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Simple
Simple
7:30 AM
$\cap\,G_k$ is a borel set since its complement is a union of countable closed sets
 
Simple
Simple
7:55 AM
Suppose $X$ is a set and $E_1,E_2,\dots,$ is a disjoint sequence of subsets of $X$ such that $\bigcup_{k=1}^{\infty}E_k=X$. Let $\mathcal{S}=\{\bigcup_{k\in\,K}E_k\,:\;K\subset\mathbb{Z}^{+}\}$.\\
(a) Show that $\mathcal{S}$ is a $\sigma$-algebra on $X$.\\
(b)Prove that a function from $X$ to $\mathbb{R}$ is $\mathcal{S}$-measurable if and only if the function is constant on $E_k$ for every $k\subset\mathbb{Z}^{+}$.
To show $S$ is a $\sigma$ algebra, we need to show it satisfy the three properties
it has empty set, completment is in $S$ and countability. So, $S$ has the empty set, for $\cup_{k\in\,K}E_k\in\,S$, $X\setminus\cup_{k\in\,K}E_k=\cup\,X\setminus\,E_k$ which is in $S$, and $\cup_{k=1}^{\infty}E_k=X$ is in $S$
 
Simple
Simple
8:18 AM
for part $b$, suppose $f:X\to\mathbb{R}$ is $S$ measurable, then $f^{-1}(B)=\cup\,E_k$ for some Borel set $B$ in $R$ and some $k$. each $E_k$ is $S$ measurable. Since each $E_k$ is disjoint, we can conclude that f(E_k) is constant for each $k$
I am not so sure about my argument
 

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