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The book "A Primer of Lebesgue Integration" by H.S. Bear defines Lebesgue integration through lower and upper sums $L(f,P) = \sum m_i\mu(E_i)$ and $U(f,P)=\sum M_i\mu(E_i)$ where infinite countable partitions are allowed.

The typical definition of Lebesgue integration that one encounters involves the supremum of simple functions. These simple functions are like lower sums. However, they are finite linear combinations and not infinite linear combinations.

When approaching Lebesgue integration through upper and lower sums why is it necessary to consider infinite countable partitions when simply functions seem to deal with only finite partitions?

Martin Sleziak
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user782220
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  • Could you add a short outline of how the construction proceeds in this book? If the integral is defined by first splitting $f$ into its positive and negative parts, then both the upper and lower sums are converge monotonly, and whether you allow arbitrary finite partitions or infinite partitions shouldn't thus make much of a difference. The former amounts to doing $\lim_{n\to\infty} \sum \ldots$ while the latter kind of pulls the limit into the sum itself. – fgp May 25 '13 at 22:51
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    What are simple functions if not a partition? – Asaf Karagila May 25 '13 at 23:58
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    @AsafKaragila I think the point is that usually simple functions are finite sums $\sum \alpha_i \chi_i$, not countable infinite sums. – fgp May 26 '13 at 00:48

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After thinking about my own question I think the example $f=1/x^2$ highlights the problem with using simple functions for upper sum approximations. There is no problem in using simple functions from below to approximate $\int_1^\infty 1/x^2dx$ from below. But a simple function $\phi\ge f$ approximating from above will necessarily have infinite integral because $\phi$ will not be able to take on arbitrarily small values to well approximate $1/x^2$ for large $x$.

user782220
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