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.

measure theory – Which part of Fubini’s hypothesis is violated?

Let $f(x, y)= (4xy−x^2−y^2)(x+y)^{−4}$ for $x$, $y$ both positive, and let $f (x, y) = 0$ elsewhere. Integration first with respect to $x$ on $(0,∞)$ then with respect to $y$ on $(0,∞)$ yields $0$. Also integration first with respect to $y$ on $(0,∞)$ then with respect to $x$ on $(0,∞)$ yields $0$. But integration first with respect to $x$ on $(0,a)$ then with respect to $y$ on $(0,a)$ yields $frac14$ which stays $frac14$ when $a to ∞$. Which part of Fubini’s hypothesis is violated? To me it seems none!

probability theory – Interchange of Expectation and integral by Fubini’s theorem

I read a condition that if we have $E[int_0^1 |X_t|dt] < infty$, then we may apply Fubini’s theorem and we can have $E[int_0^1 X_tdt] = int_0^1 E[X_t]dt$.

Though I am not sure. Is it right? And please refer to a textbook where I can get this form of Fubini’s theorem. Thanks a lot.