Borel Sets of Infinite Binary Sequences

Note: The content of this blog is an expansion of some results in Chapter 2 of Parthasarathy's Book on Probability measures on Metric Spaces.

$M=\{0,1\}^\infty$ is a compact metric space as shown previously. Let $X,Y$ be two metric spaces. Borel sets $B_1 \subset X, B_2 \subset Y$ are said to be isomorphic if there exists a bijective map, $\phi : B_1 \to B_2$ such that both $\phi, \phi^{-1}$ are measurable. This is denoted as $B_1 \sim B_2$.

Here the sigma algebra on $B_1$ is $B_1 \cap \mathcal{B}_X$ and similarly on $B_2$. Measurability of $\phi,\phi^{-1}$ should be thought of in this sense. (Note that $B_1 \cap \mathcal{B}_X = \mathcal{B}_{B_1}$). Throughout, $\mathcal{B}_X$ represents the $\sigma$-algebra generated by the open sets in $X$.

There exists a Borel set $E \subset M$ such that $E \sim [0,1]$

Define $\tau : M \to [0,1]$ by $\tau(\{x_1,x_2,\cdots\}) = \sum_{i=1}^\infty \frac{x_i}{2^i}$. Then $\tau$ is continuous from the uniform limit theorem. $\tau$ is also onto since every number in $[0,1]$ has a binary expansion. Let $E$ be set of all binary sequences which contain infinitely many zeros. Then, $M - E$ is countable. As each singleton in a metric space is a closed set, $M-E$ is a Borel set. Hence, $E$ is also Borel.

The restriction of $\tau$ to $E$ is a bijective map as every number in $[0,1]$ has exactly one binary expansion which contains infinitely many zeros. As $\tau$ is continous, it is measurable map from $M \to [0,1]$ and hence $\tau : E \to [0,1]$ is also measurable.

It needs to be shown that $\tau^{-1}$ is also measurable. $\tau^{-1} : [0,1] \to E$ and it needs to be shown that $\tau(A)$ is Borel in $[0,1]$ for all $A$ Borel in $E$ i.e. $A \in \mathcal{B}_E$. As $\mathcal{B}_E = E \cap \mathcal{B}_M$ and the fact that $\mathcal{B}_M$ is generated by the sets $F_j = \{ \{x_1,x_2\cdots\} | x_j = 1\}$ it is enough to show that $\tau(A)$ is Borel for all $A$ of the form $E\cap F_j$.

To understand what $\tau(E\cap F_j)$ is, it helps to fix the first $j-1$ entries and let the entries from $j+1$ to $\infty$ be arbitrary. Each such set gives rise to finite union of intervals whose end points are dyadic rationals. Hence $\tau(E\cap F_j)$ is Borel.

$M$ is homeomorphic to $M^\infty$

Let $x$ be an infinite binary sequence. From this we construct a sequence of infinite binary sequences as follows. From the given sequence $x$ choose elements alternatingly to obtain the first sequence. After removing this subsequence from the original sequence, the procedure is repeated ad infinitum. This can be written mathematically as $\{x_1,x_2,x_3,\cdots,\} \to \{x^1,x^2,x^3,\cdots\}$ where each $x^j \in M$ and $i$th element of the $j$th sequence $x^j_i = x_{2^{j-1}2(i-1)}$. Clearly, this map is a bijection. \begin{array} dd(\{x^1,x^2,x^3,\cdots\}, \{y^1,y^2,y^3,\cdots\}) &= \sum_{j=1}^\infty \frac{1}{2^j} \sum_{i=1}^\infty \frac{1}{2^i} \frac{d_M(x^j_i,y^j_i)}{1+d_M(x^j_i,y^j_i)} \\ &= \sum_{j=1}^\infty \frac{1}{2^j} \sum_{i=1}^\infty \frac{2^{2^{j-1} + (2i-1)}}{2^i} \frac{1}{2^{2^{j-1} + (2i-1)}} \frac{d_M(x_{2^{j-1} + (2i-1)},y_{2^{j-1} + (2i-1)})}{1+d_M(x_{2^{j-1} + (2i-1)},y_{2^{j-1} + (2i-1)})} \\ &\geq d_M(x,y) \end{array} The other direction is not too hard to show.

There exists a Borel set $E_1 \subset M$ such that $E_1 \sim I^\infty$

Let $\tau,E$ be the same as above. Define $\tau^\prime : M^\infty \to I^\infty$ by $\tau^\prime(x^1,x^2,\cdots) = (\tau(x^1),\tau(x^2),\cdots)$ where $x^i = \{x^i_1,x^i_2,\cdots\} \in M$. We show that $\tau^\prime$ is continuous.

\begin{array} dd_{I^\infty}(\tau^\prime(x),\tau^\prime(y)) &= \sum_{i=1}^\infty \frac{1}{2^i}\frac{|\tau(x^i) - \tau(y^i)|}{1+|\tau(x^i) - \tau(y^i)|} \\ &= \sum_{i=1}^\infty \frac{1}{2^i}\frac{|\sum_{j=1}^\infty \frac{1}{2^j}(x^i_j - y^i_j)|}{1+|\sum_{j=1}^\infty \frac{1}{2^j}(x^i_j - y^i_j)|} \\ &\leq \sum_{i=1}^\infty \frac{1}{2^i} \sum_{j=1}^\infty \frac{1}{2^j} {|x^i_j - y^i_j|} \\ &\leq d_{M^\infty}^0(x,y) \end{array} where $d_{M^\infty}^0$ is equivalent to $d_{M^\infty}(x,y) = \sum_{i=1}^\infty \frac{1}{2^i}\frac{d_M(x^i,y^i)}{1+d_M(x^i,y^i)}$. Hence $\tau^\prime$ is continuous and therefore measurable. Now, it will be shown that $\tau^\prime : E^\infty \to I^\infty$ is an isomorphism.

• $\tau^\prime$ is bijective as $\tau : E \to I$ is bijective.
• $E^\infty$ is a measurable set in $\mathcal{B}_{M^\infty}$
• $(\tau^\prime)^{-1} : I^\infty \to E^\infty$ is measurable. $\mathcal{B}_{E^\infty} = E^\infty \bigcap \mathcal{B}_{M^\infty} = \sigma(\{E^\infty \cap \rho_n^{-1}(U_{ni}), U_n \in \mathcal{D}\})$ where $\mathcal{D}_i$ is a denumerable base for $M$. However, each $U_n$ can in-turn be written as a countable union of finite intersection of sets of the form $V_{m,j}=\{\gamma_m^{-1}(\{j\}), j \in \{0,1\}\}$ (where $\gamma_m : M \to \{0,1\}$ is the co-ordinate projection). Hence, $\mathcal{B}_{E^\infty} = \sigma(\{E^\infty \cap \rho_n^{-1}(V_{mj})\})$. $\tau^\prime(E^\infty \cap \rho_n^{-1}(V_{mj})) = \tau^\prime(E \times E \times \cdots \times E\cap V_{mj} \times E \times E \cdots) =\tau(E) \times \tau(E) \cdots \times \tau(E \cap V_{mj}) \times \tau(E) \times \tau(E) \cdots$. Therefore, $(\tau^\prime)^{-1}$ is also measurable.

As $M$ and $M^\infty$ are homeomorphic, they are also isomorphic (in the sense defined initially) and it follows that $E^\infty$ is isomorphic to some $E_1 \subset M$. Therefore, we get that $E_1 \sim I^\infty$.