TITLE: Isoperimetric-like inequality for non-connected sets
QUESTION [17 upvotes]: The classical isoperimetric inequality can be stated as follows: if $A$ and $B$ are sets in the plane with the same area, and if $B$ is a disk, then the perimeter of $A$ is larger than the perimeter of $B$.
There are several ways to define the perimeter. Here is a unusual one: if $A \subset \mathbb{R}^2$ is a convex set, the 
Cauchy-Crofton formula says that the perimeter of $A$ equals the measure of the set of lines that hit $A$, or
$$ p(A) = \frac{1}{2} \int_{S^1} \lambda(P_{\theta} A) d\theta, $$
where $P_\theta$ is the orthogonal projection in the direction $\theta \in S^1$, and $\lambda$ the Lebesgue measure on any line.
Now, this definition of $p(A)$ makes sense for non-necessarily convex sets, excepts that it is not the usual notion of perimeter, so let's call it rather "mean shadow". My question if whether the isoperimetric inequality holds for the mean shadow instead of perimeter: if $A,B$ are (open, say) subsets of the plane with equal area, and if $B$ is a disk, is the mean shadow of $A$ larger that the mean shadow of $B$ ?
The inequality is true if $A$ is convex, and we can assume that $A$ is a disjoint union of convex sets (since taking the convex hull of a connected set does not change the mean shadow).

REPLY [3 votes]: I believe that the answer is positive. If $A$ is connected, then it has the same mean shadow as its convex hull $CH(A)$ so the isoperimetric inequality for $CH(A)$ shows that the mean shadow of $A$ is larger than the mean shadow of $B$.
If $A$ is not connected I believe the same inequality holds. I'll sketch a proof when $A$ has finitely many connected components $A_1, \cdots, A_n$, the general case then follows by an approximation argument. Choose a point $x\in CH(A)$ and define a 1-parameter family of
deformations of $A$ by making a parallel translate of each connected component $A_i$ so that its barycenter moves towards $x$, say at constant speed to reach it at time $t=1$. Stop this 1-parameter family of deformation as soon as a contact occurs between two connected components, then merge those two connected components and repeat. 
The point is that the area of $A$ does not change under this deformation, however the mean shadow is non-increasing -- actually the size of the shadow is non-increasing in every direction. At the end of this deformation one obtains a connected set to which the usual isoperimetric inequality can be applied, and the same inequality then also applies to $A$.