I've been nerdsniped by a friend of mine trying to solve a sample analysis qualifying exam, and one of the problems I'm trying to figure out is the following:

Let $E \subseteq \mathbb{R}^2$ be a convex bounded set. Determine $$\int_{\mathbb{R}^2} e^{-\mathrm{dist}(\mathbf{x},E)}\, d\mathbf{x}$$ in terms of the Lebesgue measure $\mathcal{L}^2(E)$ and the Hausdorff measure $\mathcal{H}^1(\partial E)$.

I've figured out the first step, breaking apart $\mathbb{R}^2 = \bar{E} \cup \mathbb{R}^2\setminus \bar{E}.$ Then, in $\bar{E}$, $\mathrm{dist}(\mathbf{x},E) = 0$ so that bit of the integral just becomes $\mathcal{L}^2(\bar{E}) = \mathcal{L}^2(E\cup \partial E) = \mathcal{L}^2(E)$ since $E$ is convex and hence $\mathcal{L}^2(\partial E) = 0$. It's the integral over the complement that I'm now stuck on.

I'm not looking for a complete solution - I'd like to figure as much of this out on my own as I can, but any hints towards the next steps would be very much appreciated. Thanks very much!

  • $\begingroup$ This reminds me of Putnam A5, 1976; have a look at it and its solutions, maybe something generalizes. $\endgroup$
    – Integrand
    May 19, 2023 at 19:07

3 Answers 3


I assume that $E$ is closed, hence compact since it is a bounded subset of $\mathbb{R}^2$. This is not a true restriction since the distance to any subset of a metric space equals the distance to its closure.I also assume that $E$ has a non-empty interior.

The key fact is the equalities for every $r>0$, $$\{\mathbf{x} \in \mathbb{R}^2\mathrm{dist}(\mathbf{x},E) \le E\}\ = E + B_r,$$ where $B_r$ is the closed Euclidean ball centered at $0$ with radius $r$, and $$\mathcal{L}^2(E+B_r) = \mathcal{L}^2(E) + \mathcal{H}^1(\partial E)r + \pi r^2.$$
This is a particular case of the formula giving the volume of a Minkovski sum of convex bodies, see https://en.wikipedia.org/wiki/Mixed_volume

As a result, \begin{eqnarray*} \int_{\mathbb{R}^2} e^{-\mathrm{dist}(\mathbf{x},E)}\, d\mathbf{x} &=& \int_{\mathbb{R}^2} \Big( \int_0^\infty e^{-r} 1_{\mathrm{dist}(\mathbf{x},E) \le r} \, dr \Big)\, d\mathbf{x} \\ &=& \int_0^\infty \Big( \int_{\mathbb{R}^2} e^{-r} 1_{\mathrm{dist}(\mathbf{x},E) \le r} \, d\mathbf{x} \Big) \, dr \\ &=& \int_0^\infty e^{-r} \mathcal{L}^2(E+B_r) \, dr \\ &=& \int_0^\infty e^{-r} \big(\mathcal{L}^2(E) + \mathcal{H}^1(\partial E)r + \pi r^2 \big) \, dr \\ &=& \mathcal{L}^2(E) + \mathcal{H}^1(\partial E) +2\pi. \end{eqnarray*}

  • 2
    $\begingroup$ +1 for the smart use of indicator function. BTW, when $E$ has empty interior (ie. $E$ is a line segment), shouldn't there be an extra factor $2$ in front of $\mathcal{H}^1(\partial E)$? $\endgroup$ May 20, 2023 at 21:05
  • 1
    $\begingroup$ Yes. I assume that $E$ has non-empty interior (I add it in the assumptions), otherwise $\mathcal{H}^1(\partial E)$ should be counted twice. $\endgroup$ May 21, 2023 at 6:36
  • $\begingroup$ +1 ; it's worth comparing to Putnam A5, 1976: Let $R$ be the set of points inside and on a convex polygon in the plane. For a point $p=(x,y)$ in the plane, let $D(x,y)$ be the distance from $p$ to the closest point of $R$. The problem is to evaluate $\int _{-\infty}^{\infty}\int _{-\infty}^{\infty} e^{-D(x,y)} \,dydx$. It's been a while since I pencilled it out, but I think the answer is $A + L + 2\pi$, where $A$ is the area and $L$ is the length (perimeter) of the figure. My solution required breaking into $n$ regions, the number of sides, and integrating each, which was not terrible. $\endgroup$
    – Integrand
    May 22, 2023 at 19:52

Even though I don't know about nerdsniping, I'll try to help with a hint.

I think the thing you should have in mind is that there is a formula for $\mathcal{L}^2(E+ \overline{B_\delta})$, and therefore for $\mathcal{L}^2(a<d(x,E)\leq b)$. The Minkowski-Steiner formula gives you a nice formula for annuli around $E$.

Edit: Using that you can express the integral as a limit of Riemann sums for $f(x)=e^{-x}$ and partitions $A_{k,n}:=\left\{\mathbf x\in \mathbb{R}^2: \frac{k}{n} \leq d(\mathbf x,E) < \frac{k+1}{n}\right\}$. More precisely,

$$\int_{\mathbb R^2} e^{-\text{dist}\left(\mathbf x, E\right)}\mathrm d \mathbf x = \mathcal{L}^2(E) + \lim_{n\to \infty} \sum_{k=1}^\infty f\big( d(\zeta_{k,n},E)\big) \mathcal{L}^2(A_{k,n}) \quad \text{for} \quad \zeta_{k,n}\in A_{k,n}.$$


Suggestion. Let's first assume that $E$ has a smooth boundary. In this case, the exterior points $\vec x\notin E$ can be parametrized by a collection of normal rays extending from points $\vec p$ on the boundary, and the points $\vec p\in \partial E$ can be parametrized by arclength along the boundary curve. That is, we have a two-variable parametrization of all exterior points as $\vec x(s,t)= \vec p(s) +t \vec n(s)$ where

$\vec p(s)\in \partial E$ is a parametrized boundary point,

$\vec n(s)=<\cos \theta(s), \sin(\theta(s)>$ is the unit outward normal vector at $p$,

$t>0$ describes the distance from $\vec p\in \partial E$ in the direction $\vec n(s)$,

and $s$ is arclength on $\partial E$.

Compute the Jacobian determinant $J=1+ t \frac{d\theta}{ds}$ associated to this change of variables.

The integral over the exterior region is thus $\int _{t>0} \int _{s} e^{-t}(1+ t \frac{d\theta}{ds}) ds \ dt$ which is easily evaluated since the inner integral in the variable $s$ is ${\cal H}^1(\partial (E)) + 2 \pi t$.

Presumably this argument can be extended to cover the nonsmooth case as well, since the boundary of the convex set $E$ is rectifiable and can be regarded locally as the graph of a convex function that can be approximated by e.g piecewise-linear convex functions, but unless the qualifying exam is intended to probe advanced topics in real analysis I doubt that the problem is really intended to be that hard.

  • $\begingroup$ How can we pass from general convex and bounded $E$ to ones with a smooth (really we should only need $C^1$) boundary? I don't really know my stuff when it comes to approximating things with smoother things $\endgroup$
    – FShrike
    May 19, 2023 at 21:22

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