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$M$ 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 on $M\times N$ which is invariant under $F$, such that $\pi_*m=\mu$, where $\pi$ 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.

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