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Let f be a bounded measurable function on E. Show that there are sequences of simple functions E, {$\phi_n$} and {$\psi_n$}, such that {$\phi_n$} is increasing and {$\psi_n$} is decreasing and each of these sequences converge to f uniformly on E.

I know by simple approximation lemma, for all $\epsilon$ > 0, there exists simple functions $\phi_n$, $\psi_n$, defined on E such that $\phi_n$< f < $\psi_n$ and 0 < $\psi_n$ - $\phi_n$< $\epsilon$ on E. How do I show that these functions are uniformly convergent?

Vinny Chase
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1 Answers1

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The usual approximation (from below) has $$ \phi_n := \sum_{k=0}^{n2^n-1}k2^{-n}1_{E_{n,k}} + n1_{F_n}, $$ where $$ E_{n,k} := f^{-1}([k2^{-n},(k+1)2^{-n})) $$ is defined for $n=0,1,\ldots$ and $k=0,1,\ldots,2^n-1$ and $$ F_n:=f^{-1}([n,\infty)) $$ is defined for $n=0,1,\ldots.$

Note that if $f$ is bounded, then $F_n$ is empty for $n$ large enough. Since $f-\phi_n<2^{-n}$ on $F_n^c$ (which is the whole space for large enough $n$), we conclude that the convergence is uniform. This holds similarly for the approximation from above.

John Griffin
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