Absolute continuity of pushforward measure Problem: Let $\newcommand{\IR}{\mathbb{R}}\newcommand{\IL}{\mathcal{L}}\phi: \IR \times \IR^{n} \to \IR^n$ be $\IL^{n+1}$-measurable and satisfy for every $\IL^{n}$-nullset $A \subset \IR^n$ and $\IL^{1}$-almost every $t \in \IR$
$$
\IL^n(\{x: \phi(t,x) \in A\}) = 0.
$$
Show that then for all $\IL^{n+1}$-nullset $M \subset \IR^{n+1}$ it holds that
$$
\int \chi_M(t,\phi(t,x)) d\IL^{n+1} = 0.
$$
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 $M$.
 A: 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{equation}
\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}
\end{equation}
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.
A: The Question is indeed not well-posed. The problem is in the first line already. It should be formulated as follows to be precise:
Let $\newcommand{\IR}{\mathbb{R}}\newcommand{\IL}{\mathcal{L}}\phi: \IR \times \IR^{n} \to \IR^n$ be $\IL^{n+1}$-measurable. Suppose there exists an $\IL^{1}$-nullset $N$ such that for all $t \in \IR\setminus N$ it holds that
$$
\IL^{n}(A) = 0 \qquad\Longrightarrow\qquad \IL^n(\{x: \phi(t,x) \in A\}) = 0.
$$
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:
Let $\newcommand{\IR}{\mathbb{R}}\newcommand{\IL}{\mathcal{L}}\phi: \IR \times \IR^{n} \to \IR^n$ be $\IL^{n+1}$-measurable. Suppose that for every $\IL^{n}$-nullset $A$ there exists an $\IL^{1}$-nullset $N_\phi(A)$ such that
$$
t \not\in N_\phi(A) \qquad \Longrightarrow \qquad \IL^n(\{x: \phi(t,x) \in A\}) = 0.
$$
If somebody can disprove (or prove) the assertion under this weaker assumption, I would be glad to know the counterexample (or the proof).
