"Is not Lebesgue integrable" 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?
 A: 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.
A: 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.
