Does it make sense to talk about the integral of measurable functions that are not absolutely integrable? Suppose $f$ is a real-valued (possibly infinite-valued) function on some measure space $(X, \Sigma, \mu)$, and suppose that it is measurable.  Note that $f$ is not necessarily nonnegative.
Does it make sense to talk about $\int \limits_{X} f(x) \,d\mu$ if $f \not \in L^{1}(d\mu)$ (i.e., $\int \limits_{X} |f| \,d\mu \not < \infty$)?
If this is true, I think this is a very subtle point.  I was always under the impression that any measurable function can be integrated, but now I doubt that this is true if the function is not nonnegative.
 A: One can define a meaningful integral as
$$
\int f \, d\mu = \int f_+ \, d\mu - \int f_- \, d\mu,
$$
As long as at least one of the two integrals on the right hand side is finite.
Here, $f_+ (x) = \max \{0, f(x)\}$ is the positive part of $f$ and the negative part is defined analogously.
A: This question is basically answered in comments, but one could remark that there
are ways to extend the integral beyond "quasi-integrable" functions (those for which at least one of $f_+$ or $f_-$ has finite integral).  Namely, improper integrals
give a way to do this.
Suppose that we write our measure space $X$ as the increasing union of subsets $A_n$,
often assumed to be of finite measure.  Then we can consider $$\lim_{n \to \infty} \int_{A_n} f d\mu,$$
and if this exists, call it the improper integral of $f$.  
A common case is when $X = [0,\infty)$ and $A_n = [0,n].$  Another
possibility (I think called the Cauchy principal value) is $X = [-1,1]$ (or some
other interval containing the origin) and $A_n = [-1,-1/n] \cup [1/n,1]$ (whose
union is not quite all of $X$ --- their union is $X \setminus \{0\}$ --- but has
full measure in $X$).
The theory of improper integrals is like the theory of conditionally convergent series --- it relies on cancellations to obtain a well-defined limit when the function is not absolutely integrable.   Of course not all functions have a well-defined improper integral either.  And in general the improper integral can depend on the choice of sets $A_n$.  
