Lebesgue integral and absolute value I wonder why we say that $f$ is integrable iff $\int|f|\,d\mu$ is finite? Why we use absolute value? Won't it be enough to have that $\int f\, d\mu$ is finite to call $f$ integrable? Are there functions $f$ for which $\int f \ d\mu$ is finite but $\int |f| \ d\mu$ is not (like conditional convergence for series)?. But then why we would call such $f$ not Lebesgue integrable?
 A: It's very much like conditional convergence for series.  A conditionally convergent series (i.e. a convergent series whose partial sums of the positive terms diverge to $+\infty$ and those of the negative terms to $-\infty$) converges to a different thing if you suitably rearrange its terms.
Consider the integrals
$$
\int_0^1\left(\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2}\, dy\right)\,dx \quad\text{and} \quad \int_0^1 \left(\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2}\, dx\right)\,dy.
$$
You can show by freshman calculus methods that one of these is $\pi/2$ and the other is $-\pi/2$.  Rearranging it has changed the sum.
A simpler example is
$$
\lim_{a,b\to\infty\text{ ???}} \int_b^a \frac{x\,dx}{1+x^2}.
$$
If you evaluate
$$
\lim_{a\to\infty}\int_{-a}^a \frac{x\,dx}{1+x^2}
$$
you get $0$, but
$$
\lim_{a\to\infty}\int_{-a}^{2a}\frac{x\,dx}{1+x^2}
$$
is a non-zero number.  Try it and see.
A: As far as why we define integrability this way, one reason that I can think of is that we cannot always assign a numerical value to $\int f d\mu$, but with $\int |f| d\mu$, we can - it's either finite or infinite. That is, $\int f d\mu$ is not always well-defined. However, if we know that $\int |f| d\mu < \infty$, then $\int f d\mu$ is well-defined since 
$$\int |f| d\mu = \int f^+ d\mu + \int f^- d\mu < \infty$$ which is true only when both positive and negative parts of the function are both finite. Thus,
$$\int f d\mu = \int f^+ d\mu - \int f^- d\mu $$ is finite as well.
