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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.

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What is your definition of a step function? And how does it differ from a simple function? If your definition of step functions coincide with that of Wikipedia's, then the theorem is correct. – T. Eskin Apr 18 '12 at 12:41
@ThomasE. Yes my definition is the same as that of wikipedia, i.e. a finite linear combination of indicator functions for intervals. Can you elaborate a bit on why that is correct? – Vokram Apr 18 '12 at 12:45
The proof is quite involved in the general case. In W. Rudin, Real and Complex analysis, 1970, theorem 1.17, you can find a construction of such a sequence of functions. In general $X$ can be any measure space and $f:X\to [0,\infty]$ a measurable function, and the sequence $\{\phi_{n}\}$ is chosen so that $\phi_{1}\leq ...\leq \phi_{n}\leq \phi_{n+1}\leq ... \leq f$, and the convergence is point-wise (not a.e.). A measurable function $f:X\to [-\infty,\infty]$ you can compose as $f=f^{+}-f^{-}$ where both of the functions $f^{+},f^{-}$ are non-negative, and apply the theorem in Rudin's book. – T. Eskin Apr 18 '12 at 13:00
@ThomasE. I think Vokram knows how to approximate $f$ by simple functions (i. e. finite linear combinations of indicator functions of sets with finite measure), as one has in the general case. The point here is IMO to use intervals as "steps". – martini Apr 18 '12 at 13:03
@martini. True, I see your point now. Rudin's proof uses a construction where, for a fixed $\phi_{n}$, the "steps" are pre-images of intervals in $\mathbb{R}$. This doesn't seem to work well if the function is very messy. – T. Eskin Apr 18 '12 at 13:18
up vote 4 down vote accepted

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.

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Thank you martini! – Vokram Apr 18 '12 at 16:42
Did the proposition also hold in the case that $f$ is measurable on $R^1$ rather than the bounded set $[0,1]$? see… – van abel Apr 15 '13 at 14:27

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