Let $(X,\Sigma)$ be a measure space and $f:X\rightarrow \mathbb{R}$ be measurable and bounded. Then there exists a sequence of simple measurable functions $f_n$ such that $f_n\rightarrow f$ uniformly, ie: $$\forall\epsilon>0 \mbox{ } \exists N \mbox{ } s.t. \forall x \in X \mbox{ } \forall n\geq N, |f_n(x)-f(x)|<\epsilon $$
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Sorry, for some reason I completely missed the word "uniformly" in your question. I have to think about this a little more. I'm not sure my answer was completely right. – layman Mar 21 '15 at 01:39
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I'm on mobile right now, so I can't type out a full answer. I'll give a hint, though. Chop up the range of f into little bits, say, disjoint intervals $[a_i,b_i)$ of length $2^{-n}$. Then $f_n = \sum_i a_i \chi(f^{-1}([a_i, b_i)))$ should do what you want. – Josh Keneda Mar 21 '15 at 01:48
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Hi could you explain a little more? Is it using the monotone convergence theorem? – Daniel Mar 22 '15 at 15:50