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Wheeden and Zygmund's book "Measure and Integral" gives an interesting characterization of the Lebesgue Integral that is reminiscent of the Riemann Integral. If $E$ be a Lebesgue measurable set, then then the Lebesgue integral of $f$ on $E$ is equal to $\sup\Sigma_{k=1}^n (\inf_{E_K}f)\lambda(E_k)$, where the supremum is taken over all finite partitions $E_1,...,E_n$ of $E$ into Lebesgue measurable sets. (This remains true for arbitrary measure spaces.)

But Wheeden and Zygmund say that if you switch the supremum and infimum, this need not hold true. So my question is, what is an example of a Lebesgue integrable function $f$ such that the Lebesgue integral of $f$ on $E$ is not equal to $\inf\Sigma_{k=1}^n (\sup_{E_K}f)\lambda(E_k)$?

Also, is there a subset of the set of Lebesgue integrable functions for which the switched version does hold true?

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Let $E = (0,1)$ and $f(x) = \frac{1}{\sqrt{x}}$. Then $\int f < \infty$ but if $E = \sqcup_{k=1}^n E_k$ is a finite partition, then for some $k$, we have $\sup_{E_k} f = \infty$ and $\lambda(E_k) > 0$.

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  • $\begingroup$ So is there a subset of the set of Lebesgue integrable functions for which it does hold true? $\endgroup$ – Keshav Srinivasan Nov 21 '18 at 3:10
  • $\begingroup$ @KeshavSrinivasan I don't understand that question. Just take the subset of the set of Lebesgue integrable functions for which it does hold true. $\endgroup$ – mathworker21 Nov 21 '18 at 7:23
  • $\begingroup$ Yeah, but my question is what is that subset? Is there some property of functions which will guarantee that it holds? $\endgroup$ – Keshav Srinivasan Nov 21 '18 at 23:01
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Let $E=[0,1]$ and let $f$ be the characteristic function of $E \cap \mathbb Q$.

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  • $\begingroup$ What partition will give you a nonzero value? $\endgroup$ – Keshav Srinivasan Nov 20 '18 at 12:44
  • $\begingroup$ For an arbitrary partition $E_1,....,E_n$ we have $ \sup_{E_k}f=1$, thus $\Sigma_{k=1}^n (\sup_{E_K}f)\lambda(E_k)= \Sigma_{k=1}^n \lambda(E_k) = \lambda(E)=1$, therefore $\inf\Sigma_{k=1}^n (\sup_{E_K}f)\lambda(E_k)=1 \ne 0 = \int_E f(x) dx.$ $\endgroup$ – Fred Nov 20 '18 at 12:53
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    $\begingroup$ @Fred that is not true. we are partitioning into arbitrarily Lebesgue measurable sets. What if $E_1 = E\cap \mathbb{Q}$ and $E_2 = E\setminus E_1$? Then $\sup_{E_2} f = 0$ and $\lambda(E_1) = 0$, so $\sum (\sup_{E_k} f) \lambda (E_k) = 0$ $\endgroup$ – mathworker21 Nov 20 '18 at 15:52

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