A question about a measure defined using two CDF's I know that the sum of measures is a measure (see this question). However, if $X_1$ and $X_2$ are random variables with CDF's $F_1$ and $F_2$, respectively, I know that
$$F = F_1 + F_2$$
is not even a CDF ( in fact, the CDF of $Z = X_1+X_2$ is given by the convolution of $F_1*F_2$. See this question ).
Moreover, given a CDF $F$, I also know that I can define measures, for example, as
$$\nu(E):= \int_E h(x) dF(x), \quad E \,\, \hbox{any borelian.}$$
In this case I have
\begin{equation}\label{I}\tag{I}
d\nu(x)=h(x)dF(x)
\end{equation}
So I can 'change' the measures:
$$\int g(x)dF(x) = \int \frac{g(x)}{h(x)} d\nu(x)$$
So, my question is related to the following measure:
\begin{equation}\label{kasjkjas}\tag{M}
\nu(E) = \sum_{i=1}^2 \int_E |x|^2 dF_i(x)
\end{equation}
So I'm dying to interchange the integral and the sum:
\begin{equation}\label{II}\tag{II}
\nu(E) =  \int_E |x|^2 dF(x), \quad dF(x):= \sum_{i=1}^2 dF_i(x)
\end{equation}
But I don't know if this is mathematically rigorous. The impression I get is that I'm treating $F_1 + F_2$ as a CDF, which is not true. Moreover, as in (\ref{I})
$$d\nu(x)= |x|^2 dF(x) = |x|^2 [dF_1(x)+dF_2(x)]$$
So is it correct to define $dF(x)$ as in (\ref{II}) even though $F_1+F_2$ is not well defined?
If this is not correct, how can we rewrite the following integral in terms of $dF_1(x)$ and $dF_2(x)$:
$$\int f(x) \frac{1}{|x|^2} d\nu(x)\,\,\,\,?$$
(If (\ref{II}) is correct, ignore this last question )
Attempt after comments
According to Snoop's comments, setting $\quad dF(x):= \sum_{i=1}^2 dF_i(x)$ is mathematically incorrect. So, I define
$$\nu_i (E):= \int_E |x|^2 dF_i(x), \quad i =1,2$$
It implies
$$d\nu_i(x)= |x|^2 dF_i(x), \quad i =1,2$$
So substituting in (\ref{kasjkjas}), we have:
$$\nu(E)= \int_E d\nu_1(x)+ \int_E d\nu_2(x)= \nu_1(E)+\nu_2(E)$$
Moreover $d\nu= d(\nu_1 + \nu_2)= d\nu_1 + d\nu_2$. So,
$$\int f(x) \frac{1}{|x|^2} d\nu(x) = \int f(x) \frac{1}{|x|^2} (d\nu_1(x) + d\nu_2(x)) = \int f(x) \frac{1}{|x|^2} d\nu_1(x) + \int f(x) \frac{1}{|x|^2}  d\nu_2(x)$$
So we conclude
$$\int f(x) \frac{1}{|x|^2} d\nu(x) = \sum_{i=1}^2 \int f(x) dF_i(x) $$
 A: Let $\mu,\nu$ be finite measures on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$ and define $\rho:=\mu+\nu$. (1): We prove that $\rho$ is a finite measure on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$ by checking $\rho(\emptyset)=0,\rho(\mathbb{R})<\infty$ and countable additivity; this is straightforward. (2): let $u \geq 0$ and $u \in L^1(\mu)\cap L^1(\nu)$; we have a sequence of nonnegative simple functions $u_n\uparrow u$ so
$$\begin{aligned}\int u d\mu+\int ud\nu&=\lim_{n\to \infty}\bigg(\int u_nd\mu+\int u_nd\nu\bigg)=\\
&=\lim_{n\to \infty}\bigg(\sum_{k\leq N_n}\phi_{k}^n(\mu(A_k^n)+\nu(A_k^n))\bigg)=\\
&=\lim_{n\to \infty}\sum_{k\leq N_n}\phi_{k}^n\rho(A_k^n)=\\
&=\lim_{n\to \infty}\int u_nd\rho=\\
&\stackrel{\textrm{MCT}}{=}\int ud\rho\end{aligned}$$
This implies $u \in L^1(\rho)$. In your case, $\mu,\nu$ are probability measures with respective cdfs $F_1,F_2$ and $|x|^2 \in L^1(\mu)\cap L^1(\nu)$ is a reasonable further assumption.
A: $F_1+F_2$ is not a cdf (since the measure of the whole space is two) but it is a distribution function (right continuous, non decreasing).
Every right continuous, non decreasing function $F$ defines a unique (Lebesgue Stieltjes) measure $\mu_F$ on ${\cal B}$, the Borel sets, such that $\mu_F((a,b]) = F(b)-F(a)$.
It follows that $\mu_{F_1+F_2} = \mu_{F_1} + \mu_{F_2}$.
The usual sequence (indicator functions, simple functions, integrable limits) shows that
$\int f d\mu_{F_1+F_2} = \int f d  \mu_{F_1} + \int f d \mu_{F_2}$.
Addendum:
Given the definition of $\nu_k$ above, we have
$\int g d \nu_k = \int g(x) |x|^2 d\mu_{F_k} (x)$ and so (assuming integrability as appropriate) we have
$\int f(x) {1 \over |x|^2} d  \nu_k(x) = \int f(x) d\mu_{F_k}(x)$. Summing gives the desired result.
