When one talks about Lebesgue integrability or lack thereof, what does one mean, exactly? Does it mean we cannot possibly take a Lebesgue integral of that function, because it simply isn't well-defined ... or does it mean the integral is finite? So if the latter, even a function like $f(x) = c$ for $c > 0$ is not Lebesgue integrable?
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This is a great question. There's no clear answer as it depends on what the definition of "Lebesgue integrable" is. I have seen the definition of a function $f : X \to \Bbb C$ being Lebesgue integrable as meaning $\int \limits_{X} |f| ,d\mu < \infty$, so if that is the definition, then "not Lebesgue integrable" means the absolute value does not have a finite integral. In that case, $f(x) = 1$ is not integrable over $\Bbb R$, as you pointed out. But some others might say Lebesgue integrable means that the Lebesgue integral exists (even if it is not finite). So you really have to either... – layman Sep 26 '15 at 21:59
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...look through the text you're reading to find the definition of Lebesgue integrable, or ask whatever source you were discussing this topic with what they take "Lebesgue integrable" to mean. – layman Sep 26 '15 at 22:00
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A function which is not Lebesgue integrable has (at least) one of the following properties:
It is not (Lebesgue) measurable. In this case, essentially all bets are off. Nonmeasurable functions can only be constructed using some form of the axiom of choice,so that all functions appearing in practice (piecewise continuous, monotonic,...) are Lebesgue measurable.
Note that it can happen for $|f|$ to be measurable (even integrable), although $f$ is not measurable. An example is $f=1_A - 1_{A^c}$, where $A$ is not measurable.
$f$ is measurable, but $\int |f|\,dx =\infty$. This means that $f$ is "too large" to admit a finite integral. In this case,it can still happen that $f$ is quasi integrable which means $\int f_+ \, dx <\infty$ or $\int f_-\, dx <\infty$, where $f_+,f_-$ denote the positive and negative parts of $f$. In these cases, we have $\int f \,dx =-\infty$ or $\int f \,dx =\infty$, respectively.
Finally, if $f$ is integrable, this means that $f$ is measurable and $\int |f|\,dx <\infty$, which then implies that $\int f \,dx$ is a well defined real number.
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@user46944:Yes, in principle it could. But in my answer, I am (implicitly) assuming that $f$ is real valued. Otherwise, $f_+$ and $f_-$ do not make much sense either. – PhoemueX Sep 27 '15 at 16:03
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How can one prove that
$f = 1_A - 1_{A^c}$is not measurable ? When one proves a function is measurable one takes preimages of measurable sets and checks if these are measurable. I will appreciate to see how one would do this in this particular case where A is not measurable. How this translates into making$|f|$measurable ? Thanks. – user996159 Oct 07 '22 at 09:46 -
1@user996159: Consider $f^{-1}({1})$. Finally $|f| \equiv 1$, which easily implies that $|f|$ is measurable. – PhoemueX Oct 07 '22 at 12:27
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I do not see why $f^{-1}({1})$ is non-measurable. Even though no space is given where $f$ is defined, if we assume the following $f: (\mathbb{R}, \mathcal{P}(\mathbb{R})) \rightarrow (\mathbb{R},\mathbb{B(\mathbb{R})})$, then ${1}$ is measurable and $f^{-1}({1}) = \mathbb{R}$ which is measurable. Given that we assume $A$ to be non-measurable, I considered here the power set on $\mathbb{R} $ as a sigma algebra. – user996159 Oct 07 '22 at 13:33
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You cannot use the power set as a sigma algebra, since then there are no non-measurable sets. You have $f^{-1}({1})=A$, which is non-measurable by choice of $A$. Or rather, given the context, you should consider the Lebesgue sigma algebra on the domain of $f$, $\Bbb{R}$. – PhoemueX Oct 07 '22 at 13:49
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Thanks. But something remains unclear. I have learned that in order to exclude non-measurable sets, one has to take smaller sigma algebras than the power set on $\mathbb{R}.$ For example the Borel sigma algebra. Now you are saying the opposite, 'You cannot use the power set as a sigma algebra, since then there are no non-measurable sets'. I will appreciate a further comment. – user996159 Oct 07 '22 at 14:16
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It means it doesn't satisfy the requirements to be Lebesgue integrable.
Expanding on that
If I say something isn't X it means it didn't satisfy the definition of X (or something logically equivalent to it)
if $P\implies $is lebesugue integrable then showing not $P$ isn't sufficient. If Lebesugue integrable $\implies P$ then $¬P\implies$ not lebesgue integrable.
As usual.
