Is the integral of a continuous functions along fibers again continuous?

Question

Let $X,Y$ be metric spaces (say subsets of euclidean spaces if that should be necessary), $s>0$, and $F:X\to Y$ continuous. If $\phi: X\to\mathbb{F}$ is continuous, is the pushforward $$F_{!}(\phi): Y\to\mathbb{R}, y\mapsto \int_{F^{-1}(y)} \phi(x) \textrm{d}H^s$$ also continuous? Here $H^s$ denotes the $s$-dimensional Hausdorff measure.

I know it is measurable under some mild assumptions and I know it's smooth if $F$ and $\phi$ are smooth and $F$ is a submersion for example, but I wasn't able to prove continuity.

Answer 1

Not at all. The level sets of a continuous function can be very ugly. For simplicity, let $\phi\equiv 1$; then you are asking if $H^s(F^{-1}(y))$ is continuous with respect to $y$. It's not in general.

Let $K\subset \mathbb{R}$ be a fat Cantor-type set, so that $H^1(K)>0$. Let $F(x) = \operatorname{dist}(x,K)$; this is a Lipschitz continuous function. Its level sets $F^{-1}(y)$ are empty for $y<0$, yet the $H^1$ measure of $F^{-1}(0)$ is positive.

Additionally,

• $F^{-1}(y)$ is countable for $y>0$, because $K^c$ consists of countably many intervals, each of which has at most two points of the level set.
• One can use the standard Cantor set instead, letting $s=\log 2/\log 3$ in that case.
• $F$ be replaced by a $C^\infty$-smooth function (this is a special case of Whitney's theorem). The function will not be an immersion, of course.