I could use some help for the problem:
Let $f$ be a non-negative real-valued function defined on $[0,1]$.
True or False? $f$ is measurable if and only if there is a sequence $\{E_n\}$ (finite or infinite) of measurable subsets of $[0,1]$ and a sequence of non-negative constants $\{c_n\}$ such that $f(x) = \sum_n c_n \chi_{E_n}(x)$ for every $x\in[0,1]$, where $\chi_{E_n}$ is the the characteristic function of $E_n$ for each $n$.
I think the statement is true. For the direction $(\Longleftarrow$), since $f(x) = \sum_n c_n \chi_{E_n}(x)$ for every $x\in[0,1]$, we see that $\{\sum_n^k c_n \chi_{E_n}\}_k$ is a sequence of simple functions on $[0,1]$ converging pointwise to $f$ and $|\sum_n^k c_n \chi_{E_n}|\le |f|$ on $[0,1]$ for all $k$. By simple approximation theorem, $f$ is measurable.
But I got stuck on the forward ($\Longrightarrow$) direction. Also, if the above proof is not right, please point it out. Thank you!
The forward part is also true. Let $f_n$ be simple functions increasing to $f$. Then $f=f_1+(f_2-f_1)+(f_3-f_2)+\cdots$. Writing $f_{n+1}-f_n=\sum_{j=1}^{k_n} a_{j,n} I_{E_{jn}}$ (with $a_{j,n}$'s non-negative) we see that $f$ is a countable sum of terms which are constant times characteristic functions.