# Pre-image of set of measure zero

I am trying to prove that if $X$ and $Y$ are manifolds with the same dimension and $f:X\rightarrow Y$ is a local diffeomorphism, then the pre-image of a set of measure zero in $Y$ is a set of measure zero in $X$. Here, we assume the ambient space is $\mathbb{R}^N$, for some $N$.

Let $A\subseteq Y$ be a null-set. Do you think the best approach is to go by definition and fix a chart $\phi: U\subset X\rightarrow\mathbb{R}^n$ and prove that $B=\phi(f^{-1}(A)\cap X)$ has measure zero? I think maybe it is not difficult to find a family of "small" cubes which cover $B$ but this seems very measure-theoretic and I do not have a lot of practice with Measure Theory exercises and this is a Geometry module.

On the other hand, I have Sard's Theorem as a tool but I do not see how it can be used here.

Can anyone give me a hint on how to start or a hint on which approach to choose?

Thank you very much!

I believe that in order for the claim to be true some topological conditions must be imposed upon $X$. In the following, I shall presume $X$ to be separable, i.e. there exist a countable, dense, $A \subset X$. Almost surely this (or the even stronger property of being second countable) was assumed in your definition of the concept of manifold.

What is a null set $T$ in an arbitrary smooth manifold $X$? It is a subset $T$ such that for every chart $(U,h)$ of $X$, $h(U \cap T)$ is a null set in $\Bbb R^n$.

Let $S \subset Y$ be a null set and let $T = f^{-1}(S)$. Let $(U,h)$ be an arbitrary chart of $X$. For every $a \in A$, let $a \in U_a \subseteq X$ be an open subset such that $f \big| _{U_a} : U_a \to f(U_a)$ is a diffeomorphism. Let $V_a = U_a \cap U$ and notice that $f \big| _{V_a} : V_a \to f(V_a)$ remains a diffeomorphism and that, if $B = A \cap U$, then $U = \bigcup _{a \in B} V_a$, so that $T \cap U = \bigcup _{a \in B} (T \cap V_a)$.

Let us show that $T \cap V_a$ is a null set, $\forall a \in B$. Since $f \big| _{V_a} : V_a \to f(V_a)$ is a diffeomorphism (therefore bijective), it follows that $T \cap V_a = f \big|_{V_a} ^{-1} (S) \cap f \big|_{V_a} ^{-1} \big( f(V_a) \big) = f \big|_{V_a} ^{-1} \big( S \cap f(V_a) \big)$. If $(V,k)$ is any other chart of $X$, then:

• if $V \cap V_a = \emptyset$, then $k \big( V \cap (T \cap V_a) \big) = \emptyset$ which is clearly a null set in $\Bbb R^n$;

• if $V \cap V_a \ne \emptyset$, then

$$k \big( V \cap (T \cap V_a) \big) = k \big|_{V_a} \big( (V \cap V_a ) \cap (T \cap V_a) \big) = k \big|_{V_a} \Big( f \big| _{V_a} ^{-1} \big( f \big| _{V_a} (V \cap V_a) \big) \cap f \big|_{V_a} ^{-1} \big( S \cap f(V_a) \big) \Big) = \\ \big( k \big|_{V_a} \circ f \big|_{V_a} ^{-1} \big) \Big( \big( f \big| _{V_a} (V \cap V_a) \big) \cap \big( S \cap f(V_a) \big) \Big)$$

which is clearly a null set in $\Bbb R^n$ because $\big( k \big|_{V_a} \circ f \big|_{V_a} ^{-1}, f \big| _{V_a} (V \cap V_a) \big)$ is a chart for $Y$ and $S \cap f(V_a)$ is a null set in $Y$ (being a subset of $S$, which is a null set).

This shows that $T \cap V_a$ is a null set, $\forall a \in B$, according to the definition.

Remembering that $B$ is at most countable (being a subset of $A$, which is countable), and denoting by $m$ the Lebesgue measure in $\Bbb R^n$, we may then write

$$m \big( h (T \cap U) \big) = m \Big( h \big( \bigcup _{a \in B} (T \cap V_a) \big) \Big) \le \sum _{a \in B} m \big( h (T \cap V_a) \big) = \sum _{a \in B} 0 = 0 ,$$

which shows that $h (T \cap U)$ is a null set in $\Bbb R^n$ and, since $(U,h)$ was an arbitrary chart, implies that $T = f^{-1} (S)$ is a null set in $X$.

• Thank you very much for having the patience to write such a complete answer! – user194469 Jan 26 '17 at 20:29

Notice that it suffices to consider sets $A$ of zero measure where the Jacobian of $f$ is bounded from above and below. By the change of variable formula the measures of $A$ are $f^{-1}A$ are the same (because of the hypothesis on the Jacobian). For the general case, write $A$ as a countable union of sets of zero measure where the Jacobian of $f$ is bounded from above and below.

Indeed, for such a set $A$ we have $$\int_{f^{-1}A}1=\int_A |J_f|$$ and so $$c_1\int_A 1\le\int_{f^{-1}A}1\le c_2\int_A 1$$ for some constants $c_1,c_2>0$. Hence, $A$ has measure zero if and only if $f^{-1}A$ has measure zero.

• Thank you for your answer. Can you give some more details on why it suffices to consider that case? I don't think I understand why – user194469 Jan 20 '17 at 20:25
• Oh uau, that seems very nice to me! I still have some questions. I do not have an integration formula for manifolds, just the one for $\mathbb{R}^n$. Can I reduce the problem to the case where $A\subset\mathbb{R}^n$? If I can, then I understand your proof. Where does it use the fact that $f$ is a local diffeomorphism? – user194469 Jan 20 '17 at 20:42
• Sure, you can use it in $\mathbb R^n$ (the measure on the manifold is induced from the measure on $\mathbb R^n$). If it weren't a local diffeomorphism, $|J_f|$ could vanish at some points and thus you would need to take $c_1=0$. – John B Jan 20 '17 at 20:45
• Thank you very much for the patience! I'll think about it! :) – user194469 Jan 20 '17 at 20:46
• Why can we write $A$ as a countable union of sets of measure zero where the Jacobian is bounded? – user194469 Jan 22 '17 at 22:25