Need to show that sets belongs to Borel sigma algebra

Question

I have two sets $$A_1=\{x=(x_1,x_2,...): \sup x_n>1\}$$ and $$A_2=\{x=(x_1,x_2,...): \liminf_{n \to \infty} x_n>1\}$$

I need to show that theese sets belongs to Borel sigma algebra in space $\mathbb{R}^\infty$.

Honestly, I have no Idea how to start and what to do.

Answer 1

Although I like the Answer from Kavi Rama Murthy, here is a direct (naive) way:

(1).For $n\in \Bbb N$ let $A_{1,n}=\{(x_j)_{j\in \Bbb N}: x_n>1\}.$ Each $A_{1,n}$ is open in the Tychonoff product topology. So $A_1=\cup_{n\in \Bbb N}A_{1,n}$ is Borel.

(2). For $(k,n)\in \Bbb N^2$ let $A_{2,k,n}=\{(x_j)_{j\in \Bbb N}: \forall j\geq n\,(x_j> 1+\frac {1}{k})\}.$ Then $A_2 =\cup_{(k,n)\in \Bbb N^2}A_{2,k,n}$ so it suffices to show that each $A_{2,k,n}$ is Borel.

For $k,n \in \Bbb N$ and $m\in \{0\}\cup \Bbb N$ let $A_{2,j,n,m}=$ $=\{(x_j)_{j\in \Bbb N}: (n\leq j\leq n+m\implies(x_j>1+\frac {1}{k})\}.$ Each $A_{2,k,n,m}$ is open so $A_{2,k,n}=\cap_{m=0}^{\infty}A_{2,k,n,m}$ is Borel.

Remark. In (2) we use the fact that $(x_j)_{j\in \Bbb N}\in A_2$ iff there exists $k\in \Bbb N$ such that $\{j:x_j\leq 1+\frac {1}{k}\}$ is finite.

Answer 2

The coordinate maps $p_n:\mathbb R^{\infty} \to \mathbb R$ defined by $p_n(x)=x_n$ are continuous, hence Borel measurable. Supremum and $\lim \inf$ of a seqeuence of measurable functions is measurable. Hence $\{x: \sup p_n ^{-1} (1,\infty)\}$ is measurable. Similarly $\{x: \lim \sup p_n ^{-1} (1,\infty)\}$ is measurable.