Level set of a continuous function strictly increasing in each argument Let $F : \mathbb{R}^d \to [0,1]$ be absolutely continuous and strictly increasing in each argument. Is it true that the boundary of the set $\{ \boldsymbol{x} \in \mathbb{R}^d: F(x) \geq \alpha \}$ for $\alpha \in (0,1)$ is equal to the $\alpha$-level set of $F$, i.e. to the set $\{ \boldsymbol{x} \in \mathbb{R}^d: F(x) = \alpha \}$?
 A: We can prove much more general topological fact. Let $X$ be a topological space and $F:X\to\Bbb R$ be a continuous function without local minima, $\alpha\in\Bbb R$, $A=\{x\in X: F(x) \geq \alpha \}$,  and $B=\{x\in X: F(x) =\alpha \}$. Then $\partial A=B$. Indeed, since the function $F$ is continuous the sets $A=F^{-1}[\alpha,\infty)$  and $B=F^{-1}(\alpha)$ are closed. By definition, $\partial A=\overline{A}\setminus\operatorname{int} A=A\setminus\operatorname{int} A $. Let $x\in B$ be an arbitrary point and $O_x$ be an arbitrary neighborhood of the point $x$. Since the function $F$ has no local minima, there exists a point $y\in O_x$ such that $F(y)<F(x)=\alpha$. Therefore $y\not\in A$ and so $x$ is not an interior point of the set $A$. Thus $B\subset A\setminus\operatorname{int} A=\partial A$. Conversely, let $x\in\partial A$ be an arbitrary point. Suppose that $F(x)>\alpha$. Since the function $F$ is continuous, there exists an open neighborhood $O_x$ of the point $x$ such that $F(y)>\alpha$ for each point $y\in O_x$. Then $x\in O_x\subset\operatorname{int} A$. But $\partial A\cap\operatorname{int} A=\varnothing,$ a contradiction. Hence $F(x)=\alpha$, and thus $x\in B$. 
