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
Best Answer
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.