# Absolute continuity of pushforward measure


Remark: The last statement can be written as $(\operatorname{id},\phi)_\#\IL^{n+1} \ll \IL^{n+1}$ where we denote by $f_\#\mu$ the pushforward measure of $\mu$ under $f$.

What I tried so far: I tried to use some standard Fubini argument. By setting $$M_t = \{x: (t,x) \in M\}$$ we know by Fubini that, for $\IL^{1}$-almost every $t \in \IR$, we have $\IL^{n}(M_t)=0$. And thus, for each of these $t$, we get for $\IL^{1}$-almost every $s \in \IR$ $$\IL^n(\{x: \phi(s,x) \in M_t\}) = 0.$$ But I want to know something about the diagonal, i.e. $s=t$. I need that for $\IL^{1}$-almost every $t \in \IR$ we have $\IL^n(\{x: \phi(t,x) \in M_t\}) = 0$. Obviously we get this whenever $M$ can be written as $I \times M'$ for some $I \subset \IR$ and some $M' \subset \IR^n$. Anyhow, I don't know how this could help in getting the statement for general sets $M$.

If we define $\phi_t(x)=\phi(t,x)$ then your condition: $$\IL^n(\{x: \phi(t,x) \in A\}) = 0.$$ is equivalent to $(\phi_t)_\#\IL^{n} \ll \IL^{n}$ for almost every t. Let $I=\pi_1(M)$ the projection on the first variable, then we see the following: $$\begin{split} (\operatorname{id},\phi)_\#\IL^{n+1}(M) &=\int_M\mathrm{d}\left((\operatorname{id},\phi)_\#\IL^{n+1}\right) =\int \chi_M(t,\phi(t,x)) \,\mathrm{d}\IL^{n+1}\\ &=\int \chi_M(t,\phi_t(x)) \,\mathrm{d}\IL^{n+1} =\int_{I}\int\chi_M(t,\phi_t(x))\,\mathrm{d}\IL^n(x)\,\mathrm{d}\IL(t)\\ &=\int_{I}\int\chi_{M_t}(\phi_t(x))\,\mathrm{d}\IL^n(x)\,\mathrm{d}\IL(t)\\ &=\int_{I}\int_{M_t}\,\mathrm{d}((\phi_t)_\#\IL^n(x))\,\mathrm{d}\IL(t) \end{split}$$ Then we have that $(\operatorname{id},\phi)_\#\IL^{n+1}=\IL^1\otimes(\phi)_\#\IL^n\ll\IL^1\otimes\IL^n=\IL^{n+1}$ and we conlcude.

• Yes, thanks. As you can see from my answer, I confused myself with the formulation. When thinking about this problem, I always assumed that the set of $t$s for which the condition holds depends on the set $A$. And this obviously wouldn't imply $(\phi_t)_\#\IL^{n} \ll \IL^{n}$. – thomas Jul 26 '14 at 19:00
• It doesn't really matter because: $$0=\IL^n(\{x: \phi(t,x) \in A\})=\IL^n(\phi_t^{-1}(A))$$ And so your condition could be written as: $$\int_{\mathbb{R}}\IL^n(\{x\,:\,\phi(t,x)\in A\})\,\mathrm{d}t =\int_{\mathbb{R}}(\phi_t)_\#\IL^n(A)\,\mathrm{d}t=0$$ So we should be able to use the same argument. – Lolman Jul 26 '14 at 19:11
• I don't understand the problem at all. It doesn't matter once you fix the null set $A$. – Lolman Jul 26 '14 at 19:28
• Yes, for fixed $A$ everything is fine. But actually we're considering a family of null sets $M_t$. So we're interested in the expression $$\int_{\mathbb{R}}(\phi_t)_\#\mathcal{L}^n(M_t)dt.$$ This is fundamentally different from what you say. – thomas Jul 26 '14 at 19:32

The Question is indeed not well-posed. The problem is in the first line already. It should be formulated as follows to be precise:


Using this formulation, the statement is true. And the proof is an easy application of Fubini: \begin{aligned} \int \chi_M(t,\phi(t,x)) d\IL^{n+1} &= \int_{\IR\setminus N} \int_{\IR^n} \chi_{M_t}(\phi(t,x)) d\IL^{n}(x)d\IL^{1}(t) \\ &= \int_{\IR\setminus N} \IL^{n}(\{x: \phi(t,x) \in M_t\})d\IL^{1}(t) = 0. \end{aligned}

If, however, we formulate the first two sentences in a slightly different manner, we get a completely different statement. I even presume that it's wrong under these assumptions:


If somebody can disprove (or prove) the assertion under this weaker assumption, I would be glad to know the counterexample (or the proof).