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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.2012-04-18
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    @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?2012-04-18
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    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.2012-04-18
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    @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".2012-04-18
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    @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.2012-04-18
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    @Vokram Using [Lusin's theorem](http://en.wikipedia.org/wiki/Lusin%27s_theorem) one can show that every measurable $f$ is the almost everywhere limit of continuous functions. As Every continuous function on $[0,1]$ is the uniform limit of a sequence of step functions, your result follows.2012-04-18
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    @martini: I guess that using intervals it will not work. If you can approximate $1_E$ by step functions for any $E$ in the $\sigma$-algebra, then $E$ would always be written as the limit of intervals, that is, there exists a sequence of intervals $I_n$ such that $E = \limsup I_n (= \liminf I_n)$. But that would imply that every measurable set is of the form $\bigcap_{n=1}^\infty \bigcup_{j=n}^\infty I_n$ for some sequence of intervals $I_n$.2012-04-18
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    @AndréCaldas One point here is "a. e." not "pointwise", see above.2012-04-18
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    @martini I also tried using Lusin's theorem before, but the approxiamtion fails on a very small set of _positive_ measure instead of almost every where. And that turns out to be very problematic.2012-04-18
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    @martini: Oops... :-)2012-04-18
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    It seems to me that they are clearly dense in $\mathbb L^2$, since e.g. their closure must contain any uniformly continuous fctn, and if you have an $\mathbb L^2$ convergent sequence, then a subsequence converges pt-wise.2012-04-18

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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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    Did the proposition also hold in the case that $f$ is measurable on $R^1$ rather than the bounded set $[0,1]$? see also:http://math.stackexchange.com/questions/361233/how-to-show-that-a-measurable-function-on-rd-can-be-approximated-by-step-func2013-04-15