# How to show that this function is continuous (Geometric Measure Theory)

I want to prove that the function $$F: \mathbb{R}_+ \to \mathbb{R}$$ defined by $$F(t)=\int_{\{d=t\}} g \, d\mathcal{H}^{n-1}$$ is continuous if $$g:\Omega \subset \mathbb{R}^n\to \mathbb{R}$$ is lipschitz on the bounded open set $$\Omega$$ and $$d:\Omega \to \mathbb{R}$$ is the euclidean distance from $$\Gamma \subset \overline{\Omega}$$. In my problem, $$\Gamma$$ is an (n-1)-dimensional submanifold of $$\mathbb{R}^n$$. Does this result still hold if $$g$$ is only continuous, $$d$$ is replaced by a lipschitz function, and $$\Omega$$ unbounded? (partial answers are welcome)

• You may have all sorts of trouble. Let's say $\Gamma\subset\mathbb R^2$ includes two close parallel intervals. Then, when $t$ reaches the half-distance between them, your measure suddenly collapses (because two local equidistants become one). There are other bad scenarios too, so some additional assumptions are needed (or you may ask for less). – fedja Dec 1 '18 at 13:06
• This doesn't happen if you are close to $\Gamma$. Suppose you can project to the manifold $\Gamma$ in a tubular neighbourhood of $\Gamma$ so that every point near $\Gamma$ admits a unique projection onto $\Gamma$ (this is indeed true, and the projection is also lipschitz continuous if you are close enough - in general with lipschitz constant greater than 1, but as close to 1 as you wish, up to restricting to smaller and smaller tubular neighbourhoods). So for example there is no way to project the center of a circumference, but you can project the other points as they are close enough. – HighLiuk Dec 9 '18 at 14:47
• Then, if you are interested in small $t$ only and the manifold is $C^2$, you can locally make a change of variables sending it to a hyperplane. However, you still may have a problem with the boundary of $\Omega$: imagine the scenario in $\mathbb R^2$ when $\Gamma$ is a line and $\partial\Omega$ contains intervals parallel to $\Gamma$ and arbitrarily close to it. Each such interval will create a jump. So, you need to impose some kind of transversality condition for $\Gamma$ and $\partial\Omega$ too. – fedja Dec 9 '18 at 21:48
• I'm not sure to have understood what you mean. If you can be more explicit in describing the situation, I would be glad. Anyway, let's also suppose $\mathcal{H}^{n-2}(\Gamma \cap \partial \Omega) < \infty$. Can this help? Please also be more precise in describing the change of variables, as you then have to deal with the Jacobian inside. Thanks in advance – HighLiuk Dec 16 '18 at 10:32
• What I meant was the following. Let $\Gamma=[0,1]$ and let $\Omega=\{|y|<f(x)\}$ where $f(x)=\frac 1n$ when $\frac 1{n+1}<x\le \frac 1n$. The condition $\mathcal H^{n-2}(\Gamma\cap\partial\Omega)=0$ won't help for $n>2$. The intersection may be just one point and this effect will still happen. – fedja Dec 16 '18 at 15:21