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Let $(I,B,\mu)$ be the standard probability space with $I=[0,1]$ , $B=$ Borel sets, and $\mu$ is Lebesgue measure. Now, for each $n\in N$, let $F_n$ be the rotation $F_n(x)=x+1/n \pmod 1$. Is it correct that for every Borel set $E$ , $ \mu (F_nE\;\triangle\; E)\to0$ as $n\to \infty$?

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1 Answer 1

Yes. This is a consequence of the fact that translation (and also rotation) is continuous with respect to $L^1$. See this question and its answer.

Edit: Another way to write this is: let $f = 1_E$ be the indicator function of $E$. We want to show $$\int |f(F_n(x)) - f(x)| d\mu \to 0$$ (you can check that the integrand is the indicator of $F_n E \triangle E$). In other words, we want to show $f \circ F_n \to f$ in $L^1(\mu)$.

The proof, as outlined in the answer I linked, goes as follows. First show that if $g$ is a continuous function, we have $g \circ F_n \to g$ in $L^1$. (Continuity of $g$ shows that $g \circ F_n \to g$ pointwise, and dominated convergence does the rest.) Then choose a continuous $g$ with $||g-f||_{L^1} < \epsilon$ (which is possible since $C(I)$ is dense in $L^1(\mu)$). Now $$||f \circ F_n - f|| \le ||f \circ F_n - g \circ F_n|| + ||g \circ F_n - g|| + ||g-f||.$$ The third term is less than $\epsilon$, and so is the first (since $\mu$ is invariant under $F_n$), and the second term goes to 0. So by taking the limsup of both sides we have $$\limsup_{n \to \infty} ||f \circ F_n - f|| \le 2 \epsilon.$$ But $\epsilon$ was arbitrary.

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Hi, thanks for your answer, but I didn't quite get it. Could you please give a more detailed explanation. –  user25640 Feb 24 '12 at 23:57
    
@user25640: Done. –  Nate Eldredge Feb 25 '12 at 0:22

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