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The question is:

Show that a positive function $ f $ on $ E $ is $ \varepsilon $-measurable if and only if it has the form $$ f = \sum_{n=1}^{\infty} a_{n} I_{A_{n}} $$ for some sequence $ \{ a_{n} \}_{n=1}^{\infty} \subseteq \overline{\mathbb{R}}_{+} $ and some sequence $ \{ A_{n} \}_{n=1}^{\infty} \subseteq \varepsilon $.

I know that the absolute-value function from $ \overline{\mathbb{R}} $ to $ \overline{\mathbb{R}}_{+} $ is $ \varepsilon $-measurable when $ \varepsilon $ equals the Borel $ \sigma $-algebra on $ \overline{\mathbb{R}} $. But it doesn’t have the form $$ f = \sum_{n=1}^{\infty} a_{n} I_{A_{n}}, $$ since its image is uncountable. What am I getting wrong?

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Dear Yangzhe, I edited your question to make it clearer. I hope you don’t mind! :) –  Haskell Curry Feb 28 '13 at 5:05
    
@HaskellCurry: Thank you very much for your help. –  Yangzhe Lau Feb 28 '13 at 6:17

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The range of $x\mapsto \sum_n a_n \, I_{A_n}(x)$ can be uncountable. Given $x\in E$, each choice of whether $x$ does or does not belong to $A_1$, whether $x$ does or does not belong to $A_2$, $\dots$, produces a different subset of the $a_n$s to sum. Since there are countably infinitely many choices, each of which can be made in two different ways, this permits the range of $x\mapsto \sum_n a_n\, I_{A_n}(x)$ to have cardinality up to $2^{\aleph_0}$. This is the cardinality of all of $\overline {\Bbb R}_+$, so there is no cardinality restriction on the range which prevents you from getting all measurable functions in the given way.

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