identifying the measure $\lambda f^{-1}$ on the interval $[0,1]$

Question

Let $X_i=\{0,1\}$ be the space equipped with the measure $\mu$ s.t. $\mu(\{0\})=\mu(\{1\})=\frac{1}{2}$. Now define $\Omega$ to be the product space of $X_i$'s with the product $\sigma$-field and the product measure $\lambda$. Consider the map $$f:\Omega\to[0,1]$$ $$\omega=(x_1,\ldots,x_n,...)\mapsto\sum_{j=1}^{\infty}\frac{x_j}{2^j}\in[0,1]$$

My aim is to identify the measure $\lambda f^{-1}$ on the interval $[0,1]$.

First, I take an example. I take $E=(\frac{3}{4},\frac{7}{8})$, which is a dyadic interval. With the binary expansion defined, we see that $f^{-1}(E)=\{1\}\times\{1\}\times\{0\}\times\ldots$, a cylinder with volume $\frac{1}{8}$. Hence, $(\lambda f^{-1})(E)=\lambda(f^{-1}(E))=\frac{1}{8}$.

We can say $\lambda f^{-1}(E)=m(E)$, where $m$ is the Lebesgue/Borel measure, for every dyadic interval. We can conclude that $\lambda f^{-1}$ is just the standard Borel measure on $[0,1]$.

Details added: Let $E=\left(\frac{k}{2^j},\frac{k+1}{2^j}\right)$ with $n\in\mathbb{N}$ and $0\leq k<2^j$. Let $x=x_1\ldots x_j$ be the binary expansion, with two exceptions $x=\frac{k}{j}$ and $x=\frac{k+1}{j}$. Hence $f^{-1}(E)=F\setminus\{p,q\}$, where $F$ consists of all sequences that start with $x$ and $p=(x,0,0,\ldots)$ and $q=(x,1,1,\ldots)$. It is clear that $\lambda(F)=2^{-j}$ by definition of the product measure, and $\lambda(\{p\})=\lambda(\{q\})=0$. Hence $\lambda\left(f^{-1}(E)\right)=2^{-j}$, which is the Borel measure of $E$. Since the dyadic intervals generate $\mathcal{B}$, $\lambda\left(f^{-1}(E)\right)=m(E)$ for any measurable $E$, and $m$ is the Borel measure on $[0,1]$. Does this complete the proof for dyadic intervals?

I think my statement is correct, but I need a proof to generalize it, instead of just taking dyadic intervals. Here is a post regarding a similar problem as mine:identify the interval $[0, 1]$ with the Lebesgue measure to the probability space for tossing a fair coin. The result is that $f(\omega)$ is almost bijective, meaning that $f(\omega)$ is a bijection except at countably many points $x\in[0,1]$ that have two inverse images; $f(\omega)$ is measure-preserving. Are these two results from this post helpful for writing a rigorous proof regarding my statement? And how can I do that? Thank you.

Answer 1

If you can prove the statement for all open dyadic intervals it would be already very useful. (I assume that by dyadic interval you mean an interval whose endpoints are of the type $k2^{-n}$ for suitable integers $k,n$.)

If you have $\lambda f^{-1}(E)=m(E)$ for open dyadic intervals $E$, then one can show that $\lambda f^{-1}(E)=m(E)$ also holds for all (non-dyadic or dyadic) open intervals $E\subset [0,1]$. This can be done by approximating the open intervals by dyadic intervals from inside: If you have real numbers $a,b\in [0,1]$ with $a<b$, then there exist sequences $k_n,l_n\in\Bbb N$ such that $x_n:= k_n2^{-n}$ converges from above to $x$ and $y_n:=l_n2^{-n}$ converges from below to $b$. For large $n$, the sequences $k_n,l_n$ can be chosen such that $a\leq x_n\leq a+2^{-n} < b-2^{-n} \leq y_n \leq b$ is satisfied. Since the interval $(x_n,y_n)$ is a dyadic interval, we have $\lambda f^{-1}((x_n,y_n))=\mu((x_n,y_n))=y_n-x_n$. Using the properties of a measure (like continuity from below) it follows that $\lambda f^{-1}((a,b))=\mu((a,b))=y_n-x_n$ holds for all real numbers $a,b\in [0,1]$.

