Fubini's theorem for Hausdorff measures $B\subset \mathbb{R}^2$ is a Borel set. Define the slices $B_x:= \{y \in \mathbb{R}: (x,y) \in B \}$.
If $\lambda$ denotes the Lebesgue measure on $\mathbb{R}$, presentations of Fubini's theorem often include that fact that the function $\lambda(B_x)$ is measurable.
Question: If $H^s$ denotes the $s$th Hausdorff measure, how do I see that the function $H^s(B_x)$ is measurable? $(\star)$
I came across this while looking at Marstrand's slice theorem in a book. The authors suggest to use a Monotone class argument wherein one would have to show the following

*

*If $B=U\times V$ then, $H^s(B_x)= \mathbb{1}_U(x)\cdot H^s(V)$ which is measurable.

*If $B$ is a finite union of disjoint rectangles then again $H^s(B_x)$ is measurable.

*If $B_n$ is an increasing family of sets each of which satisfies $(\star)$ then $H^s\left((\cup B_n)_x\right) = H^s(\cup (B_n)_x) = \lim H^s((B_n)_x)$ which is again measurable.

*If $B_n$ is a decreasing family of sets each of which satisfies $(\star)$, one would like to show the same for $H^s((\cap B_n)_x)$. However, this is equal to $H^s(\cap (B_n)_x)$. This in general won't be $\lim H^s((B_n)_x)$ since we don't know if any of the terms has finite $H^s$ measure.

I don't see how to prove the last point in the absence of $\sigma$-finiteness. Am I missing something easy?
 A: If $s>1$, then clearly $H^s(B_x)=0$ so there is nothing to do. If $s=1$, $H^1$ is just the Lebesgue measure so measurability follows. If $0<s<1$ the situation is a way more complicated, but the answer is "yes" if $H^{1+s}(B)<\infty$ and it follows from the following result due to Federer [F, Theorem 2.10.25], commonly known as the Eilenberg inequality. See also [F, $\S$2.10.26] where the measurability of the integrand is addressed explicitly.
A metric space is boundedly compact if bounded and closed sets are compact.

Theorem. (Eilenberg inequality)
Let $\Phi:X\to Y$ be a Lipschitz mapping between boundedly compact metric spaces.
Let $0\leq m\leq n$ be real numbers (not necessarily integers).
Assume that $E\subset X$ is $H^n$-measurable
with $H^n(E)<\infty$. Then

*

*$\Phi^{-1}(y)\cap E$ is $H^{n-m}$-measurable for $H^m$-almost all $y\in Y$.


*$y\mapsto H^{n-m}(\Phi^{-1}(y)\cap E)$ is $H^m$-measurable.
Moreover
$$
\int_Y H^{n-m}(\Phi^{-1}(y)\cap E)\, dH^m(y)\leq
(\operatorname{Lip}(\Phi))^m
\frac{\omega_m\omega_{n-m}}{\omega_n}\, H^n(E).
$$

How take $\Phi:\mathbb{R}^2\to\mathbb{R}$, $f(x,y)=x$, $m=1$, $n=1+s$, $E=B$. If $H^{1+s}(B)<\infty$, then the function
$$
x\mapsto H^{n-m}(\Phi^{-1}(x)\cap E)=H^{s}(B_x)
$$
is $H^m$ measurable. Since $H^m=H^1$ is the Lebesgue measure, it is Lebesgue measurable.
The above argument is true under the assumption that $B$ is $H^{1+s}$-measurable with $H^{1+s}(B)<\infty$. If you know that $B$ is Borel, it is a much stronger condition than
just measurability and I believe it is true without the assumption that $H^{1+s}(B)<\infty$, but I have no clue how to prove it and I do not even know if it is true. In fact I believe that the function $x\mapsto H^s(B_x)$ is measurable with respect to the $\sigma$-algebra generated by analytic sets. A comment that I learned from Pertti Mattila [EH, Remark 1.2] suggests that.
[EH] B. Esmayli, P. Hajłasz, The coarea inequality. Ann. Acad. Sci. Fenn. Math. (To appear).
[F] H. Federer, Geometric measure theory. Die Grundlehren der mathematischen Wissenschaften,
Band 153 Springer-Verlag New York Inc., New York 1969.
