Fubini’s theorem for Hausdorff measures

$Bsubset mathbb{R}^2$ 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 $B=Utimes V$ 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 $H^sleft((cup B_n)_xright) = H^s(cup (B_n)_x) = lim H^s((B_n)_x)$ 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?

Also posted on mse.