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Simple
Simple
5:30 AM
@MartinSleziak can you give me some direction for part $b$ the forward direction, I have stuck on it for hours
 
Martin Sleziak
Martin Sleziak
6:20 AM
For b, trying to prove it by contradiction seems like a reasonable approach.
We assume that $f$ is not constant on $E_k$.
That means, there are $x_{1,2}\in E_k$ such that $f(x_1)\ne f(x_2)$.
Let us denote $f(x_1)=y_1$ and $f(x_2)=y_2$.
There are disjoint open sets $U_{1,2}$ such that $y_1\in U_1$, $y_2\in U_2$.
If you look at $f^{-1}(U_1)$, you should be able to show that this set is not measurable. (And thus $f$ is not measurable.)
BTW for complements, you could have written is explicitly using $\mathbb Z^+\setminus K$.
Also the argument for countable union is problematic - you do not know that $\bigcup_{k=1}^\infty E_k=X$ for every sequence of sets $E_k\in\mathcal S$.
So probably you could have a look also at the part a - to see whether the arguments you suggested can be cleaned up.
I see that this is discussed also in the main chatroom, starting here: chat.stackexchange.com/transcript/message/53330861#53330861
in Mathematics, 4 hours ago, by Simple
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 $\mathcal{S}=\{\bigcup_{k\in\,K}E_k\,:\;K\subset\mathbb{Z}^{+}\}$. 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 $k\in\mathbb{Z}^{+}$
 
Simple
Simple
6:41 AM
@MartinSleziak thank you so much
 

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