# If $f∈C^1$ and $\{∇f=0\}$ has Lebesgue measure $0$, then $\{f∈B\}$ has Lebesgue measure $0$ for all Borel measurable $B⊆ℝ$ with Lebesgue measure $0$

Let $$d\in\mathbb N$$ and $$f\in C^1(\mathbb R^d)$$. Assume $$\left\{\nabla f=0\right\}$$ has Lebesgue measure $$0$$.

How can we conclude that $$\left\{f\in B\right\}$$ has Lebesgue measure $$0$$ for all Borel measurable $$B\subseteq\mathbb R$$ with Lebesgue measure $$0$$?

The claim can be found in an answer on mathoverflow.

The author writes that the claim "is true locally, in a neighborhood of each point where $$\nabla f\ne0$$, due to the implicit function theorem". Honestly, I don't even understand what exactly he's meaning.

Let $$a\in\mathbb R^d$$ with $$\nabla f(a)\ne0$$. Then surely, by continuity of $$\nabla f$$ at $$a$$, there is an open neighborhood $$N$$ of $$a$$ with $$\nabla f(x)\ne0\;\;\;\text{for all }x\in N\tag1.$$ But how do we need to apply the implicit function theorem and what's the resulting "local" conclusion? Maybe that $$N\cap\left\{f\in B\right\}$$ has Lebesgue measure $$0$$?

• Yes you need to prove that $N\cap\{f\in B\}$ has measure $0$. If you show this the result follows, because there is a countable family of open sets $\{B_i\}$ covering $\Bbb R^d$ such that $B_i\cap\{f\in B\}$ has measure $0$. Then $\mu(\{f\in B\})=\sum_i \mu(\{f\in B\}\cap B_i)=0$. – Adam Chalumeau May 6 at 15:57
• @AdamChalumeau Thank you for the clarification. I'm struggling to understand how we need to apply the implicit function theorem (I'm sure it's easy). – 0xbadf00d May 6 at 16:47

Non-singular maps. A map $$f\,:\,\mathbb{R}^N\rightarrow\mathbb{R}^M$$ whose inverse image preserves null-sets, -- i.e., $$\mu(f^{-1}(B))=0$$ for any null-set $$B$$, -- is often referred to as a non-singular map. The question is about showing that the class of $$\mathcal{C}^1$$ maps with non-singular Jacobian (or more specifically, non-zero gradient) almost everywhere is contained in the class of non-singular maps.

Remark. $$c\in\mathbb{R}$$ is called a regular value of $$f\in\mathcal{C}^{1}(\mathbb{R}^d)$$, if $$\nabla f(x)\neq 0$$ for all $$x\in f^{-1}(c)$$. The Implicit Function Theorem (IFT) asserts that $$f^{-1}(c)$$ is a (d-1)-dimensional submanifold of class $$\mathcal{C}^1$$ -- for any regular value $$c$$. Hence, $$f^{-1}(c)$$ is a null-set.

Let $$\widetilde{N}\overset{\Delta}=\left\{\nabla f\neq 0\right\}$$ (which is open).

From the IFT, we have that $$\widetilde{N}\cap f^{-1}(c)$$ is a (d-1)-submanifold of class $$\mathcal{C}^1$$.

Now, you have $$\widetilde{N}\cap \left\{f\in B\right\}=\bigcup_{t\in B}\widetilde{N} \cap f^{-1}(t)$$, where $$\widetilde{N} \cap f^{-1}(t)$$ is a null-set for all $$t$$ from the above remark (since it is a $$\mathcal{C}^1$$ submanifold from the IFT). Therefore, when $$B$$ is countable, the referred set is a null-set.

When $$B$$ is uncountable, it follows from Fubini's Theorem that $$\bigcup_{t\in B}\widetilde{N}\cap B_r \cap f^{-1}(t)$$ is a null-set for any bounded open ball $$B_r$$.

