# Another question about the generated $\sigma$-field

Suppose I have a probability space $(\Omega,\mathcal{F},P)$ and the set $$\mathcal{C}:=\{F=\sum_{i=0}^nf_i\mathbf1_{(t_i,t_{i+1}]}|n\in\mathbb{N},f_i:\Omega\to\mathbb{R} \mbox{ measurable and bounded}\}$$ In fact the sets of all $F=f_0\mathbf1_{0}+f_1\mathbf1_{(0,t_1]}+\dots+f_n\mathbf1_{(t_{n-1},t_n]}$ for any $t\in (0,\infty)$ and any finite subdivison of $[0,t]$, i.e $\{0=t_0<t_1<\dots<t_n=t\}$.

What is the generated $\sigma$-field of this set? It should be the product $\sigma$-field on $\Omega\times (0,\infty)$. How can I show this?

hulik

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 What is the underlying set? $\Omega\times\mathbb{R}$? – Michael Greinecker Jul 13 '12 at 13:57 what do you mean by underlying set? – hulik Jul 13 '12 at 14:00 You want to have a $\sigma$-algebra on what set? In particular, what is the domain of the indicator functions used in the definition? – Michael Greinecker Jul 13 '12 at 14:02 As I wrote, $t_i\in[0,\infty)$, so $\mathbf1_{(t_i,t_{i+1}]}(\omega)$, where $\omega\in\Omega$ – hulik Jul 13 '12 at 14:06 There is a verb missing in the sentence. I can certainly have an indicator function $1_{(1,2]}$ defined on all of $\mathbb{R}$, so that doesn't really help. – Michael Greinecker Jul 13 '12 at 14:08
If $X$ is a set and $\big(f_i\big)_{i\in I}$ is a family of functions from $X$ to $\mathbb{R}$, we usually take the $\sigma$-algebra generated by $\big(f_i\big)_{i\in I}$ to be the coarsest $\sigma$-algebra on $X$ that makes all $f_i$ measurable. It is generated by the family of sets $$\Big\{f_i^{-1}(B):i\in I\text{ and }B\text{ is a Borel subset of }\mathbb{R}\Big\}.$$ As specified in the comments, the underlying set $X$ is suposed to be $\Omega\times[0,\infty)$ in this problem. We will show that the $\sigma$-algebra generated by the family of functions $\mathcal{C}$ is $\mathcal{F}\otimes\mathcal{B}_+$. That is, the smallest $\sigma$-algebra that contains all sets of the form $F\times B$ with $F\in\mathcal{F}$ and $B$ being a Borel subset of $[0,\infty)$.
We first show that every set of the form $F\times B$ is in the generated $\sigma$-algebra, which we denote by $\mathcal{P}$. We first show that this is the case when $B$ is a half-open interval of the form $(a,b]$. Let $n=1$, let $f_0$ bet the function that is constant $0$ and let $f_1$ be the indicator function of $F$. Let $0=t_0$, $t_1=a$, $t_2=b$. Then $$\sum_{i=0}^1 f_i\mathbf1_{(t_i,t_{i+1}]}$$ is the indicator function of $F\times (a,b]$. So every set of the form $F\times (a,b]$ is in the generated $\sigma$-algebra. Now the half-open intervals generate the Borel $\sigma$-algebra on $[0,\infty)$, which shows that $F\times B\in\mathcal{P}$ in general and therefore $\mathcal{F}\otimes\mathcal{B}_+\subseteq\mathcal{P}$.
To prove now that $\mathcal{P}\subseteq\mathcal{F}\otimes\mathcal{B}_+$, it is enough to show that every function in $\mathcal{C}$ is already $\mathcal{F}\otimes\mathcal{B}_+$-measurable. So let $f:\Omega\times[0,\infty)$ be of the form $\sum_{i=0}^n f_i\mathbf1_{(t_i,t_{i+1}]}$. Since sets of the form $(-\infty,r)$ generate the Borel $\sigma$-algebra on $\mathbb{R}$, it is enough to show that $f^{-1}\Big((-\infty,r)\Big)\in\mathcal{F}\otimes\mathcal{B}_+$ for every real number $r$. Now for $r<0$, $$f^{-1}\Big((-\infty,r)\Big)=\bigcup_{i=0}^n f_i^{-1}\big((-\infty,r)\big)\times (k_i,k_{i+1}],$$ which lies in $\mathcal{F}\otimes\mathcal{B}_+$. For $r\geq 0$ we have $$f^{-1}\Big((-\infty,r)\Big)=\Omega\times [0,\infty)\Big\backslash\bigg(\bigcup_{i=0}^n f_i^{-1}\big([r,\infty)\big)\times (k_i,k_{i+1}]\bigg),$$ which lies again in $\mathcal{P}=\mathcal{F}\otimes\mathcal{B}_+$. Hence $\mathcal{P}=\mathcal{F}\otimes\mathcal{B}_+$.