How to check whether a set belongs to a $\sigma$-algebra? Not understanding "a Borel function"

Question

Hello fine ladies and gentlemen,

Now, I have never been very good with measure theory and I am struggling to be able to do the exercise.

My definition of a sigma-algebra is the following. (Definition) Let $X$ be a set. A collection of $\Sigma$ subset of $X$ is called a sigma-algebra if

(i) $\phi \in \Sigma$

(ii) $E \in \Sigma$ implies $X \setminus E \in \Sigma$

(iii) $E_n \in \Sigma, n\geq 1$ implies $U_{n=1}^{\infty}E_n \in \Sigma$.

I am not sure about what comes beneath the line "As a $\sigma$-algebra $\mathcal{T}( \mathbb{R}_{\geq 0}, \mathbb{R}^d)$ is also generated by the following families...

And I am struggling with the exercise! Also, I would like a more intuitive understanding to explain the natural events part.

Answer 1

By $X=\mathcal A([0,1],\Bbb R)$ and $\sigma=\mathcal T([0,1],\Bbb R)$ I shall understand the natural counterparts of notions $\mathcal A(\Bbb R_{\ge},\Bbb R)$ and $\mathcal T(\Bbb R_{\ge},\Bbb R)$ defined above. That is $X$ is a set of all functions from $[0,1]$ to $\Bbb R$ and $\Sigma$ is a $\sigma$-algebra on $X$ generated by respective cylindrical sets.

A key point to the solution is the following easy

Claim. Each set $E\in\Sigma$ depends on countable many coordinates. Namely, there exists a countable subset $T$ of $[0,1]$ such that for each $f\in X$ holds $f\in E$ iff there exists a function $g\in E$ such that $g|T=f|T$.

Proof. To show Implication $\Rightarrow$ it suffices to take $g=f$. Denote by $\Sigma’$ the family of all subsets of $X$ satisfying Implication $\Leftarrow$. It is easy to check that $\Sigma’$ is closed with respect to complements and countable unions, so conditions (ii) and (iii) from the definition of a $\sigma$-algebra imply that $\Sigma’$ is a $\sigma$-algebra. It is easy to see that $\Sigma’$ contains all cylindrical subsets of $X$, so it is a $\sigma$-algebra containing $\Sigma$. $\square$

Let $E$ be the set form the first part of Exercise. Suppose to the contrary that $E\in\Sigma$. Pick a countable set $T\subset [0,1]$ satisfying Claim for the set $E$. Pick an arbitrary $t’\in [0,1]\setminus T$ and define a function $f\in X$ by putting $f(t’)=1$ and $f(t)=0$ for all other $t\in [0,1]$. Since $\sup_{t\in [0,1]} f(t)=f(t’)=1$, $f\not\in E$. On the other hand, let $g\equiv 0$ be the zero function on the set $[0,1]$. Since $g\in E$ and $f(t)=g(t)$ for each $t\in T$, $f\in E$, a contradiction.

Let $E$ be the set form the second part of Exercise. Suppose to the contrary that $E\in\Sigma$. Pick a countable set $T\subset [0,1]$ satisfying Claim for the set $E$. Pick an arbitrary $t’\in [0,1]\setminus T$ and define a function $f\in X$ by putting $f(t’)=0$ and $f(t)=1$ for all other $t\in [0,1]$. Since $f(t’)=0$, $f\in E$. On the other hand, let $g\equiv 1$ be the unit function on the set $[0,1]$. Since $g\not\in E$ and $f(t)=g(t)$ for each $t\in T$, $f\not\in E$, a contradiction.

Answer 2

For the first part of your question, the Borel sets are just the $\sigma$-algebra generated by open sets (or some define as the ones generated by compact sets. These are the same in $\mathbb{R}^n$). If you don't understand what is meant by the $\sigma$-algebra generated by some set, you should prove as an exercise that the intersection of $\sigma$-algebras is again a $\sigma$-algebra and that the power set of a set is a $\sigma$-algebra (the $\sigma$-algebra generated by a set is then just the intersection of all $\sigma$-algebras containing the set). Knowing this, I think you should be able to understand the first part. Note that it's quite difficult to determine whether or not a general set is actually in the Borel sets, so don't try to get some sort of nice form for them. Hopefully this is also enough to get you started on the exercise?