To see this latter claim, we can resort to a more specialized form of Fubini tailored to our case (referred to as co-area formula),

$$\int_{\widetilde{N}\cap B_r} g\left|\nabla f\right| d\mu = \int_{\mathbb{R}} \left(\int_{f^{-1}(t)\cap\widetilde{N}\cap B_r} g(x)d\mu_{d-1}(x)\right) dt$$.

Take $$g$$ to be the indicator of the foliation $$\bigcup_{t\in B}\widetilde{N}\cap B_r \cap f^{-1}(t)$$ and note that

$$\int_{\mathbb{R}} \left(\int_{f^{-1}(t)\cap\widetilde{N}\cap B_r} g(x)d\mu_{d-1}(x)\right) dt=\int_{B} \left(\int_{f^{-1}(t)\cap\widetilde{N}\cap B_r} d\mu_{d-1}(x)\right) dt=0$$,

where the last identity holds since $$B$$ is a null-set. Thus,

$$\int_{\widetilde{N}\cap B_r} g\left|\nabla f\right| d\mu=0$$ and therefore $$g\left|\nabla f\right|=0$$ almost everywhere in $$\widetilde{N}\cap B_r$$. Since, $$\left|\nabla f\right|\neq 0$$ almost everywhere, it follows that $$g(x)=0$$ almost everywhere in $$\widetilde{N}\cap B_r$$. In other words,

$$\mu\left(\bigcup_{t\in B}\widetilde{N}\cap B_r \cap f^{-1}(t)\right)=\int g d\mu =0$$.

Update. For the sake of completeness, I am adding the general statement.

Theorem 1. Let $$f\,:\,\mathbb{R}^N\rightarrow \mathbb{R}^M$$ be smooth (i.e., $$f\in\mathcal{C}^1$$). If the set of critical points of $$f$$ is a null-set, i.e.,

$$\mu\left(\left\{x\in\mathbb{R}^N : \text{rank} \left(Df(x)\right)<\min\left\{M,N\right\}\right\}\right)=0,$$

then, $$\mu\left(f^{-1}(B)\right)=0$$ for any null-set $$B$$.

The proof follows from the IFT and Fubini (or, more precisely, the co-area formula) just as done before.

Update 2. I am adding a Corollary.

Definition. [Null-sets on manifolds] Let $$\mathcal{V}$$ be a smooth manifold of dimension $$d$$ with smooth structure $$\left\{U_{\alpha},\varphi_{\alpha}\right\}$$. $$A\subset \mathcal{V}$$ is called a null subset of $$\mathcal{V}$$ if $$\mu\left(\varphi_{\alpha}(U_{\alpha}\cap A)\right)=0$$ for all $$\alpha$$.

Relevant Property. If $$\mu(\widehat{A})=0$$ with $$\widehat{A}\subset \mathbb{R}^d$$ then,
$$\varphi^{-1}_{\alpha}(\widehat{A})$$ is a null-set for any $$\alpha$$. This follows by observing that $$\varphi_{\beta}\left(U_{\beta}\cap\varphi^{-1}_{\alpha}(\widehat{A})\right)=\varphi_{\beta}\circ \varphi^{-1}_{\alpha}(\widehat{A})$$ is necessarily a null-set, for any $$\beta$$, since $$\widehat{A}$$ is a null-set and $$\varphi_{\beta}\circ \varphi^{-1}_{\alpha}$$ is a diffeomorphism -- hence, from Theorem 1, $$\varphi_{\beta}\circ \varphi^{-1}_{\alpha}(\widehat{A})$$ is a null-set.

In the next corollary, we assume that the manifolds admit countable atlas -- i.e., are separable.

Corollary 1. Let $$f\,:\,\mathcal{M}\rightarrow \mathcal{N}$$ be a smooth map between two smooth separable manifolds $$\mathcal{M}$$, $$\mathcal{N}$$ of dimensions $$M$$ and $$N$$, respectively. If the set of critical points of $$f$$ is a null-set, then $$f^{-1}(B)$$ is a null-set for any null-set $$B$$.

