Does $\mu_k(U \cap \mathbb{R}^k)=0$ for all affine embeddings of $\mathbb{R}^k$ in $\mathbb{R}^n$ imply $\mu_n(U)=0$? Is the following true:
We write $\mu_n$ for the Lebesgue measure on $\mathbb{R}^n$. Let $U \subset \mathbb{R}^n$, $U$ measurable and $k \leq n$. Say for every affine embedding $i \colon \mathbb{R}^k \hookrightarrow \mathbb{R}^n$ we have $\mu_{k}(U \cap i(\mathbb{R^k}))=0$. Does this imply $\mu_n(U)=0$?
 A: Since you changed your question from linear embeddings (which was a really interesting question, to which I only know the (positive) answer in the case $k=1$) to affine mappings, the answer is easy:
Simply apply Fubini's Theorem:
\begin{eqnarray*}
\mu_n(U) & =& \int_{\Bbb{R}^n} \chi_U (x) \, d\mu_n (x) = \int_{\Bbb{R}^{n-k}} \int_{\Bbb{R}^k} \chi_U (x,y) d\mu_k (y) d\mu_{n-k} (x)  \\
&=& \int_{\Bbb{R}^{n-k}} \mu_k (\{y \in \Bbb{R}^k \mid (x,y) \in U)\}) d\mu_{n-k}(x) \\
&=& \int_{\Bbb{R}^{n-k}} \mu_k (U \cap \iota_x (\Bbb{R}^k) )\,  d\mu_{n-k}(x) = 0,
\end{eqnarray*}
where $\iota_x (y) = (x,y)$ is an affine embedding of $\Bbb{R}^k$ into $\Bbb{R}^n$.
In the above calculation, the last line might not be completely clear. The problem here (as in your question) is that the $k$-dimensional Lebesgue-measure $\mu_k$ is actually only defined on $\Bbb{R}^k$ and not on $\Bbb{R}^n$. Here, I interpret
$$
\mu_k (U \cap \iota (\Bbb{R}^k)) := \mu_k (\iota^{-1}(U)) = \mu_k (\iota^{-1}(U \cap \iota(\Bbb{R}^k))).
$$
This makes the last line true by definition. Also, the actual choice of the embedding $\iota$ does not matter (i.e. the aove definition only depends on the set $\iota(\Bbb{R}^k)$), as long as you are only interested in null-sets, basically because there is (up to multiplication by a constant) only one translation-invariant measure (Haar-measure) on $\iota(\Bbb{R}^k)$ (when $\iota(\Bbb{R}^k)$ is (by translation) considered as a vector space).
