Let $\Sigma$ be an embedded smooth surface in $\mathbb{R}^3$, and let $f:\Sigma\to\mathbb{R}$ be a smooth function. Suppose $f$ is square-integrable on $\Sigma$, with \begin{align} 0<\int_{\Sigma}f^2d\mu<\infty \end{align} where $d\mu$ is the area element of $g$, the induced metric on $\Sigma$ from the flat metric of $\mathbb{R}^3$.

Denote the Euclidean open ball of radius $r>0$ with centre $x\in\mathbb{R}^3$ by $B_r(x)$. In this question I'm interested in

the continuity of the function $r\mapsto\displaystyle\int_{\Sigma\cap B_r(x)}f^2d\mu$

At first this seems to be continuous for any $x\in\mathbb{R}^3$. However, a simple counterexample can be found: If $\Sigma$ is a round sphere of radius $R>0$ with centre $0\in\mathbb{R}^3$, then \begin{align} \Sigma\cap B_r(0)=\left\{ \begin{array}{ccl} \emptyset & \text{if} & r\leq R \\ \Sigma & \text{if} & r>R \end{array} \right. \end{align} and so we have a discontinuity at $r=R$. More generally, whenever $\Sigma$ has a region which is a spherical cap, we also have such discontinuity for some $x$.

However, in the counterexample above, it seems that the continuity only fails for a single choice of $x$. Moreover, at the discontinuity $r=R$, we still have a left-continuity. I wonder if this is true in general:

Is it true that $r\mapsto\displaystyle\int_{\Sigma\cap B_r(x)}f^2d\mu$ is continuous (or at least left-continuous) for almost every $x\in\mathbb{R}^3$?

If yes, how should we prove it? And if no, what counterexample can we construct? Also, I would be glad to know the best result that we can have in this direction.

Any comment or answer is greatly welcomed and appreciated.

  • $\begingroup$ What do you mean by almost every? Any monotonous function has at most countably many discontinuities.. $\endgroup$ Apr 4, 2020 at 20:19
  • $\begingroup$ @IlyaBogdanov By "almost every" i mean it in the usual measure-theoretical sense; that is, there exists a set $Z\subseteq\mathbb{R}^3$ of measure zero such that for all $x\in\mathbb{R}^3\setminus Z$, the function I'm interested is continuous. $\endgroup$ Apr 5, 2020 at 7:16
  • $\begingroup$ @IlyaBogdanov I'm aware of the result that monotonous function has at most countably many discontinuities, which is standard in real analysis. However, here my functions have input $r\in(0,\infty)$, while being a family of functions 'parametrized' by $x\in\mathbb{R}^3$. My question is on whether the continuity holds for almost all $x$, not almost all $r$. $\endgroup$ Apr 5, 2020 at 7:20
  • $\begingroup$ Ah, sorry, I misunderstood that. Thanks forthe explanation! $\endgroup$ Apr 5, 2020 at 12:14

1 Answer 1


The function $\lambda:A\mapsto\int_{\Sigma}{\mathbf 1}_{A}f^2\mathrm d\mu$ is a measure on the Borel sets of $\mathbb R^3$, for ${\mathbf 1}_A$ the indicator function of $A$. The quantity you are interested in is $r\mapsto\lambda(B_r(x))$.

We can use the monotone convergence theorem to see that $$ \lim_{r\uparrow r_0}\lambda(B_r(X)) = \lambda(B_{r_0}(x)), $$ so the left continuity is true for every $x$. Using Lebesgue's dominated convergence theorem, we see that in fact $$ \lim_{r\downarrow r_0}\lambda(B_r(X)) = \lambda\big({\overline B}_{r_0}(x)\big) $$ where ${\overline B}$ denotes the closed ball, so continuity holds at $r_0$ for fixed $x$ whenever $\lambda(\partial B_{r_0}(x))=0$.

I will show that only countably many $(x,r)$ may be such that $\lambda(\partial B_{r}(x))\neq0$; in particular your function will be continuous on $\mathbb R_+$ for all but countably many $x$. I assume the surface is closed, but it is not a necessary hypothesis, as I discuss at the end.

Fix $\varepsilon,R>0$, and let $S = S_{\varepsilon,R}$ be the set of pairs $(x,r)$ such that ${\overline B}_r(x)\subset B_R(0)$ and $\Sigma\cap\partial B_{r}(x)$ has surface area larger than $\varepsilon$ (seen as a subset of $\Sigma$). For a finite collection ${(x_i,r_i)}_{0<i\leq k}$ of elements of $S$, the sum of the areas of $\Sigma\cap\partial B_{r_i}(x_i)$ (which is at least $k\varepsilon$) is the area of $\Sigma\cap\bigcup_i\partial B_{r_i}(x_i)$, because the intersection of two distinct spheres is a circle, a point or empty, hence has measure zero. In particular, the sum is less than the area of ${\overline B}_R(0)\cap\Sigma$, which is finite (because the surface is closed). This means that $S_{\varepsilon,R}$ is in fact finite.

This concludes readily: if $(x,r)$ is a point such that $\lambda(\partial B_r(x))\neq0$, then it belongs to $\bigcup_{n\geq1}S_{1/n,n}$, which is countable as a countable union of finite sets.

If $\Sigma$ is not a closed surface (I imagine this means that it is not embedded, so it goes beyond your question), one may, in the definition of $S_{\varepsilon,n}$, replace $\partial B_n(0)\cap\Sigma$ by a countable collection of compacts subsets $K_n$ of $\Sigma$ whose interiors increase to $\Sigma$. Then if $(x,r)$ is a point such that the area of $\partial B_r(x)\cap\Sigma$ is larger than $\varepsilon$, by inner regularity there exists a compact subset $K\subset\Sigma\cap\partial B_r(x)$ with area larger than $\varepsilon$ as well. This compact is included in one of the interiors of the $K_n$, so $(x,r)$ belongs to the corresponding $S_{\varepsilon,n}$.

  • $\begingroup$ Thanks so much for the detailed answer. One soft question: Is this particular trick of considering $S_{\epsilon, R}$ (or its variations) commonly used in analysis? I feel that it may be useful in certain context.... $\endgroup$ Apr 7, 2020 at 8:10
  • 1
    $\begingroup$ I don't know that it's commonly used. If the sum of non-negative terms over an arbitrary set is convergent, then only countably many of them are positive (S = indices of terms $>\varepsilon$); a compact operator has at most countably many eigenvalues, with at most one limit point (S = eigenvalues with modulus $>\varepsilon$); a discrete closed set of $\mathbb R^d$ is countable (S = elements with norm $\leq R$). I guess it is useful enough to be found here and there, but not sophisticated enough to have a name. $\endgroup$
    – Pierre PC
    Apr 7, 2020 at 14:28

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