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On the Real Analysis - Modern Techniques and Their Application (second edition) by Gerald Folland, page 47 i found this theorem: "Let $f$ a measurable function. Then exists a sequence $(\phi_n)_{n \in \mathbb{N}}$ of simple functions such that: (a) $\phi_n \to f$ pointwise; (b) $\phi_n \to f$ uniformly on any set on which $f$ is bounded".

I need a proof step by step of this theorem. Can someone suggest me a more didactic book?

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    _Real Analysis_ by N.L. Carothers provides a proof of (a), but leaves a few steps to the reader. _Measure and Integral_ by Zygmund and Wheeden is more explicit, I think, but also more formal. Neither book provides a proof that $\phi_n \rightarrow f$ uniformly on a set on which $f$ is bounded, but it's pretty clear from the construction of $\phi_n$. Think about this: for any $n$, what is $\sup_x |f(x) - \phi_n(x)|$ if $f$ is bounded?2011-11-03

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This doesn't really answer the question, but it might help you understand Folland's proof:

First note that the sets $E_n^k$ and $F_n$ are measurable since $f$ is measurable. Next, decipher the definition of $\varphi_n$: it turns out that $\varphi_n(x)$ simply equals $2^n$ if $x > 2^n$, and equals $f(x)$ truncated down to the nearest multiple of $1/2^n$ if $x \le 2^n$. The strange way of writing $\varphi_n$ using characteristic functions is just to show that $\varphi_n$ is measurable (using that the sets $E_n^k$ and $F_n$ are measurable).

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    @Katy23: Obviously you need to be a little more subtle than throwing in $2^n$ here. ;-) If $f(x) \le 2^n$, then $x$ belongs to exactly one of the sets $E_n^k$. And for all points in that set, you know the exact value of $\varphi_n$ and tight bounds on the value of $f$. Does that help? (This is really just a technical way of stating the simple fact from my previous comment.)2011-11-03