Borel measurable functions $f:[0,1]\to[0,\infty)$ which cannot be expressed as pointwise limit of nondecreasing sequence of step functions 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.
 A: Let $f:[0,1]\to[0,\infty)$ be given by
$$f(x)=\begin{cases} 0 &; x\in\mathbb{Q} \\ 1 &; x\notin\mathbb{Q}\end{cases}$$
If $0\leq s\leq f$ is a step function, then $s$ is non-zero for at most finitely many points. Hence, any well-defined limit $\lim_{n\to\infty}s_n$ of step functions $0\leq s_n\leq f$ is non-zero for at most countably many points and hence is not equal to $f$.
A: 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.
A: Let $f: [0,1] \to [0, \infty)$ be the characteristic function of the set of irrationals in $[0, 1]$. Assume there exists a nondecreasing sequence of step functions $\{s_n\}$ converging pointwise to $f$. Without loss of generality we can assume that $n \ge 0$. Since rationals are dense and $\{s_n\}$ is a nondecreasing sequence of step functions, $s_n(x)$ might be positive only for finitely many points $($irrationals$)$. Hence $\lim_{n \to \infty} s_n(x)$ might be positive only for countably many points, which is a contradiction since irrationals are uncountable.
