$B\subset \mathbb{R}^2$ 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?