1
$\begingroup$

Let $f: \mathbb{R}^n \rightarrow \mathbb{R}^m $ be Lipschitz and $n \geq m$. A version of the coarea formula says:

$$ \int_A g(x) J_m f(x) d \mathcal{L}^n (x) = \int_{\mathbb{R}^m } \int_{ A \cap f^{-1}(t) } g(p) d \mathcal{H}^{n-m}(p) d \mathcal{L}^m (t) $$

for each Lebesgue measurable subset $A \subset \mathbb{R}^n $ and an integrable function $g: A \rightarrow \mathbb{R} $.

In the proof process, the following map $T$ can be shown to be $\mathcal{L}^m$ measurable:

$$ T(t) = \mathcal{H}^{n-m}( A \cap f^{-1}(t) ) $$

Now consider the case when $A$ is open and the restriction of the map $T$ to $f(A)$: $ T: f(A) \longrightarrow \mathbb{R}^{+} $. I'm wondering if there are extra conditions we can impose on $f$, say it is $C^1$ with Lipschitz Jacobian so that we can estimate $T$'s modulus of continuity (on $f(A)$)?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

The answer is in the negative. Let $Q=(0,1)\times (0,1)$. Let $I_i$, $i\in\mathbb{N}$ be open intervals in the complement of the ternary Cantor set $C$, and let $$ A=Q\cup\bigcup_{i=1}^\infty I_i\times [1,3/2). $$ That means on top of the square we add rectangles of height $1/2$ with base in the complement of the Cantor set $C$.

Let $f:\mathbb{R}^1\to\mathbb{R}$, $f(x,y)=x$. Then $f(A)=(0,1)$. If $t\in C\cap (0,1)$, then $T(t)=\mathcal{H}^1(f^{-1}(t))=1$ while for $t\in (0,1)\setminus C$, $t\in I_i$ for some $i$ and hence $T(t)=\mathcal{H}^1(f^{-1}(t))=3/2$. Thus the function $T$ is discontinuous at the points of $C$.

Adding more rectangles of height $1/4$ on top of each rectangle $I_i\times [1,3/2)$ and then adding rectangles of height $1/8$ on top of these rectangles and continuing this construction infinitely many times we can produce a bounded open set $A$ such that the function $T$ is discontinuous at almost every point of $(0,1)$.

Improve this answer
$\endgroup$
5
  • $\begingroup$ Sorry for the confusion. Your example is not in the intended setting I was thinking about. I have updated the question accordingly. In particular, we consider the restriction of $T$ to $f(A)$. $\endgroup$
    – random_shape
    Commented Mar 13, 2018 at 0:40
  • $\begingroup$ Does my example answer your question? $\endgroup$
    – Piotr Hajlasz
    Commented Mar 26, 2018 at 21:02
  • $\begingroup$ Thanks for your answer and sorry for the delayed reply. Your answer helps me think carefully about the needed conditions to well pose the question in my mind. I think your answer is instructive, but I'm not sure if it answers my real question. I will need to think more about it and maybe edit the question accordingly. If you don't mind, please leave your answer here. Thank you again. $\endgroup$
    – random_shape
    Commented Mar 26, 2018 at 22:22
  • $\begingroup$ If you e-mail me (you [email protected]) I can send you a picture for the domain in my answer so it will help you understand my construction. $\endgroup$
    – Piotr Hajlasz
    Commented Mar 26, 2018 at 22:26
  • $\begingroup$ It is very kind of you! I kind of get the construction, the question that bothers me now is how much does the continuity of $T$ depend on the domain $A$? If $A$ is a nice shape like a ball or rectangle, can we restrict $f$ so that $T$ is continuous? Obviously, the induced map $T$ depends on both $f$ and $A$. Your construction shows its strong dependence on $A$. $T$'s dependence on $f$ or lack of it remains unclear - which is what I'm initially interested in - but I naively ignored the importance of $A$. $\endgroup$
    – random_shape
    Commented Mar 27, 2018 at 17:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .