# 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?

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 $$0 the situation is a way more complicated, but the answer is "yes" if $$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.

• Thanks so much! Will go through this soon. May 23, 2021 at 0:22
• Thanks, this clarifies the $H^{1+s}(B) < \infty$ case. And this is good enough for me since I'm learning this stuff casually. However, I should add that the book (Fractals in Probability and Analysis by Bishop-Peres) does not make such an assumption (page 26). Perhaps later I will email the authors and edit the question if I get an answer. May 24, 2021 at 16:21
• @HWPolice Peres is very active on MathOverflow so you may ask him directly. Just put a comment in one of his posts so he will get attention to your question. May 24, 2021 at 16:25
• (The paper [EH] is at arxiv.org/abs/2006.00419 on the Arxiv, if anyone else was looking) May 28, 2021 at 3:10