For the proof, one just needs to notice that any local coordinate representation of $$f$$ fulfills the conditions of Theorem 1.

• It's a bit unfortunate that you redefined the symbol $N$ which is already used in the question. – 0xbadf00d May 6 at 18:29
• Something is wrong in your post: (a) $f$ is a real-valued function. So, the notation $f^{-1}(c)$ (which I suppose is a shorthand for $f^{-1}(\{c\})$) is undefined. (b) Isn't a regular value defined to be the opposite of a critical/stationary point? (So, $a\in\mathbb R^d$ is a regular point of $f$ iff $\nabla f(a)\ne 0$. In this case, as mentioned in the question, there is a neighborhood $N$ of $a$ with $(1)$.) I guess you confused this with the notion of a regular value of $f$ which is precisely defined as the image of $f$ at a regular point. – 0xbadf00d May 6 at 18:49
• If I am not missing anything, the $N$ you have in your question, is contained in the $N$ I have in my answer; (a) $f^{-1}(c)=\left\{x\in\mathbb{R}\,:\,f(x)=c \right\}$ refers to the inverse image of $c\in\mathbb{R}$ by $f$. Yes, one should rather write $f^{-1}(\left\{c\right\})$; (b) what is relevant is that $f^{-1}(c)$ is a surface whenever $c$ fulfills the above characterizing property (that I am referring in this case as regular). – Augusto Santos May 6 at 19:01
• Yes, sure, but if I'm reading other questions I find it highly confusing if symbols are defined more than once. – 0xbadf00d May 6 at 19:06
• Could you elaborate on how exactly we obtain that $f^{-1}(c)$ is a $(d-1)$-dimensional submanifold by the IFT? Since this is the only crucial part I don't understand (and actually the reason why I asked this question). – 0xbadf00d May 6 at 19:09

I don't know about the implicit function theorem, but you can a one related theorem, the Local Submersion Theorem. With your notations, locally around $$a$$, $$f$$ looks like a projections onto the first coordinate. You are left to prove that $$p:(x_1,\dots,x_d)\in\Bbb{R}^d\mapsto x_1$$ has the property that $$(B\text{ has measure zero})\Longrightarrow (\{p\in B\}\text{ has measure zero}).$$ But $$\{p\in B\}=B\times\Bbb R^{d-1}$$ so you can conclude.

• You have to be careful with the degrees of smoothness: Usually, submersions are defined in the $C^\infty$ category, while OP is only assuming $C^1$. But Submersion theorem locally is the same as the Implicit Function Theorem. – Moishe Kohan May 6 at 19:25
• I don't know the "local submersion theorem". However, the following is clear to me: If $M$ is a $k$-dimensional submanifold of $\mathbb R^d$ and $a\in M$, there is an open neighborhood $U_a$ of $a$ and a $C^1$-diffeomorphism $\varphi_a$ from $U_a$ to an open subset $V_a$ of $\mathbb R^d$ with $$\varphi_a(U_a\cap M)=V_a\cap(\mathbb R^k\times\{0\})$$ and hence $U_a\cap M$ is Borel measurable and has Lebesgue measure $0$ (assuming $k<n$). – 0xbadf00d May 7 at 15:50
• Moreover, it's clear to me that if $c\in\mathbb R$ is a regular value of $f$ (i.e. $c=f(a)$ for some $a\in\mathbb R^d$ with $\nabla f(a)\ne 0$), then $M_c:=\{f=c\}$ is a $(d-1)$-dimensional submanifold of $\mathbb R^d$. But I don't know how we can conclude from that for the same reason I don't see why we can conclude in the other answer. It would be clear if $\{f\in B\}=\bigcup_{c\in B}M_c$ would be a countable union. – 0xbadf00d May 7 at 15:50