# Continuity of $r\mapsto\int_{\Sigma\cap B_r(x)}f^2d\mu$

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.

• What do you mean by almost every? Any monotonous function has at most countably many discontinuities.. Apr 4, 2020 at 20:19
• @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. Apr 5, 2020 at 7:16
• @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 $r$. Apr 5, 2020 at 7:20
• Ah, sorry, I misunderstood that. Thanks forthe explanation! Apr 5, 2020 at 12:14

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 $${(x_i,r_i)}_{0 of elements of $$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 $$S_{\varepsilon,n}$$.

• 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.... Apr 7, 2020 at 8:10
• 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. Apr 7, 2020 at 14:28