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There is a standard means of approximating a bounded nonnegative function from below in a measure theoretic setting, which is

$$f_n=2^{-n}\lfloor{2^nf}\rfloor\wedge n=2^{-n}\sum_{j=0}^{n2^n}j\mathbf{1}_{A_{n_j}}$$

where $A_{n_j}=f^{-1}[\frac{j}{2^n},\frac{j+1}{2^n})$ for $j\neq n2^n$ and $A_{n_j}=f^{-1}[n,\infty)$ for $j=n2^n$.

I see intuitively (by drawing pictures) why this is a uniform approximation, and where the second equality comes from. However I can't see how to prove these rigorously in a clean way. Does anyone have a clean and insightful proof of (a) the second equality and (b) that the $f_n$ uniformly approximate $f$?

Many thanks!

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1 Answer 1

up vote 1 down vote accepted

Since $f$ is bounded, for every $n$ large enough $f\leqslant n$ everywhere hence the part $\wedge n$ of the formula defining $f_n$ does not modify the function.

Then $f_n(x)=j/2^n$ if and only if $j/2^n\leqslant f(x)\lt (j+1)/2^n$ hence $f\leqslant f_n\lt f+1/2^n$ uniformly.

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Thanks for the pointers - I've edited the question so that the expression is now right. So you are saying that the $\wedge n$ part of the formula is unnecessary then? Presumably it is necessary if $f$ is unbounded, in which case this still works, right? –  Edward Hughes May 24 '12 at 13:21
    
Unnecessary: for every $n$ large enough, yes. // Unbounded case: yes. –  Did May 24 '12 at 13:26

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