To construct an invariant measure on the product space $M$ is a separable complete metric space. $N$ is a compact metric space. $F$, defined by $F(x,y)=(f(x),g(x,y))$, is a continuous transformation from $M\times N$ to itself. We know there is a measure $\mu$ on the space $M$ which is invariant under $f$. Then, the problem asks me to show there is a measure $m$ on $M\times N$ which is invariant under $F$, such that $\pi_*m=\mu$, where $\pi : M \times N \to M$ is the projection map. 
For each $x\in M$, I try to define an invariant measure $\nu_x$ which is invariant for $g_x(y)=g(x,y)$, and then glue all these measures together. But I can not proceed any further, partially due to the fact that since $f$ is not invertible, it is very difficult to make the measure thus constructed really invariant globally. 
I was told that the following theorem should be used. 
A set of probability Borel measures on X, $\Gamma$, is called tight, if for any $\varepsilon$, there is some compact subset $K$ such that $\mu(K)>1-\varepsilon$ for any $\mu\in \Gamma$. The theorem asserts that if $\Gamma$ is tight, then any sequence of measures $\mu_n$ in $\Gamma$ admits some convergent subsequence, with respect the weak$^*$ topology. 
So what we need is to find a sequence of measures which are "quasi invariant", and then to take the limit to make things work. But I am still stuck here. Could anyone help me with this? Thank you very much.
 A: One way to procede is to adapt the usual proof of the Bogolyubov-Krylov theorem to this non-compact case. In the following, I will assume that the given $f$-invariant measure $\mu$ is finite (and then without losing any generality that it is a probability). I don't know if this assumption is necessary, but it will be essential for my argument. In any case, this is a common assumption in the context of dynamical systems and ergodic theory.
Recall the usual proof of the Bogolyubov-Krylov theorem: to find a $F$-invariant measure we start with an arbitrary probability measure $\nu$ and we look at the sequence of iterates
$$
m_n = \frac{1}{n} \sum_{i=0}^{n-1} (F_*)^i \,\nu
$$
where $F_*$ denotes the push-forward by $F$. By the Prokhorov theorem, if the sequence $m_n$ is tight it must have a weakly converging subsequence, and then we can show that the limit must be $F$-invariant. In the compact case, any set of probability measures is tight so we are done.
In the non-compact case, this argument fails because mass can escape to infinity and the sequence won't necessarily be tight. However, the same argument will work if we can manage to restrict everything to a suitable subspace where we are guaranteed that the measures $m_n$ will converge weakly. 

With this in mind, let $\mathcal{P}( M \times N)$ be the space of probability measures on $M \times N$ endowed with the weak topology, and let $\mathcal{P}_{\mu}$ be the subspace of measures that satisfy $\pi_* m = \mu$, where $\pi: M \times N \to M$ is the projection. $\mathcal{P}_{\mu}$ is non-empty since it contains, for example, the product measures $\mu \times \nu$, where $\nu \in \mathcal{P}(N)$. 
The crucial facts that we need to make the Bogolyubov-Krylov argument work is that $\mathcal{P}_{\mu}$ is a convex, compact and $F_*$-invariant subset of $\mathcal{P}( M \times N)$. Here is a proof:


*

*$\mathcal{P}_{\mu}$ is convex.


If $\alpha$, $\beta \in \mathcal{P}_{\mu}$ and $s \geq 0$, then
$\pi_* [ (1-s) \alpha +s \beta ] = (1-s) \pi_* \alpha + s \pi_* \beta = (1-s) \mu + s \mu = \mu$. Thus $(1-s) \alpha +s \beta \in \mathcal{P}_{\mu}$.


*$\mathcal{P}_{\mu}$ is closed.


$\mathcal{P}_{\mu}$ is the preimage of the closed set $\{ \mu \}$ under the map $\pi_* : \mathcal{P}( M \times N ) \to \mathcal{P} (M)$, which is continuous. (See this answer for details). 


*$\mathcal{P}_{\mu}$ has compact closure (hence is compact by 2.).


By the Prokhorov theorem, it suffices to check that $\mathcal{P}_{\mu}$ is tight. This will follow from the fact that $\mu$, being a probability measure on a Polish space, is itself tight. Indeed, this means that for any given $\varepsilon >0$ there is a compact set $K \subset M$ with $\mu(K) \geq \mu(M) - \varepsilon$. Then $K \times N$ is compact and for any $m \in \mathcal{P}_{\mu}$ we have:
\begin{align}
m( K \times N) 
= m( \pi^{-1} (K) ) 
\stackrel{*}{=} \mu(K) 
\geq \mu(M) - \varepsilon 
\stackrel{*}{=} m( \pi^{-1}(M)) - \varepsilon 
= m( M \times N) - \varepsilon
\end{align}
where the equalities marked with $*$ follow from $\pi_* m = \mu$. Thus we have $m( K \times N) \geq m( M \times N) - \varepsilon$ for every $m \in \mathcal{P}_{\mu}$ and the set is tight.


*$\mathcal{P}_{\mu}$ is $F_*$-invariant. 


This will follow from the fact that $\mu$ is $f$-invariant and the special form of the map $F$. Indeed, by definition of $F$ we have $\pi \circ F = f \circ \pi$. Therefore, for any $m \in \mathcal{P}_{\mu}$ we have:
$$
\pi_* ( F_* m ) = ( \pi \circ F)_* m = (f \circ \pi)_* m = f_* ( \pi_* m)
=f_* \mu = \mu
$$
Hence $F_* m  \in \mathcal{P}_{\mu}$. 

Now that everything is on the table, the argument goes exactly as in the Bogolyubov-Krylov theorem. Choose an arbitrary starting measure $\nu \in \mathcal{P}_{\mu}$ (recall that $\mathcal{P}_{\mu}$ is non-empty). Then define the iterates
$$
m_n = \frac{1}{n} \sum_{i=0}^{n-1} (F_*)^i \,\nu
$$
Then $m_n \in \mathcal{P}_{\mu}$ for all $n$ by the $F_*$-invariance and convexity of $\mathcal{P}_{\mu}$. Since $\mathcal{P}_{\mu}$ is compact, the sequence has a subsequence converging to some $m \in \mathcal{P}_{\mu}$. The proof that $m$ is $F$-invariant goes through without change.
