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Firstly,

Definition 1: function f is measurable if we have a sequence of simple function $s_n$ such that $s_n \to f$.

Definition 2: a set $A$ is measurable if characteristic function $\chi_A$ is measurable.

but by these definition, I very confuse when I prove that continuous function then measurable. Can anyone give me a proof?.

PS: only use these definition.

Secondly, I have Another question: show that if $A_n$ are measurable sets then $\bigcap_{n=1}^{\infty}A_n$ also measurable set.

My solution: Beacause $A_n$ are measurable sets, characteristic functions $\chi_{A_n}$ measurable, or exist sequence of simple finction $s_n$ s.t $s_n \to \chi_{A_n}$. So $$s=\prod_{n=1}^{\infty}s_n \to \prod_{n=1}^{\infty}\chi_{A_n}=\chi_{\bigcap_{n=1}^{\infty}A_n}$$ and since $s$ also simple function, we have $\chi_{\bigcap_{n=1}^{\infty}A_n}$ measurable. Therefore $$\bigcap_{n=1}^{\infty}A_n$$ measurable. QED

I still feel something is wrong. Can anyone help me check my solution?. Thankful

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How are simple functions defined? I think simple functions must be defined using the definition of measurable (of sets) –  Amr Nov 8 '12 at 11:45
    
Here, I know this definition very confuse. But it is definition. –  Firmino Nov 8 '12 at 13:32

1 Answer 1

Your approach of intersection using the product of simple functions is almost perfect. You only miss the sequence: $$s'_k:=\prod_{i=1}^k s_i$$ Then $s'_k\to \chi_{\bigcap A_i}$. Note that their limit, $s$ itself may not be already simple.

For a given continuous function $f$, first over a compact interval, and for an $n\in\Bbb N$, cut the range of $f$ (which is now bounded) in $n$ equal parts, and consider the corresponding simple function.

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Here, $\bigcap_{n=1}^{\infty}A_n$. Then $\prod_{n=1}^{\infty}s_n$ also simple function, true or false? –  Firmino Nov 8 '12 at 13:28
    
I don't understand your opinion about continuous function then measurable –  Firmino Nov 8 '12 at 13:31

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