In the post Does $\mu_k(U \cap \mathbb{R}^k)=0$ for all affine embeddings of $\mathbb{R}^k$ in $\mathbb{R}^n$ imply $\mu_n(U)=0$?, the OP originally posed (essentially) the following question:
Fix $1 \leq k < n$. If $U \subset \Bbb{R}^n$ is Lebesgue-measurable (w.l.o.g. compact) such that $U \cap \iota(\Bbb{R}^k)$ is a $k$-dimensional null-set for every linear embedding $\iota : \Bbb{R}^k \to \Bbb{R}^n$, i.e. such that $$\lambda_k (\iota^{-1}(U)) = 0$$ for all such $\iota$, does it follow that $U$ is a null-set?
He later revised the question to require the above for every affine embedding, so that the answer is "yes", using a Fubini-argument.
In the case $k=1$, the answer to the above question is also "yes", as one can see using integration in Polar coordinates:
\begin{eqnarray*} \int_{\Bbb{R}^n} \chi_U \, dx & = & \int_{S^{n-1}} \int_0^\infty r^{n-1} \cdot \chi_U (r\cdot \xi) \, dr\, d \mathcal{H}^{n-1}(\xi) =0, \end{eqnarray*}
because $\chi_U (r \cdot \xi) = \chi_{\iota_\xi^{-1}(U)} (r) = 0$ for almost every $r \in \Bbb{R}$, because $\iota_\xi : \Bbb{R} \to \Bbb{R}^n , t\mapsto t \cdot \xi$ is a linear embedding.
My question is: Is the above (with linear embeddings) also true for $1 < k < n$?
After some additional thinking I came upon the following proof showing that the statement is indeed correct:
We use induction on $\ell:=n-k\in\mathbb{N}$. Let us first establish the case $\ell=1$, i.e. $k=n-1$. Here, we make use of the diffeomorphism $$ \varphi:\mathbb{R}^{n-2}\times\mathbb{R}^{\ast}\times\mathbb{R}\to\mathbb{R}^{n-2}\times\mathbb{R}^{\ast}\times\mathbb{R},\left(\begin{matrix}x\\ s\\ t \end{matrix}\right)\mapsto\left(\begin{matrix}x\\ s\\ ts \end{matrix}\right) $$ with inverse map $$ \varphi^{-1}:\mathbb{R}^{n-2}\times\mathbb{R}^{\ast}\times\mathbb{R}\to\mathbb{R}^{n-2}\times\mathbb{R}^{\ast}\times\mathbb{R},\left(\begin{matrix}x\\ s\\ y \end{matrix}\right)\mapsto\left(\begin{matrix}x\\ s\\ y/s \end{matrix}\right). $$ The important properties of $\varphi$ for us are that
$\varphi$ maps the set $M:=\mathbb{R}^{n-2}\times\mathbb{R}^{\ast}\times\mathbb{R}$ of full measure in $\mathbb{R}^{n}$ diffeomorphically onto the set $M$ (which is again of full measure).
For each fixed $t\in\mathbb{R}$, the map $$ \mathbb{R}^{n-2}\times\mathbb{R}^{\ast}\to\mathbb{R}^{n},\left(\begin{matrix}x\\ s \end{matrix}\right)\mapsto\varphi\left(\begin{matrix}x\\ s\\ t \end{matrix}\right)=\left(\begin{matrix}x\\ s\\ ts \end{matrix}\right) $$ is (or can be extended in an obvious way to) a linear embedding of $\iota_{t}:\mathbb{R}^{n-1}\to\mathbb{R}^{n}$.
