets of finite (and therefore equal) measure. Then both these subsets are isomorphic as measure spaces just to an interval endowed with the Lebesgue measure. Now it remains to "rearrange" this interval in such a way that your roof function becomes monotone (presuming that all level sets have measure 0, it is done just by its distribution function; if not, one has to take care of these level sets individually).

Answer 2

This is not true as stated, since nothing prevents us from modifying $f$ on a null set so that it takes values which $g$ does not.

For a simple example, take $n=1$, $g(x)=x^2+1$, and $$f(x) = \begin{cases} x^2+1, & x \ne 0 \\ 0, & x=0. \end{cases}$$ Then $\{f < \lambda\} = \{g < \lambda\} \cup \{0\}$, so the distribution functions agree. But if $\phi : \mathbb{R} \to \mathbb{R}$ is any bijection at all, then letting $x_0 = \phi^{-1}(0)$ we have $f(\phi(x_0)) = 0 \ne g(x_0)$.