A function $g\colon M\to M$ is called to conserve the volume if for each Jordan-measurable subset $J\subset M$ it is $\text{vol}(g^{-1}(J))=\text{vol}(J)$.
Resurrection Theorem by Poincaré: Let $M\subset\mathbb{R}^N$ be a Jordan-measurable set and $g\colon M\to M$ a bijective, continious function that conserves the volume and which inverse function is continious, too. Then in each open Jordan-measurable set $U\subset M$ there is a point $x\in U$ and a number $n\geqslant 1$ such that $g^n (x)\in U$.
Proof:
Because $M$ is Jordan-measurable it is $\text{vol}(M)<\infty$. Furthermore it is $$ \text{vol}(M)=\text{vol}(g^n(U))~\forall~n\in\mathbb{N}.~~~~~~~~~~~(*) $$ So because of these both facts, it is impossible that infinite many $g^n(U),n\geqslant 1$ are pairwise disjoint. So there are $j>k\geqslant 1$ with $$ g^j(U)\cap g^k(U)\neq\emptyset. $$ From this follows $$ \emptyset\neq g^{-k}(g^j(U)\cap g^k(U))=U\cap g^{j-k}(U). $$ So for $n=j-k\geqslant 1$ there is a point $x\in U$ with $g^n(x)\in U$.
I have two questions concerning this proof.
1.) I do not understand the equation $(*)$. Can you please explain it to me: Why is $$ \text{vol}(M)=\text{vol}(g^n(U))~~~\forall n\in\mathbb{N}? $$
2.) Why is it impossible that infinite many $g^n(U)$ are disjoint? Do not understand it.
After reading more carefully, this is not Poincare's recurrence theorem, it is much weaker. One can prove a much stronger statement in that the set of points which do not come back to $U$ infinitely often has measure 0. The claim here is simply that one point comes back to $U$ sometime (which immediately implies it comes back infinitely often by repeating the argument).
A function $g\colon M\to M$ is said to conserve the volume if for each Jordan-measurable subset $J\subset M$ it is $\text{vol}(g^{-1}(J))=\text{vol}(J)$.
Proof:
Because $M$ is Jordan-measurable we have $\text{vol}(M) < \infty$ (this is important, we use it strongly, in fact the result is not true if $vol(M) = \infty$, consider translation on $\mathbb{R}$, a small interval does not have a recurrent point). Furthermore, since $g$ conserves volume, we have $$ \text{vol}(U)=\text{vol}(g^{-n}(U))~\forall~n\in\mathbb{N}.~~~~~~~~~~~(*) $$
Suppose by way of contradiction that $U,g^{-1}(U) , g^{-2}(U) \ldots$ are pairwise disjoint (since $g$ is continuous these sets are all measurable). Since $g$ preserves volume, all of these sets all have the same measure. Note that $V_m = vol(\cup_{i=1}^{m} g^{-i}(U))$ is open (so measurable) since it is the finite union of open sets (using that $g$ is continuous and $U$ is open). Then $$vol(V_m) = vol(\cup_{i=1}^{m} g^{-i}(U)) = \sum_{i=1}^{m} vol(g^{-i}(U)) = \sum_{i=1}^{m} vol(U) = mvol(U).$$ Why is this impossible? We have that $vol(V_m) = m vol(U)$ and $V_m \subset M$ (so $vol(V_m) \leq vol(M)$). Taking $m$ large, this is a contradiction, using that $vol(U) >0$ (the result is false if we do not assume this, take $U$ to even be the empty set).
So there are $j>k\geqslant 0$ with $$ g^{-j}(U)\cap g^{-k}(U)\neq\emptyset. $$ From this it follows, using that $g$ is a bijection, that $$ \emptyset\neq g^{k}(g^{-j}(U)\cap g^{-k}(U))=g^{k-j}(U) \cap U. $$ So for $n=k-j \geqslant 1$ there is a point $x\in U$ with $g^n(x)\in U$.
From here I would recommend looking at any ergodic theory book (I recommend the one by Paul Halmos), finding Poincare's recurrence and reading it. It will be enlightening, even if you do not understand fully the definitions, it will show you that this result above is strengthened in a big way, by a (much) better idea then to just look at the $V_m$.