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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!

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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 $X$.

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

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  • $\begingroup$ 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 $\endgroup$ – user194469 Jan 20 '17 at 20:25
  • $\begingroup$ 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? $\endgroup$ – user194469 Jan 20 '17 at 20:42
  • $\begingroup$ 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 $c_1=0$. $\endgroup$ – John B Jan 20 '17 at 20:45
  • $\begingroup$ Thank you very much for the patience! I'll think about it! :) $\endgroup$ – user194469 Jan 20 '17 at 20:46
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    $\begingroup$ Why can we write $A$ as a countable union of sets of measure zero where the Jacobian is bounded? $\endgroup$ – user194469 Jan 22 '17 at 22:25

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