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In these notes I was reading, it is stated that for a measurable function $f:X\to [0,\infty]$, its Lebesgue Integral is $$ \int_E f \ d\mu =\sup\left\{\int_E s(x) \ d\mu\mid 0\le s\le f, s \ \text{simple}\right\}. $$ In this statement, are we just assuming that $f$ is Lebesgue Integrable? Would you not have to first verify that the upper and lower lebesgue integrals are equivalent? Does measurability imply Lebesgue Integrability? As I read in this question, the function $1/x$ is a counterexample to this statement, so I cannot see what justifies the statement. Is it just a definition?

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    Yes. It's a definition. – Sean Roberson Jan 30 '18 at 02:01
  • @SeanRoberson But, we still need to know that $f$ is Lebesgue integrable, correct? –  Jan 30 '18 at 02:39
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    No. They're defining what it means for $f$ to be integrable. – Sean Roberson Jan 30 '18 at 04:10
  • I Lebesgue integration there is a difference between f being integrable and $\int f d\mu$ being defined. Integral of f is being defined above but there is no assertion about finiteness of the integral. Basically one is afraad of infinity in Calculus but not in measure theory! So you should use the term 'integrable' carefully. – Kavi Rama Murthy Jan 30 '18 at 06:25

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