I'm having trouble judging whether this statement is correct:
For an arbitrary bounded measurable function $f$ defined on $[0,1]$, $\exists{}\ $a sequence of step functions $\{\phi_n\}$, such that $\{\phi_n\}$ converges to $f$ pointwisely a.e. on $[0,1]$.
By the Simple Approximation Theorem, this is true if we are allowed to use simple functions. But I am curious whether this still holds when we restrict ourselves to step functions only.
I have a feeling that this may not be true because for a measurable function, its domain may be too "broken up" to be fitted by step functions. But I don't know how to find a counter-example…
So can anybody help me find a counterexample or confirm that this is correct?
Thank you very much!
Edit: By a step function I mean a (finite) linear combination of indicator functions for intervals.
Best Answer
so, I'll try to write down what I remember. Let $n \in \mathbb N$. By Lusins theorem there is a $H_n \subseteq [0,1]$ closed with $\lambda([0,1] - H_n) < \frac 1n$ and $f|_{H_n}$ continuous. By the Tietze extension theorem there is a continuous $f_n\colon [0,1] \to \mathbb R$ with $f_n|_{H_n} = f$. Now for every $k$ \[ \lambda\left(|f_n - f| \ge \frac 1k\right) \to 0, \quad n \to \infty \] We can therefore choose a subsequence $(f_{n_k})$ with $\lambda(|f_{n_k} - f| \ge \frac 1k) < 2^{-k}$ for each $k$. Let $N = \bigcap_k \bigcup_{\ell \ge k} \{|f_{n_\ell} - f| \ge \frac 1\ell\}$. Let $x \in [0,1]\setminus N$, then there is an $k$ such that for $\ell \ge k$ we have $|f_{n_\ell}(x) - f(x)| < \frac 1\ell$ and therefore $f_{n_k}(x) \to f(x)$. But $\lambda(N) \le \sum_{\ell \ge k} 2^{-\ell}$ for each $k$, i. e. $\lambda(N) = 0$, therefore $f_{n_k} \to f$ a. e. on $[0,1]$.
Since each $f_{n_k}$ is continuous we can choose a step function $s_k$ with $\|f_{n_k} - s_k\|_\infty \le \frac 1k$. But then for $x \not\in N$ \[ s_k(x) = s_k(x) - f_{n_k}(x) + f_{n_k}(x) \to 0 + f(x) = f(x) \] i. e. $f$ is the almost everywhere limit of step functions.