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. 



Continuity of $r\mapsto\int_{\Sigma\cap B_r(x)}f^2d\mu$
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 $S\_{\varepsilon,n}$.


