Let $(\mathbb{R},\mathcal{B}(\mathbb{R}),\mu)$ be a probability space, and let $F$ be the cumulative distribution function of the probability measure $\mu$ (that is, $F(x)=\mu((-\infty,x])$ and $\mu((a,b])=F(b)-F(a)$ for all $-\infty<a \leq b<\infty$).
Using the Fubini/Tonelli Theorem for nonnegative functions, it may be shown that
$\displaystyle2\int_{\mathbb{R}}F(x)\mu(dx)=1+\int_{\mathbb{R}}\mu(\{x\})\mu(dx)=1+\sum_\limits{x \in \mathbb{R}}\mu(\{x\})^2$, where $\sum_\limits{x \in \mathbb{R}}$ is interpreted as integration with respect to the counting measure.
I'm very confused as to how Fubini/Tonelli comes into play in this scenario; is there a way to manipulate the left hand side into iterated integrals? I should also note that we also have $\mu(\{a\})=F(a)-\lim_\limits{x\rightarrow a^-}F(x)$ and that $\lim_\limits{x \rightarrow \infty}F(x)=\mu(\mathbb{R})=1$, though I don't think introducing limits into the work would help at all. Any help or insight would be greatly appreciated!
Note $F(x)=\mu(-\infty,x]=\int_{\Bbb R} \chi_{(-\infty,x]}(y)\text{d}\mu(y)$, where the integrand is measurable and positive, so applying Tonelli's Thm. gives: $$\begin{align} \int_{\Bbb R} F(x)\text{d}\mu(x) &= \iint\chi_{(-\infty,x]}(y)\text{d}\mu(y)\text{d}\mu(x)= \iint\chi_{(-\infty,x]}(y)\text{d}\mu(x)\text{d}\mu(y) \\ &=\iint\chi_{[y,\infty)}(x)\text{d}\mu(x)\text{d}\mu(y)=\int\mu[y,\infty)\text{d}\mu(y)=\int1-\mu(-\infty,y)\text{d}\mu(y) \\ &= 1-\int\mu(-\infty,x)\text{d}\mu(x) \end{align}$$ Now add $\int_{\Bbb R} F(x)\text{d}\mu(x)$ to both sides to get $$\begin{align} 2\int_{\Bbb R} F(x)\text{d}\mu(x) &= 1+\int_{\Bbb R}\mu(-\infty,x]-\mu(-\infty,x)\text{d}\mu(x)=1+\int_{\Bbb R}\mu(\{x\})\text{d}\mu(x) \end{align}$$ The claim about the counting measure is just by definition.