13
$\begingroup$

Recall that we say that a bounded measurable set $S\subset\mathbb R^n$ is said to be Caccioppoli if the indicator function $1_S$ is BV, and we set $$ \operatorname{perim}(S)=\| \nabla 1_S\|_{TV} $$ where $\|\cdot\|_{TV}$ denotes the total variation. So, if $S$ and $T$ are Caccioppoli sets, is it known whether $S\cap T$ is Caccioppoli?

$\endgroup$
  • 3
    $\begingroup$ You might be interested in the following excellent book where you can find the affirmative answer to your question (Lemma 12.22) and much more. Francesco Maggi: Sets of finite perimeter and geometric variational problems, 2012. $\endgroup$ – Manfred Sauter Mar 28 '18 at 0:15
12
$\begingroup$

That is true. Caccioppoli sets are also known as sets of finite perimeter.

Theorem. Suppose $f\in L^1(\mathbb{R}^n)$ vanishes outside the unit cube $[0,1]^n$. For $i=1,2,\ldots,n$ consider the function $V_if(x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_n)= V_0^1f(x_1,\ldots,x_{i-1},\cdot,x_{i+1},\ldots,x_n)$, i.e. the (essential) variation of the one dimensional sections. Then $f\in BV(\mathbb{R}^n)$ if and only if for every $i$, $V_i\in L^1([0,1]^{n-1}$).

This is Theorem 5.3.5 in:

W. P. Ziemer, Weakly differentiable functions. Sobolev spaces and functions of bounded variation. Graduate Texts in Mathematics, 120. Springer-Verlag, New York, 1989.

Basically it is a characterization of the functions of bounded variation by one dimensional slices.

If $h_1,h_2$ are characteristic functions of sets in $[0,1]$, then $V_0^1(h_1h_2)\leq V_0^1h_1+V_0^1 h_2$, see the proof of Theorem 2 in http://mathonline.wikidot.com/multiples-and-products-of-functions-of-bounded-variation. Now if $S$ and $T$ are Caccioppoli sets cotained in the unit cube, the characteristic functions $\chi_S$ and $\chi_T$ have bounded variation and the one dimensional result mentioned here shows that $$ V_i(\chi_S\chi_T)\leq V_i(\chi_S)+V_i(\chi_T)\in L^1([0,1]^{n-1}). $$ That implies that $\chi_S\chi_T=\chi_{S\cap T}$ has bounded variation so $S\cap T$ is a Caccioppoli set.

I assumed here that the sets are contained in the unit cube, but the argument applies to any bounded set.

Another answer is provided in a comment by Manfred Sauter (see above).

$\endgroup$
3
$\begingroup$

The intersection of two Caccioppoli sets is again a Caccioppoli set: as Piotr Hajlasz points out, a Caccioppoli set is simply a set of finite perimeter therefore the (standard) proof I report below is based on the following, slightly more general, definition of the (relative) perimeter of a set.

Definition. Let $S$ a Lebesgue measurable set in $\mathbb{R}^n$. For any open subset $\Omega\subseteq\mathbb{R}^n$ the perimeter of $S$ in $\Omega$, denoted as $P(S,\Omega)$, is the variation of $\chi_S$ in $\Omega$ i.e. \begin{split} P(S,\Omega)&=\sup\left\{\int_S \mathrm{div}\varphi\,\mathrm{d}x\,:\,\varphi\in [C_c^1(\Omega)]^n, \|\varphi\|_\infty\leq1\right\}\\ & =\| \nabla (\chi_{S\cap\Omega})\|_{TV}=TV(S,\Omega) \end{split} The perimeter so defined is a lower-semicontinuous set function which is additive respect to the argument $\Omega$.

Theorem. Let $S$ and $T$ two Lebesgue measurable sets in $\mathbb{R}^n$: then, for any open set $\Omega\subseteq\mathbb{R}^n$ $$ P(S\cup T,\Omega)+P(S\cap T,\Omega)\leq P(S,\Omega) + P(T,\Omega) $$ Proof. We can choose two smooth sequences of functions $\{s_k\}, \{t_k\}$ in $C^\infty(\Omega)$ converging respectively to $\chi_S$ and $\chi_T$ in $L^1_{loc}(\Omega)$, for example by using partitions of unity, and such that $$ 0\leq s_k\leq 1\quad 0\leq t_k \leq 1, $$ and $$ \lim_{k\to\infty}\int_\Omega |\nabla s_k|\mathrm{d}x=P(S,\Omega)\quad \lim_{k\to\infty}\int_\Omega |\nabla t_k|\mathrm{d}x=P(T,\Omega) $$ Now since $s_kt_k\to\chi_{S\cap T}$ and $s_k+t_k-s_kt_k\to\chi_{S\cup T}$, the theorem follows by passing to the limit for $k\to\infty$ in the following elementary inequality $$ \int_\Omega |\nabla (s_kt_k)|\mathrm{d}x+\int_\Omega |\nabla (s_k+t_k-s_kt_k)|\mathrm{d}x \leq \int_\Omega |\nabla s_k|\mathrm{d}x+ \int_\Omega |\nabla t_k|\mathrm{d}x\quad\blacksquare $$ An immediate consequence of this theorem is the following corollary, which includes the sought for answer to the posed question.

