Mixing system implies induced transformation is mixing

I am wondering if the following is true:

Suppose $$T:X\to X$$ is a transformation with probability measure $$\rho$$. Suppose tht $$T$$ is mixing. i.e., $$|\rho(A\cap T^{-n}B)-\rho(A)\rho(B)|\to 0$$

If $$Y\subset X$$ with positive measure and $$R:Y\to \mathbb{N}$$ is the first return time to $$Y$$ such that $$T^{R(y)}(y)\in Y$$ for all $$y\in Y$$, we can define the induced transformation by $$F=T^R:Y\to Y$$.

Can we say that $$F$$ is mixing (with respect to $$\rho|_Y$$)?

• is it obvious that $F$ preserves $\rho\mid_Y$? – mathworker21 Oct 11 at 1:10
• For example, take $X = \{1,2\}$ with uniform measure, $T(1)=2,T(2)=1$, $Y=X$, $R(1) = 1, R(2)=2$. Then $F(1)=2,F(2)=2$, so $F$ is not measure-preserving. – mathworker21 Oct 11 at 4:40
• Have a look at mathoverflow.net/questions/160955/…. I edited the question to refer to the first return time. – John B Oct 11 at 15:37
• @mathworker21 there is a theorem in "Laws of Chaos" by Boyarsky and Gora which confirms $F$ preserves the measure. I can not remember the number off the top of my head. – Mr Martingale Oct 11 at 17:25
• @JohnB this is perfect, thanks! – Mr Martingale Oct 11 at 17:25