Replacing improper integral by sum of integrals In some situations we replace the improper integral by sum of integrals. I mean if $a\gt 0$ then$$\int_{-\infty}^{+\infty}f(x)dx = \sum_{n = -\infty}^{+\infty}\int_{na}^{(n+1)a}f(x)dx$$
Intuitively, it seems clear because we are  decomposing the interval of integration but how we can justify this? What are the conditions for $f(x)$ to satisfy this relation? Divergence of one side implies the divergence of the other side? Also what should we do if there are infinite limits for $f(x)$ (for example  $\lim_{x\to b^{-}} f(x) = \pm\infty$)?
 A: If the left hand side converges, then the right hand side converges.  But not the converse.  For example, if $f(x) = \sin(x)$ and $a = 2\pi$, then the right hand side is $0$, and the left hand side is undefined.  If $f$ has a finite number of infinite values, you can handle them one at a time, and what you do on the left hand side is the same as what you do to the right hand side.
If the left hand side converges, we have
$$ \lim_{R \to \infty, S \to -\infty} \int_S^R f(x) \, dx = L .$$
So we can restrict $R$ and $S$ to multiples of $a$ to see
$$ \lim_{n \to \infty, m \to -\infty} \int_{am}^{an} f(x) \, dx = L ,$$
and
$$ \int_{am}^{an} f(x) \, dx = \sum_{k=m}^{n-1} \int_{ka}^{(k+1)a} f(x) \, dx .$$
Next, if, say, $\lim_{x\to b-} f(x) \, dx = \pm\infty$, then for the left hand side you add in
$$ \lim_{u\to b-, v\to b+} \int_S^u + \int_v^R f(x) \, dx ,$$
and for the right hand side you pick the $n$ such that $na < b < (n+1)a$, and instead of
$$ \int_{na}^{(n+1)a} f(x) \, dx $$
use
$$ \lim_{u\to b-, v\to b+} \int_{na}^u + \int_v^{(n+1)a} f(x) \, dx ,$$
or if $b = na$, then instead of
$$ \int_{(n-1)a}^{na} f(x) \, dx + \int_{na}^{(n+1)a} f(x) \, dx $$
use
$$ \lim_{u\to b-, v\to b+} \int_{(n-1)a}^u + \int_v^{(n+1)a} f(x) \, dx ,$$
A: This is true as soon as $f$ is Lebesgue integrable... which makes sense if you speak of $\int_{-\infty}^\infty f(x) dx$.
The equality is just a particular case of Lebesgue countable additivity property.
