Relationship between Borel $\sigma$-algebra on $l^2(\mathbb{N})$ and the product $\sigma$-algebra on $\mathbb{R}^\infty$

Question

Recall that $\otimes_{n=1}^\infty \mathcal{B}(\mathbb{R})$ is defined as the smallest $\sigma$-algebra on $\mathbb{R}^\infty$ that makes every coordinate projection measurable. Furthermore $\mathcal{B}(\mathbb{R}^\infty)$ is the Borel $\sigma$-algebra generated by the product topology. We also know that $\otimes_{n=1}^\infty \mathcal{B}(\mathbb{R})=\mathcal{B}(\mathbb{R}^\infty)$.

Furthermore let $l^2(\mathbb{N})$ be the space of square summable sequences indexed by $\mathbb{N}$. This is a Hilbert space with inner product $\langle x,y \rangle = \sum_n x_ny_n$. Let $\mathcal{B}(l^2(\mathbb{N}))$ denote the Borel sigma algebra induced by the natural metric on $l^2(\mathbb{N})$.

We may note that $l^2(\mathbb{N})\subset \mathbb{R}^\infty$, but does it hold that $ l^2(\mathbb{N})\in \mathcal{B}(\mathbb{R}^\infty)$ and $\mathcal{B}(\mathbb{R}^\infty) \cap l^2(\mathbb{N})=\mathcal{B}(l^2(\mathbb{N}))$ ?

The reason I'm interested in this is: I have two probability measures $P,Q$ on $(l^2(\mathbb{N}),\mathcal{B}(l^2(\mathbb{N})))$ and I would like to show that $P=Q$ by (if the above holds) showing that $$ \tilde{P}:=P(\cdot \cap l^2(\mathbb{N})), \quad \text{and} \quad \tilde{Q}:=Q(\cdot \cap l^2(\mathbb{N})), $$ are identical measures on $\mathbb{R}^\infty$ (which can be done by for example looking at the finite dimensional distribution).

Answer 1

To see that $\ell^2$ is Borel in $\mathbb{R}^\infty$, consider the function $\pi_i : \mathbb{R}^\infty \to \mathbb{R}$ defined by $\pi_i(x) = x_i$. This is a continuous function on $\mathbb{R}^\infty$ by very definition of the product topology, so it is also Borel. Thus $f : \mathbb{R}^\infty \to [0,\infty]$ defined by $f(x) = \sum_{i=1}^\infty |f_i(x)|^2$ is also Borel, being a countable sum of Borel functions. But $f$ is just the $\ell^2$ norm squared, so $\ell^2 = f^{-1}([0,\infty)$ is a Borel set in $\mathbb{R}^\infty$.

For your second question, let $\mathcal{G}$ be the $\sigma$-algebra on $l^2$ induced by the Borel $\sigma$-algebra on $\mathbb{R}^\infty$. You can show $\mathcal{G}$ equals the Borel $\sigma$-algebra on $l^2$ by showing that each $\sigma$-algebra contains a generating set for the other. In one direction, note that sets of the form $\pi_i^{-1}((a,b)) \cap l^2$ generate $\mathcal{G}$, and are open in $l^2$ (since $\pi_i$ is a continuous linear functional on $l^2$). In the other direction, the function $f$ constructed above can be used to show that every ball of $l^2$ is Borel in $\mathbb{R}^\infty$ and hence in $\mathcal{G}$.

Answer 2

Can someone please verify that this proof is correct?

It is obvious that the induced metric $d$ on $l^2(\mathbb{N})$ satisfies that for any sequence $(x_n)_{n\in \mathbb{N}}=((x_{1,n},x_{2,n},...))\subset l^2(\mathbb{N})$ and $x=(x_1,x_2,...)\in l^2(\mathbb{N})$ \begin{align*} d(x_n,x) = \Big(\sum_{i=1}^\infty (x_{i,n}-x_i)^2\Big)^{1/2} \to_n 0 \iff x_{i,n} \to_n x_i \quad \forall i\in \mathbb{N}, \end{align*} by Weierstrass M-test. Hence $l^2(\mathbb{N})$ has the topology $\tau$ of coordinate-wise convergence. We may define a metric $\tilde{d}$ on $\mathbb{R}^\infty$ given by $\tilde{d}(x,y)=\sum_{i=1}^\infty \frac{1\land |x_{i,n}-x_i|}{2^i}$ and by example 1.2 Billingsley (1999) this metric also induces the topology of coordinate-wise convergence, which we denote $\tilde{\tau}$.

Since convergence uniquely determines the topology for metrizable topologies, we get that $\tilde{\tau}$ is the product topology on $\mathbb{R}^\infty$ (since the product topology also is characterized by coordinate-wise convergence), such that $\sigma(\tilde{\tau})=\mathcal{B}(\mathbb{R}^\infty)$.

Consider the restriction $\tilde{d}|_{l^2(\mathbb{N})}$ of $\tilde{d}$ to $l^2(\mathbb{N})$ and note that it is a metric equivalent to $d$, since they have the same convergent sequences and limits. Thus $$ \mathcal{B}(l^2(\mathbb{N})) := \mathcal{B}(l^2(\mathbb{N}),d) = \mathcal{B}(l^2(\mathbb{N}), \tilde{d}|_{l^2(\mathbb{N})}) = \sigma(\tilde{\tau}\cap l^2(\mathbb{N}))$$ $$ = \sigma(\tilde{\tau})\cap l^2(\mathbb{N}) = \mathcal{B}(\mathbb{R}^\infty) \cap l^2(\mathbb{N}),$$ where we used that the restricted metric $\tilde{d}|_{l^2(\mathbb{N})}$ induces the subspace topology $\tilde{\tau}\cap l^2(\mathbb{N})$.