- (15 points) Consider the mapping \(F : [ 0 , 1 ] \to [ 0 , 1 ]\) given by \(F ( s ) = s ^ { 2 }\)
Let \(F ^ { - j } ( A )\) be the inverse image of j iterates of F applied to a measurable subset \(A \subset [ 0 , 1 ]\) . That is, if \(F = F ^ { 1 }\) and \(F ^ { j } , j = 2 , 3 , . .\) . is defined inductively as \(F ^ { j } = F ^ { j - 1 } \circ F\) ， then \(F ^ { - j } ( A ) = \{ x : F ^ { j } x = y\) , some \(y \in A \}\)
Given \(N = 1 , 2 , \dots\) show that \(\begin{array} { r } { \mu _ { N } ( A ) = N ^ { - 1 } \sum _ { j < N } | F ^ { - j } ( A ) | } \end{array}\) is a measure which is absolutely continuous with respect to Lebesgue measure. Here |B| denotes the Lebesgue measure of a measurable set.
Show that \(\mu _ { N } ( [ a , b ] ) \to 0 { \mathrm { ~ i f ~ } } 0 < a < b \leq 1\)
If f is a continuous function on \([ 0 , 1 ]\) does \[ \lim _ { N \to \infty } \int _ { [ 0 , 1 ] } f ( s ) d \mu _ { N } ( s ) \] tend to a limit? If so, what is the limit?
Solution
For each \(j \ge 0\), let \[ \nu_j(A):=|F^{-j}(A)| = |\{s\in[0,1]:F^j(s)\in A\}|. \]
a) For any measurable sets \((A_n)\) that are pairwise disjoint, \((\nu_j(\bigsqcup A_n))=|F^{-j}(\bigsqcup A_n)|=\sum |F^{-j}(A_n)|=\sum\nu_j(A_n)\). So \(\nu_j\) is a measure on Borel sets. Here \(F^j(s)=s^{2^j}\) and so \(F^{-j}(A)=\{t^{2^{-j}}:t\in A\}\), the image of \(A\) under the map \(g_j(t)=t^{2^{-j}}\). Since \[g_j'(t)=2^{-j}t^{2^{-j}-1}\in L^1[0,1],\] \(g_j\) is absolutely continuous and has Luzin’s property (N), hence \(|A|=0\Rightarrow |\{t^{2^{-j}}:t\in A\}|=0\). Therefore \(\nu_j\ll |\cdot|\).
Now \(\mu_N=\frac1N\sum_{j=0}^{N-1}\nu_j\), so \(\mu_N\) is a measure and \(\mu_N\ll\) Lebesgue.
b) Since \(F^j\) is increasing, \[ F^{-j}([a,b])=[a^{2^{-j}},b^{2^{-j}}], \] hence \[ \mu_N([a,b])=\frac1N\sum_{j=0}^{N-1}\bigl(b^{2^{-j}}-a^{2^{-j}}\bigr). \] Each term tends to \(0\) as \(j\to\infty\) (because \(0<a<b\le1\)), and the sequence is bounded. Its Cesàro mean therefore tends to \(0\).
c) For bounded measurable \(f\), \[ \int_0^1 f\,d\mu_N =\frac1N\sum_{j=0}^{N-1}\int_0^1 f\,d\nu_j =\frac1N\sum_{j=0}^{N-1}\int_0^1 f(F^j(s))\,ds \] with \[ \int_0^1 f\,d\nu_j=\int_0^1 f(s^{2^j})\,ds. \] For each \(s\in[0,1)\), \(s^{2^j}\to0\), so \(f(s^{2^j})\to f(0)\) and dominated convergence gives \[ \lim_{j\to\infty}\int_0^1 f(s^{2^j})\,ds=f(0). \] Therefore the Cesàro average of the first \(N\) terms converges to \(f(0)\): \[ \lim_{N\to\infty}\int_0^1 f\,d\mu_N=f(0). \] So the limit exists and equals \(f(0)\).