Corollary. If $S$ and $T$ are Caccioppoli sets, so are $S\cup T$ and $S \cap T$.

Notes

  • The proof is taken almost verbatim from the wonderful book of Ambrosio, Fusco and Pallara ([1], §3.3 p. 144): De Giorgi, Colombini and Piccinini ([2], §1.3 pp.18-19) prove the same result in a different way and assuming from the start that $S$ and $T$ are Caccioppoli sets.
  • A direct proof of the result $P(S),P(T)<\infty\Rightarrow P(S\cap T)<\infty$ is presented by Gurtin, Williams and Ziemer ([3], p. 6) and is based on lattice and measure theoretical considerations: in fact, it is worth noting that this property of Caccioppoli sets make them very important in the formal definition of the concept of body in the modern rational continuum mechanics.
  • It is perhaps also worth to point out that the property of BV functions used by Piotr Hajlasz is in reality their original definition as conceived by Lamberto Cesari back in 1936.

[1] Ambrosio, Luigi; Fusco, Nicola; Pallara, Diego (2000). Functions of bounded variation and free discontinuity problems. Oxford Mathematical Monographs. New York and Oxford: The Clarendon Press/Oxford University Press, New York, pp. xviii+434, ISBN 0-19-850245-1, MR1857292, Zbl 0957.49001.

[2] De Giorgi, Ennio; Colombini, Ferruccio; Piccinini, Livio (1972), Frontiere orientate di misura minima e questioni collegate [Oriented boundaries of minimal measure and related questions], Quaderni (in Italian), Pisa: Edizioni della Normale, pp. 180, MR 0493669, Zbl 0296.49031.

[3] Gurtin, M. E.; Williams, W. O. & Ziemer, W. P., "Geometric measure theory and the axioms of continuum thermodynamics", Archive for Rational Mechanics and Analysis, 1986, 92, 1-22, MR 816619, Zbl 0599.73002.

$\endgroup$
2
$\begingroup$

Why not indulge in some overkill:

Theorem (cf. Federer, Theorem 4.5.11). $\newcommand{\RR}{\mathbb{R}}A\subset\RR^n$ has finite perimeter iff $\mathcal{H}^{n-1}(\partial^* A)<\infty$.

Here $\mathcal{H}^s$ denotes the $s$-dimensional Hausdorff measure and $\partial^*A\subset\RR^n$ the measure-theoretic boundary of $A$. The latter is determined by $$ x\in\RR^n\setminus\partial^*A \quad\Longleftrightarrow\quad \lim_{r\to 0+}\frac{\mathcal{H}^n(A\cap B(x,r))}{\mathcal{H}^n(B(x,r))}\in\{0,1\}. $$ In other words, $x\in\partial^*A$ iff $A$ does not have Lebesgue density $0$ or $1$ at $x$.

By Federer's theorem it suffices to show that $\partial^*(S\cap T)\subset\partial^*S\cup\partial^*T$, or equivalently, $$(\RR^n\setminus\partial^*S)\cap(\RR^n\setminus\partial^*T)\subset\RR^n\setminus\partial^*(S\cap T).$$

The rest is elementary. Let $x$ be an element of the left hand side. Clearly, if $S$ or $T$ has density $0$ at $x$, then $S\cap T$ has density $0$ at $x$. So suppose both $S$ and $T$ have density $1$ at $x$. But as $$(S\cap B(x,r))\setminus (B(x,r)\setminus T) = S\cap T\cap B(x,r),$$ also $S\cap T$ has density $1$ at $z$. So $x$ is an element of the right hand side.

$\endgroup$
  • $\begingroup$ You should include your reference to Lemma 12.22 in the book by Francesco Maggi in your answer. Comments are not as easily seen as answers. $\endgroup$ – Piotr Hajlasz Mar 30 '18 at 19:07

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.