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.

almost every? Any monotonous function has at most countably many discontinuities.. $\endgroup$