Equivalent definitions of a measurable set in a manifold There seem to be at least two ways of defining measurability in a smooth $n$-manifold $M$, and I want to know if these definitions are equivalent.
(1) For the first one, we say $A\subseteq M$ is measurable if for each coordinate chart $(U,\varphi)$, $\varphi(A\cap U)$ is a Lebesgue measurable set in $\mathbb{R}^n$.
The second one is a little complicated and goes as follows.
(2) $A\subseteq M$ is said to be measurable if it can be written as a countable union of sets $A_i$ with each $A_i$ contained in some coordinate chart $(U_i,\varphi_i)$ and thus mapped onto a Lebesgue measurable set $\varphi_i(A_i)$.
Now I'd like to determine whether or not (1)$\Leftrightarrow$(2). To prove the first definition implies the second one, I want to exploit the notion of compactness, because every open cover of a compact set contains a finite subcover. That might be helpful if we want to break $A$ into countably many pieces. However, I don't have too much information about $A$. Maybe it is not compact at all. Then what should I do? Thank you.
 A: Item (2) should be written more carefully. The phrase "and thus" seems inappropriate: just because $A_i$ is contained in the coordinate chart $(U_i,\phi_i)$ it does follows that $\phi_i(A_i)$ is a measurable subset. In fact measurability of $\phi_i(A_i)$ should be imposed as a condition. So (2) should be instead say

(2) A subset $A \subset M$ is measurable if $A$ can be written as a countable union of sets $A_i$, and for each $i$ there is a coordinate chart $(U_i,\phi_i)$, such that $A_i \subset U_i$ and such that $\phi_i(A_i) \subset \mathbb R^n$ is Lebesgue measurable.

To prove equivalence of (1) and (2), your idea of breaking $A$ into countably many pieces is very important, but I don't think compactness is particularly relevant. On the other hand, you can apply second countability of $M$ to choose a countable collection of coordinate charts $(U_i,\phi_i)$ so that the sets $U_i$ form a countable basis of $M$. You can then quickly prove (1)$\implies$(2) by taking $A_i = A \cap U_i$.
To prove the converse direction (2)$\implies$(1), consider any coordinate chart $(U,\phi)$ as in (1). From (2) we are given a countable collection of coordinate charts $(U_i,\phi_i)$ and a subset $A_i \subset A \cap U_i$ so that $A = \cup_i A_i$ and so that $\phi_i(A_i)$ is measurable in $\mathbb R^n$. We are assuming that $M$ has a countable basis, and therefore we can choose a countable collection of coordinate charts $(V_j,\psi_j)$ such that $U = \cup_j (U \cap V_j)$, and it follows that $A \cap U =  \cup_{i,j} (A_i \cap V_j)$. For each of the countably many ordered pairs of indices $(i,j)$, notice that $U_i \cap V_j$ is open in $U_i$, so $\phi_i(U_i \cap V_j)$ is open in $\mathbb R^n$, so the set
$$\phi_i(A_i \cap V_j) = \phi_i(A_i \cap U_i \cap V_j) = \phi_i(A_i) \cap \phi_i(U_i \cap V_j)
$$
is measurable in $\mathbb R^n$. Applying smoothness of the overlap map
$$\phi \circ \phi_i^{-1} : \phi_i(U \cap U_i) \to \phi(U \cap U_i)
$$
it follows that $\phi(A_i \cap V_j)$ is measurable in $\mathbb R^n$. Taking the union over the countable set of index pairs $(i,j)$ it follows that $\phi(A \cap U)$ is measurable in $\mathbb R^n$, as required to prove (1).
