# 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.