Exercise: The measure of the boundary of a level set of a Holder continuous function is positive 
Let $M$ be a compact manifold in $R^3$ and consider a non negative
function $f\in C^{1,\alpha}(M)$ for some $0<\alpha<1$. I am interested
in showing that $|\partial \{f>0\} |>0$, where $|\cdot|$ stands for
the Hausdorff measure and $\partial A$ denotes
the boundary of $A$.

My Attempt:
By definition, $x^* \in M$ is a boundary point of $\{f>0\}$, if for every $r>0$ it holds that $B_r(x^*) \cap \{f>0\} \neq \emptyset$ and $B_r(x^*) \cap \{f=0\} \neq \emptyset$. In other words, if for every ball with center at $x^*$, I can find not only points of $\{f>0\}$ but also of its complement. Now, it is also true that $M=\{f>0\}\cup \{f=0\} $. Hence, if I could find/construct a dense subset of $\{f>0\}\cup \{f=0\}$, let's say $D$, I could obtain that $B_r(x^*) \cap D \neq \emptyset$ by density. This in particular would yield that $B_r(x^*) \cap (\{f>0\}\cup \{f=0\})=(B_r(x^*) \cap \{f>0\})\cup (B_r(x^*) \cap\{f=0\}) \neq \emptyset$. At this point, two issues arise:

*

*The latter does not imply that both intersections must be non empty, but only at least one between them.

*I am not sure if such a dense set exists.

DISCLAIMER: I have seen some similar question here in which the answer is given by a Cantor set. However, since I am not that familiar with topology, I would prefer to prove this (if possible) using the "slightest" tools.
I feel that density is a key part here but I am not able to tackle this problem properly. Any help or hint will be much appreciated!
Thanks in advance.
 A: I do not understand why the dense set would help you in this situation as nonempty sets can still have zero Hausdorff measure. In this answer I will denote the $2$-dimensional Hausdorff measure by $\mathcal{H}^2$ (using the absolute value seems very misleading to me). In the comments I gave some heuristics. Let me back this up by something a bit more rigorous.
As I pointed out in the comments it depends on the desired notion of boundary (either withing the manifold with the subspace topology or in $\mathbb{R}^3$). As you are using $\{ f> 0 \}^c = \{ f=0 \}$, you seem to take the boundary within the manifold. In this case, your claim is not true. Consider
$$\mathbb{T}^2 = \{ (x,y,z) \in \mathbb{R}^3 \ : \ (x^2+y^2-2)^2+z^2 =1 \}$$
(some 2-dimensional torus) embedded in $\mathbb{R}^3$. This is a compact, smooth manifold. Now consider the map $f: \mathbb{T}^2 \rightarrow \mathbb{R}, f(x,y,z)=z^2$. Then $f$ is a smooth, nonnegative map. However, the boundary of the zero locus is equal to two circles (1-dimensional manifolds) and thus, the $2$-dimensional Hausdorff measure will vanish.
If you were to pick the boundary in the topology of $\mathbb{R}^3$, (meaning, the balls should intersect $\{ f>0 \}$ and $\{ f>0 \}^c$), then you get $\{f>0 \} \subseteq \partial \{ f>0 \}$ (that is because any ball in $\mathbb{R}^3$ intersects $M^c$). However, $\{ f>0 \}$ is an open subset of a $2$-dimensional, $C^{1,\alpha}$ manifold and thus is a $2$-dimensional $C^{1,\alpha}$ manifold itself. However, nonempty $2$-dimensional $C^{1,\alpha}$ manifolds have positive $2$-dimensional Hausdorff measure. As $M$ is embedded, for every point $p\in M$ there exists a diffeomorphism $\varphi: B_1(0) \rightarrow \mathbb{R}^3$ such that $$\varphi((-1,1)^3) \cap \{ f>0\} = \{ \varphi(x,y,0) \ : \ (x,y,0)\in (-1,1)^3 \}$$
and such that $\varphi(0)=p$. By the monotonicity of measures, we get
$$ \mathcal{H}^2( \varphi([-0.5, 0.5] \times [-0.5,0.5] \times \{0\})) \vert \leq \mathcal{H}^2( \{ f>0 \} ) \leq \mathcal{H}^2( \partial \{ f>0 \} ). $$
However, as $\varphi$ is $C^1$, we have that its restriction to $[-0.5,0.5]^3$ is bilipschitz (lets say with the inverse having lipschitz constant equal to $L$). By Measure of image of Lipschitz function is bounded? (the $C^{1,\alpha}$ regularity actually would allow you for a simpler proof of this fact, in fact we do not really need $M$ to be more than a topological manifold, see the next paragraph) we have
$$ 0<1 = \mathcal{H}^2([-0.5,0.5] \times [-0.5,0.5]\times \{0\})=\mathcal{H}^2(\varphi^{-1} \varphi([-0.5, 0.5] \times [-0.5,0.5] \times \{0\}))
\leq L^2 \mathcal{H}^2(\varphi([-0.5, 0.5] \times [-0.5,0.5] \times \{0\})). $$
Hence, we get
$$ 0< 1/L^2 \leq \mathcal{H}^2(\partial \{ f>0\}). $$
Added: The $C^{1, \alpha}$ assumption is not really need. Everything works in the topological setting if your manifold is locally given as a graph. Then we can use the projection map. I.e. we have
$$ \mathcal{H}^2((-1,1)^2)=\mathcal{H}^2(P(\{ (x,y, g(x,y)) \in \mathbb{R}^3 \ : \ (x,y)\in (-1,1)^2 \})) \leq \mathcal{H}^2(\{ (x,y, g(x,y)) \in \mathbb{R}^3 \ : \ (x,y)\in (-1,1)^2 \})  $$
where $g$ is any map (we need no regularity, only the fact that we can write it as a graph) and $P: \mathbb{R}^3\rightarrow \mathbb{R}^2, (x,y,z)=(x,y)$ is the projection. So you only need the estimate for the projection which is much easier to show (also because it has all the regularity one can hope for).
A: I think I figured out the missing keys so I will post the whole argument here as an answer. In case, there are still mistakes, please correct me.
We know that $f \in C^{1,\alpha}(M)$ for some $0<\alpha<1$. Hence, $f \in C^{1,\beta}(M)$ for some $0<\beta<\alpha<1$. We can find a sequence of smooth functions $g_n$ converging to $f$ such that $g_n(x)>0$ for all $x\in B_{\delta_n}(x)$ and $g_n(x)=0$ otherwise (here $\delta_n$ is chosen arbitrarily small). Moreover, since $M$ is a compact manifold, there exists a countable dense subset $\{ x_n, n \in \mathbb N\}$ in $M$. Passing to a subsequence $\hat x_n$ we obtain by density that $\hat x_n \to x^* \in M$.
We can write $M=M \setminus \big (\bigcup_{n=1}^{\infty} B_{\delta_n}(x_n) \big) \cup \big (\bigcup_{n=1}^{\infty} B_{\delta_n}(x_n) \big)$. We recall that $\big (\bigcup_{n=1}^{\infty} B_{\delta_n}(x_n) \big) \neq \emptyset$ since $\{x_n\}$ is dense. At this point, we distinguish cases. If $x^*\in M \setminus \big (\bigcup_{n=1}^{\infty} B_{\delta_n}(x_n) \big) $ then $f(x^*)=0$. If $x^* \in \big (\bigcup_{n=1}^{\infty} B_{\delta_n}(x_n) \big)$ then due to $g_n$ and by passing to a subsequence, we obtain that $f(x^*)>0$. This completes the argument.
