For a function to be Riemann integrable, the upper and lower Riemann sum need to be equal. However, this no longer applies to Lebesgue integrals. Let $(\Omega,\Sigma,\mu)$ be a measure space and define for $f\in \Sigma$ that $$ \text{SF}(f)=\{s\in \Sigma: s\text{ is a simple function and } 0\leq s(x)\leq f(x), \forall x\in X\}\\ \text{SG}(f)=\{s\in \Sigma: s\text{ is a simple function and } s(x)\geq f(x), \forall x\in X\} $$ Here is my question: why are we defining $$ \int_X f d\mu=\sup_{s\in SF(f)} \int_X sd\mu$$ rather than $$ \int_X f d\mu=\inf_{s\in SG(f)} \int_X sd\mu??? $$ Of course, the later definition force $f$ to be bounded, but why aren't we using the later definition?
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I'm not sure of the answer myself, but perhaps the fact that your second definition gives when you have a measure zero set always $0$? – Keen-ameteur Oct 08 '19 at 08:09
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Note that your $f$ should be positive in order for this to be the definition. Also, don't you yourself point out the problem? Yhe latter definition is more restrictive and it's less clear how to work with unbounded positive functions, while the first one works with unbounded functions just fine. – WoolierThanThou Oct 08 '19 at 08:27
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See here – RRL Oct 08 '19 at 09:37