If $\psi$ is a nonnegative simple function check that $$\int \psi = \sup\left\{\int\phi:0\le\phi\le\psi, \phi\, \text{simple and integrable}\right\}$$ I'm not sure how to "check" this since this is just the definition of the Lebesgue integral for $f: \Bbb R \to [0, \infty]$?
If $\psi$ is a nonnegative simple function check that $\int \psi = \sup\{\int\phi:0\le\phi\le\psi, \phi\, \text{simple and integrable}\}$
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real-analysis
lebesgue-integral
lebesgue-measure
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0What is the precise definition you are using for $\int \psi?$ – 2017-02-19
1 Answers
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The book gave two definitions of the integral of a non-negative simple function. (I know this because you gave one directly in the question, and that definition references the other one, so the other one had to have been given too.) The first one given was $$\int \psi = \sum_n a_n\mu(\psi^{-1}(a_n))$$ where $\{a_n\}$ is the set of values that $\psi$ takes on. (You will have to do some interpretation to match this with what your book says, because I don't know exactly how it was expressed there. But however it was done, it is equivalent to the above.) Let's call that definition $\int^s \psi$.
The second is of course the Lesbegue integral $$\int \psi = \sup\{\int^s \phi\mid \phi \text{ is non-negative simple and } \phi \le \psi\}$$ Let's call that $\int^l \psi$.
What the question wants you to do is verify that these two definitions give the same value: $$\int^s \psi = \int^l \psi$$
And as a hint, note that $\psi$ itself is non-negative simple and $\psi \le \psi$.