Decoupling terms in integrals when measure is finite I am reading about the Levy-Khintchine formula, and a particular result says that if $\nu$ is a finite measure, i.e. $\nu(\mathbb{R})< \infty$, then one can decouple the following integral like this:
$$
\int_{\mathbb{R}}(e^{iux}-1-x_{|x|\leq1}) \nu(dx) = \int_{\mathbb{R}}(e^{iux}-1) \nu(dx)- \int_{\mathbb{R}} x_{|x|\leq1} \nu(dx)
$$
I would like show that this is in fact impossible if $\nu$ is an infinite measure. Can one show that if $\nu(\mathbb{R}) = \infty$, then in general
$$
\int_{\mathbb{R}} (f+g) \nu(dx) \neq  \int_{\mathbb{R}} f \nu(dx) + \int_{\mathbb{R}} g \nu(dx)
$$
? I know about some results which say that if $f$ and $g$ are measurable, then indeed
$$
\int (f+g)\mu = \int f\mu +\int g\mu
$$
but I don't see how the measurability of the functions are connected with the finiteness of the measure. 
 A: The formula 
$$
\int (f+g)\,\mathrm{d}\mu = \int f\,\mathrm{d}\mu +\int g\,\mathrm{d}\mu
$$
is not true if $f,g$ are only measurable, you need to add some hypotheses on the integrability of the functions. More specifically, the left hand side makes sense if either of the following holds:


*

*$f,g$ are nonnegative;

*$f+g$ is integrable, in the sense that $\int|f+g|\,\mathrm{d}\mu<+\infty$.


If $f,g$ are nonnegative, then the right hand side still makes sense (with the usual convention that some of the terms could be $+\infty$).
If however, $f+g$ are only integrable, then you may decouple the integral if and only if one of the $f,g$ is also integrable.
In your post, you ask for an example of $f,g$ such that the integrals can not be decoupled when $\nu(\mathbb{R})=+\infty$. Take $f=1$ and $g=-1$. These functions are not $\nu$-integrable, even though their sum is.
Lastly, it is clear that $\nu(\mathbb{R})<+\infty$ is a sufficient condition for decoupling. It is not necessary though. Indeed, consider the Lévy measure $\nu$ which has the following density with respect to the Lebesgue measure on $\mathbb{R}$:
$$
f_\nu(x)=e^{-x^2/2}\mathbb{1}_{\{|x|\ge1\}}+\left|\frac1x\right|\mathbb{1}_{\{|x|<1\}}.
$$
Then, $\nu(\mathbb{R})\ge2\int_{0}^1\frac1x\,\mathrm{d}x=+\infty$, and you can check that both terms on the right hand side of the Levy-Khintchine formula are $\nu$-integrable.
