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Note: A semisimple function takes on countably many values.

The hypotheses are that $f$ is measurable and real valued on a measurable set $E$. I would like to show that there is a sequence of semisimple functions $\{ f_n \}$ on $E$ that converge to $f$ uniformly on $E$.

I know that a real valued measurable function may be approximated pointwise by a sequence of simple functions on a measurable set $E$, with $$ |\phi_n|<|f| $$ My idea is to somehow partition the set $E$ into a countable number of subsets, and then maybe tweak the the approximation theorem above to get uniform convergence? However, I am unsure of how to get uniform convergence out of this.

Any help getting a start would be appreciated. Thank you!

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    What is a semisimple function? – 2017-02-15

2 Answers 2

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For positive integer $n$ define:$$f_n(x)=\frac1{n}\lfloor nf(x)\rfloor$$ Then $f_n$ is a semisimple measurable function with:$$f_n(x)\leq f(x)

This guarantees uniform convergence.

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    This is amazing – 2018-12-24
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Hint 1. Fix $n\in{\mathbb N}$. For $k\in{\mathbb Z}$, let $E_{n,k}=\{\omega: f(\omega) \in [\frac{k}{2^n},\frac{k+1}{2^n})\}$.

Hint 2.

If $\omega \in E_{n,k}$, then $0\le f(\omega) - k2^{-n}\le 2^{-n}$.