 # I am following Folland to study integration theory. It seems like we are restricted to consider only hughy46u 2022-05-20 Answered
I am following Folland to study integration theory. It seems like we are restricted to consider only measurable functions such that ${\int }_{X}|f|\phantom{\rule{thinmathspace}{0ex}}d\mu <+\mathrm{\infty }$ as integrable. However, Folland also mentioned that the definition ${\int }_{X}f\phantom{\rule{thinmathspace}{0ex}}d\mu :={\int }_{X}{f}^{+}\phantom{\rule{thinmathspace}{0ex}}d\mu -{\int }_{X}{f}^{-}\phantom{\rule{thinmathspace}{0ex}}d\mu$ is well-defined for either ${\int }_{X}{f}^{+}\phantom{\rule{thinmathspace}{0ex}}d\mu <+\mathrm{\infty }$ or ${\int }_{X}{f}^{-}\phantom{\rule{thinmathspace}{0ex}}d\mu <+\mathrm{\infty }$.
My question is: What desirable result would fail if we defined integrable function not as "absolutely integrable", but as the later definition for measurable functions such that either ${\int }_{X}{f}^{+}\phantom{\rule{thinmathspace}{0ex}}d\mu <+\mathrm{\infty }$ or ${\int }_{X}{f}^{-}\phantom{\rule{thinmathspace}{0ex}}d\mu <+\mathrm{\infty }$? It seems natural, at least at first glance, to define integrable function as in the later sense and allow infinite integrals to happen.
I can start off with a simple one: If a function ${f}_{1},{f}_{2}:X\to \mathbf{R}$ are integrable in the later sense, then it is not necessarily true that ${f}_{1}+{f}_{2}:X\to \mathbf{R}$ is integrable and so we do not have ${\int }_{X}{f}_{1}+{f}_{2}\phantom{\rule{thinmathspace}{0ex}}d\mu$ as ${\int }_{X}{f}_{1}\phantom{\rule{thinmathspace}{0ex}}d\mu +{\int }_{X}{f}_{2}\phantom{\rule{thinmathspace}{0ex}}d\mu$.
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Folland's version is standard for integrals defined by measure theory.

1. Define
${\int }_{X}f\phantom{\rule{thinmathspace}{0ex}}d\mu :={\int }_{X}{f}^{+}\phantom{\rule{thinmathspace}{0ex}}d\mu -{\int }_{X}{f}^{-}\phantom{\rule{thinmathspace}{0ex}}d\mu$
and declare $f$ integrable if both are finite.
2. If only one of those integrals is infinite, do not say that $f$ is integrable but write this anyway:
${\int }_{X}f\phantom{\rule{thinmathspace}{0ex}}d\mu :={\int }_{X}{f}^{+}\phantom{\rule{thinmathspace}{0ex}}d\mu -{\int }_{X}{f}^{-}\phantom{\rule{thinmathspace}{0ex}}d\mu$
allowing values $+\mathrm{\infty }$ and $-\mathrm{\infty }$.
3. If ${\int }_{X}{f}^{+}\phantom{\rule{thinmathspace}{0ex}}d\mu ={\int }_{X}{f}^{-}\phantom{\rule{thinmathspace}{0ex}}d\mu =\mathrm{\infty }$ do not say anything and do not use the expression ${\int }_{X}f\phantom{\rule{thinmathspace}{0ex}}d\mu$ since (in measure theory) it cannot be assigned any meaningful value.

For purposes of exposition this allows a nice theory for integrable functions, and it allows many theorems for nonintegrable functions to be stated with infinite integrals. Most good theorems will be stated for integrable functions. With some caution you can still use the infinite valued integral, but be careful. It is not worth fussing too much, just don't be seduced into false statements.
In case #3 there is no theory? Well in many cases there is, but it is outside measure theory.
Restrict attention to the real line. Now this is Lebesgue's classical theory, except he originally called "integrable" functions "summable" probably so as not to conflict with Riemann integrability.
Is there any interest in functions $f:\left[a,b\right]\to \mathbb{R}$ for which
${\int }_{a}^{b}{f}^{+}\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx={\int }_{a}^{b}{f}^{-}\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx=\mathrm{\infty }$
which is case #3 above (ignored by measure theory and by Lebesgue's original integral)?
The problem that Lebesgue gave as his motivation was to find an integral that handled all derivatives and included the Riemann integral. The Riemann integral does not integrate all derivatives.
Unfortunately neither does Lebesgue's integral. I gather this annoyed him. It handled all bounded derivatives and includes the Riemann integral.
But "most" derivatives are like this: ${F}^{\prime }\left(x\right)=f\left(x\right)$ for all $x\in \left[a,b\right]$ but
${\int }_{a}^{b}{f}^{+}\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx={\int }_{a}^{b}{f}^{-}\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx=\mathrm{\infty }.$
So very sorry. Cannot integrate derivatives. But Newton and Leibnitz could (albeit trivially). Why isn't
${\int }_{a}^{b}f\left(x\right)\phantom{\rule{thinmathspace}{0ex}}dx=F\left(b\right)-F\left(a\right)?$
Apologize. Sheepishly say, the Lebesgue integral wasn't designed for that. Or develop an entirely different method of integration that is not based on measure theory.
That was done about ten years after Lebesgue's integral by a German mathematician Perron and a French mathematician Denjoy. Their integral includes the Lebesgue integral and integrates all derivatives. So you might say their integration method explores the case #3 above that measure theory refuses to address.
The story continues. A trigonometric series can converge everywhere to a function that is not Lebesgue integrable (or even Denjoy-Perron integrable). Ideally you would want to compute the coefficients using the Fourier formulas---but what integral can you use? Time for another "nonabsolutely convergent integral" that lives in the same crack left by #3.
Want more? Probably not, but there are more.
The short answer is that the case #3 that measure theory and Lebesgue himself abandonned can be explored by other methods suitable to the situation. But measure theory is designed to leave that case alone. But the situations #1 and #2 cover most interesting and important applications anyway, so not a lot is lost.