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In Vector Integration and Stochastic Integration in Vector spaces, Nicolae Dinculeanu states the following theorem:

Given a measure space: $(S, \Sigma)$, a function $f : S —> R$ is $\Sigma$-measurable iff there is a sequence ($f_n$) of finite, real-valued, $\Sigma$ - step functions such that $f_n \rightarrow f$ pointwise.

I see that if $f$ is measurable then this implies the existence of measureable step functions that converge. (Using $f_n = 2^{-n}\lfloor 2^n f \rfloor$.) However I can't see why the converse holds.

Any help is much appreciated!

2 Answers 2

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The convers is easy, since if $f_n$ converges to $f$ pointwise, and $f_n$ are measurable, then $f$ is measurable. It's easy to prove. For example you can show that, given $f_n$ measurable functions, $\text{limsup}_n f_n$ and $\text{liminf}_n f_n$ are measurable, using the "level set" criterion

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given a sequence of measurable fucntions (fn) into R, g = sup fn is measurable because the preimage of (r, inf) is the union of the preimages for each fn of (r, inf)

once can make a similar argument for h = inf fn considering preimages of (-inf, r)

then then fn -> f simply implies lim sup fn = f , so that if gN = sup for n >= N of fn, f = inf gN, hence f is measurable