Push-forward Lebesgue measure properties for diffeomorphism on compact manifolds

Question

I am trying to solve exercice 11.1.1 from Foundations of Ergodic Theory, by Marcelo Viana and Krerley Oliveira, which reads:

Let $f : M → M$ be a local diffeomorphism in a compact manifold $M$ and $m$ be the Lebesgue measure on $M$. Check the following facts:

1. If $m(B) = 0$, then $m(f^{−1}(B)) = 0$.
2. If $B$ is measurable then $f(B)$ is measurable.
3. If $m(B) = 0$ then $m(f(B)) = 0$.
4. If $A = B$, up to zero Lebesgue measure zero, then $f(A) = f(B)$ and $f^{−1}(A) = f^{−1}(B)$, up to zero Lebesgue measure.
5. If $A$ is an invariant set, then $f(A) = A$, up to zero Lebesgue measure.

I have read some great answers on the topic here, here and here, but I couldn't addapt the arguments to this case. Any hints would be appreciated.

Here is my attempt so far:

I have tried to write inclusions on a "clever" way in order to use compactness of $M$ or the fact that $f$ is a local diffeomorphism.

1. By definition, for all $a \in M$, there exists $a \in U_{a} \subset M$ open such that the restriction ${\displaystyle f\vert _{U_a}:U\to f(U_a)}$ is a diffeomorphim (i.e., $f(U_a)$ is open, $f$ is injective and $f^{-1}$ is also class $C^1$). Write $f^{-1}(B) = \cup_{a \in M}(f^{-1}(B) \cap U_a)$ and now I would like to prove that $(f^{-1}(B) \cap U_a)$ is a null set. This is obvious for those intersections where $f^{-1}(B) \cap U_a = \emptyset$, but I am now stuck on the cases where $f^{-1}(B) \cap U_a \neq \emptyset$. $f^{-1}(B) \cap U_a = {\displaystyle f\vert _{U_a}^{-1}(B)} \cap {\displaystyle f\vert _{U_a}^{-1}(f(U_a))} = {\displaystyle f\vert _{U_a}^{-1}(B \cap f(U_a))}$ and then what? I tried to follow with $B \subseteq f(f^{-1}(B))$, unsuccessfully. Where could I use compactness of $M$?
2. It suficces to consider the Borel sets: Let $\mathcal{E}$ be the class of all sets $B \in \mathcal{B}(M)$ such that $f^{-1}(B) \in \mathcal{B}(M)$. $\mathcal{E}$ is a $\sigma$-algebra that contains all open sets, since $f$ and $f^{-1}$ are continuous, therefore they send open sets to open sets. Hence, $\mathcal{E} = \mathcal{B}(M)$.
3. I believe this follows from the "substitution rule for measures" for the characteristic function $\chi$).

I will write my remaining attempts later, but any hints and help here would be great.