Fubini type results for Hausdorf dimension? Suppose that I have a stack of hyperplanes in Euclidean space $\mathbb{R}^n$, let's call each plane $P_{a, x}=\{y\in\mathbb{R}^n\mid \langle y, x\rangle=a\}$  Suppose that a measurable subset $A$ of $\mathbb{R}^n$ intersects each of these hyper planes nonemptily for some range of $a$ values $(a_1, a_2)$ and a fixed value of $x$.  Must this set $A$ have Hausdorf dimension at least $1$?  This is a general part of a question I have in my mind which is whether or not one can use Fubini-type intuitions to count Hausdorf dimensions from intersections with lower dimensional subspaces.  The reason I call this Fubini is that if I know that $A$ intersects every such plane with full $n-1$ dimensional lebesgue measure, then by Fubini, I know $A$ is almost every point in $(a_1, a_2)\times \mathbb{R}^{n-1}$ appropriately rotated.
Viewed another way, this result, if true, would be intuitive because if $A$ intersects all hyperplanes, then you can translate the points of intersection to form a line, which has Hausdorf dimension $1$. (And then there might be more points of intersection than $1$.)
If this question is too difficult to be explained in an answer on MSE, please refer me to where I can read about this kind of question.  Thanks!

If you go back to the definition of the Hausdorff measure then
\begin{equation*}
H^{1}(A)=\lim_{\delta\to 0} (\inf \{ \sum |U_{i}|, A\subset \bigcup U_{i}, |U_{i}|<\delta \} ) 
\end{equation*}
where $|U|$ designates $U$'s diameter. If you define $A_{U}=\{a\in(a_{1},a_{2})\mid U\cap P_{a,x}\neq \emptyset\}$ then $A_{U}$ is contained in an interval of diameter $|U|$, so has one dimensional Lebesgue measure $\leqslant |U|$. Since by hypothesis the $A_{U_{i}}$ cover $(a_{1},a_{2})$ it follows that $\sum |U_{i}| \geqslant a_{2}-a_{1}$ so that $A$ has positive $H^{1}$ measure. You even know that $H^{1}(A)\geqslant a_{2}-a_{1}$.
I don't know what to say about your Fubini intuition though, or whether if you can use it to do a proof.
