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


1 Answer 1


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.

  • $\begingroup$ Thanks so much! Will go through this soon. $\endgroup$ May 23, 2021 at 0:22
  • $\begingroup$ 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. $\endgroup$ May 24, 2021 at 16:21
  • 3
    $\begingroup$ @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. $\endgroup$ May 24, 2021 at 16:25
  • 1
    $\begingroup$ (The paper [EH] is at arxiv.org/abs/2006.00419 on the Arxiv, if anyone else was looking) $\endgroup$ May 28, 2021 at 3:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.