TITLE: Borel measurable functions $f:[0,1]\to[0,\infty)$ which cannot be expressed as pointwise limit of nondecreasing sequence of step functions
QUESTION [6 upvotes]: An interval in this problem may be open, closed or half open. We regard a singleton $\{a\}$ as being an interval also.
A step function is a real valued function on $\mathbb{R}$ which is a linear combination of characteristic functions of intervals. How do I go about showing there are Borel measurable functions $f: [0,1] \to [0, \infty)$ which cannot be expressed as the pointwise limit of a nondecreasing sequence of step functions? As with progressm I know I want to show there are Borel measurable functions $f: [0,1] \to [0, \infty)$ such that there does not exist a sequence of step functions $\{s_n\}$ with $s_n(x) \le s_{n+1}(x)$ for each $x$ and $\lim_{n \to\infty} s_n(x) = f(x)$. Any help would be appreciated.

REPLY [3 votes]: Suppose that $s$ is a step function. Then for any $a \in \mathbb R$ the set $\{s > a\} = \{x \in \mathbb R : s(x) > a\}$ is either empty or a finite union of intervals. Since all intervals are $F_\sigma$ sets, so is $\{s > a\}$.
Now suppose that $s_n \nearrow f$. Then for any $a \in \mathbb R$ you have $\{f > a\} = \bigcup_n \{s_n > a\}$ so that $\{f > a\}$ is a countable union of $F_\sigma$ sets which is again an $F_\sigma$ set.
If $f$ is Borel measurable and $\{f > a\}$ is not an $F_\sigma$ set, you will have the example you need. As the other answers point out, the indicator function of the irrationals is such a function.