If two measures are equal on all open intervals, then it is known that these measures agree on all Borel measurable sets, see for example this question and its comments and answers (the fact that you use $[0,1]$ while the question uses $\mathbb R$ does not make a significant difference, the arguments work the same in both cases).

Thus we can conclude that $\lambda f^{-1}$ is just the standard Borel measure on $[0,1]$.

Answer 2

Here is another approach:

$\lambda f^{-1}$ is the standard Lebesgue measure $m$ on $[0,1]$. We just need to show that $\lambda f^{-1}\left((a,b]\right)=m\left((a,b]\right)$, for any $0\leq a<b\leq 1$, since $(a,b]$ generates $\mathcal{B}([0,1])$, so we can extend it to any measurable sets $E\in\mathcal{B}([0,1])$.

First, we should make sure that $\lambda f^{-1}$ is a measure. We need to show that $f(\omega)$ is measurable. The idea is to define the finite binary expansion $f_n(\omega)=\sum_{j=1}^{n}\frac{x_j}{2^j}$. Note that $f_n(\omega)\to f(\omega)$ pointwise, since the $(n+1)$- tail of the expansion is smaller than $\varepsilon$. Since $f_n:\prod_{j=1}^{n}X_j\to[0,1]$, and the domain of $\{f_n\}$ is the n-dimensional cylinder, composed of $n$ singletons, which is measurable. Hence $f_n(\omega)$ is measurable. Passing to the pointwise limit, $f(\omega)$ is measurable. Together with the fact that $\lambda$ is the product measure on the product space $\Omega$, we get the conclusion.

To show the equivalence of two measures on any measurable sets in $\mathcal{B}([0,1])$, the idea is to use countable additivity of measure $\lambda$ to complete the proof. Let $x=\sum_{j=1}^{\infty}\frac{x_j}{2^j}$, where $x_j\in\{0,1\}$. Construct a volume set s.t. the first $k_1-1$ terms of $x_j$'s are $0$, and $x_{k_1}$ is the first term to be $1$. Then starting from $j\ge k_1+1$, leaving the entries free, i.e. $$E_1=(0,\ldots,0)\times\{1\}\times\{x_{k_1+1}\}\times\ldots$$ We can see that $E_1\in f^{-1}\left((0,x]\right)$. Then, we proceed to construct a second volume set s.t.it takes value $1$ at the $j=k_2$, while $x_j=0$ for all $k_1+1\leq j\leq k_2-1$. Then starting from $j\ge k_2+1$, leaving the entries free, i.e. $$E_2=(0,\ldots,1,\ldots,0)\times\{1\}\times\{x_{k_2+1}\}\times\ldots$$ Also, $E_2\in f^{-1}\left((0,x]\right)$. By the process, we can construct a sequence of countably many volume sets $\{E_n\}$'s, since under binary expansion of $x$, every entry of 1 at the $k_n$-th position corresponds to one $E_n$. Hence $\bigcup_{n=1}^{\infty}E_n\subset f^{-1}\left((0,x]\right)$. To prove $\supset$, note that for $y<x$ with the expansion $y=\sum_{j=1}^{\infty}\frac{y_j}{2^j}$, there must be a $j_N$ s.t. $y_{j_N}=0$, and $x_{j_N}=1$. Otherwise, $y>x$. So $y$ must lie in one of $E_n$'s. Hence $\bigcup_{n=1}^{\infty}E_n= f^{-1}\left((0,x]\right)$. By countable additivity, $$\lambda f^{-1}\left((0,x]\right)=\lambda\left(\bigcup_{n=1}^{\infty}E_n\right)=\sum_{n=1}^{\infty}\lambda(E_n)=\sum_{n=1}^{\infty}\frac{1}{2^{j_n}}=x=m\left((0,x]\right)$$ where $m$ is the Lebesgue measure for $x\in(0,1]$. Since $(a,b]=(0,b]\setminus(0,a]$ for $0\leq a<b\leq 1$, the result also holds for $(a,b]$.