Furthermore, $$ \left(D\varphi^{-1}\right)\left(\begin{matrix}x\\ s\\ y \end{matrix}\right)=\left(\begin{matrix}1 & & & & 0\\ & \ddots & & & \vdots\\ & & & & \vdots\\ & & & 1 & 0\\ 0 & \dots & 0 & -\frac{y}{s^{2}} & 1/s \end{matrix}\right) $$ and thus $$ \left|\det\left(\left(D\varphi^{-1}\right)\left(\begin{matrix}x\\ s\\ y \end{matrix}\right)\right)\right|=\left|\frac{1}{s}\right|=\left|\left(\begin{matrix}x\\ s\\ y \end{matrix}\right)_{n-1}\right|^{-1}. $$ Using the change-of-variables formula and Fubini's theorem, we conclude \begin{eqnarray*} && \lambda_{n}\left(U\right)=\int_{\mathbb{R}^{n}}\chi_{U}\left(x\right)\, dx \\ &=& \int_{\mathbb{R}^{n-2}\times\mathbb{R}^{\ast}\times\mathbb{R}}\chi_{U}\left(\varphi\left(\varphi^{-1}\left(x\right)\right)\right)\cdot\left|\left(\varphi\left(\varphi^{-1}\left(x\right)\right)\right)_{n-1}\right|\cdot\left|\det\left(\left(D\varphi^{-1}\right)\left(x\right)\right)\right|\, dx\\ & = & \int_{\mathbb{R}^{n-2}\times\mathbb{R}^{\ast}\times\mathbb{R}}\chi_{U}\left(\varphi\left(y\right)\right)\cdot\left|\left(\varphi\left(y\right)\right)_{n-1}\right|\, dy\\ & = & \int_{\mathbb{R}}\int_{\mathbb{R}^{n-2}\times\mathbb{R}^{\ast}}\chi_{U}\left(\iota_{t}\left(\begin{matrix}x\\ s \end{matrix}\right)\right)\cdot\left|s\right|\, d\left(\begin{matrix}x\\ s \end{matrix}\right)\, dt=0. \end{eqnarray*} In the last step, we used that $\chi_{U}\left(\iota_{t}\left(\begin{smallmatrix}x\\ s \end{smallmatrix}\right)\right)=0$ for almost all $\left(\begin{smallmatrix}x\\ s \end{smallmatrix}\right)\in\mathbb{R}^{n-1}$ by assumption. Hence, $\lambda_{n}\left(U\right)=0$, where $\lambda_{m}$ is Lebesgue-measure on $\mathbb{R}^{m}$.
The general case follows by induction using the base case: Assume that the result is true for $\ell\in\mathbb{N}$ and that $1\leq k<n$ with $n-k=\ell+1$ and furthermore that $U\subset\mathbb{R}^{n}$ is measurable with $$ \lambda_{k}\left(\iota^{-1}\left(U\right)\right)=0 $$ for all linear embeddings $\iota:\mathbb{R}^{k}\to\mathbb{R}^{n}$.
Let $\kappa:\mathbb{R}^{n-1}\to\mathbb{R}^{n}$ be an arbitrary linear embedding. Then $U':=\kappa^{-1}\left(U\right)\subset\mathbb{R}^{n-1}$ is measurable (if $U$ is Borel-measurable) and for each linear embedding $\iota:\mathbb{R}^{k}\to\mathbb{R}^{n-1}$, we have $$ \lambda_{k}\left(\iota^{-1}\left(U'\right)\right)=\lambda_{k}\left(\iota^{-1}\left(\kappa^{-1}\left(U\right)\right)\right)=\lambda_{k}\left(\left(\kappa\circ\iota\right)^{-1}\left(U\right)\right)=0, $$ because $\kappa\circ\iota:\mathbb{R}^{k}\to\mathbb{R}^{n}$ is a linear embedding. By induction hypothesis (because of $\left(n-1\right)-k=\left(\ell+1\right)-1=\ell$), we conclude $$ \lambda_{n-1}\left(\kappa^{-1}\left(U\right)\right)=\lambda_{n-1}\left(U'\right)=0. $$ As this holds for all linear embeddings $\kappa:\mathbb{R}^{n-1}\to\mathbb{R}^{n}$, we can use the base case $\ell=1$ (because of $n-\left(n-1\right)=1$) established in the beginning to conclude $\lambda_{n}\left(U\right)=0$.