Computing probability as a Lebesgue-Stieltjes integral I came across an expression for a probability in terms of a Lebesgue-Stieltjes integral and I'd like to understand why it is true.  Let $T$ and $C$ be independent random variables.  Let $F$ be the cumulative distribution function of $T$ and define $S(t) = P(C > t)$.  For $t > 0$ we have
\begin{align} 
P(0 < T \leq t, T < C) = \int_0^t S(u)dF(u). \tag{1}
\end{align}
Here is my attempt at justifying $(1)$.  Let $\mu_T$ and $\mu_C$ be the pushforward measures of $T$ and $C$ on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ (where $\mathcal{B}(\mathbb{R})$ is the Borel $\sigma$-algebra on $\mathbb{R}$); i.e. $\mu_T(E) := P(T^{-1}(E))$ for $E \in \mathcal{B}(\mathbb{R})$ and similar for $\mu_C$.  Likewise let $\mu_{(T, C)}$ be the pushforward measure of the random vector $(T, C)$ on $(\mathbb{R^2}, \mathcal{B}(\mathbb{R}^2))$.  Let $A_t = \{(u, v) \in \mathbb{R^2} : 0 < u \leq t \text{ and } v > u\}$. Then
\begin{align}
\int_0^t S(u)dF(u) & = \int_{\mathbb{R}} \mathbb{I}_{(0,t]}(u) S(u) \mu_T(du) \tag{2}\\
 & = \int_{\mathbb{R}} \mathbb{I}_{(0,t]}(u) \left( \int_{\mathbb{R}} \mathbb{I}_{(u,\infty)}(v) \mu_C(dv) \right) \mu_T(du) \tag{3}\\
 & = \int_{\mathbb{R}} \left( \int_{\mathbb{R}} \mathbb{I}_{(0,t]}(u) \mathbb{I}_{(u,\infty)}(v) \mu_C(dv) \right) \mu_T(du) \tag{4}\\
 & = \int_{\mathbb{R}^2} \mathbb{I}_{(0,t]}(u) \mathbb{I}_{(u,\infty)}(v) \mu_T \times \mu_C(d(u,v)) \qquad (\text{Fubini's Theorem}) \tag{5}\\
 & = \int_{\mathbb{R}^2} \mathbb{I}_{A_t}(u, v) \mu_T \times \mu_C(d(u,v)) \tag{6}\\
 & = \int_{\mathbb{R}^2} \mathbb{I}_{A_t}(u, v) \mu_{(T, C)} (d(u,v)) \qquad (\text{Independence of $T$ and $C$)} \tag{7}\\
 & = \mu_{(T, C)} (A_t) \tag{8}\\
 & = P((T, C)^{-1}(A_t)) \tag{9}\\
 & = P(0 < T \leq t, T < C) \tag{10}.
\end{align}
My questions are: 


*

*Is this justification correct?

*Is there a better or faster way to show $(1)$?

 A: It seems right to me. A faster way to see the same would be setting $Z:=(T,C)$ and noticing that the region of integration is the set $\{(x,y)\in \Bbb R ^2:y>x\,\land\, x\in[0,t]\}$, what have a very easy geometric interpretation: is the union of the infinite box $[0,t]\times [t,\infty )$ and the triangle between the points $(0,0),\,(0,t),\, (t,t)$. 
Therefore, due to the independence of the random variables, you can integrate the region first respect to the measure of the $X$-axis and after respect to the measure of the $Y$-axis (or viceversa) because the measure on the plane $XY$ is just the product measure (because $C$ and $T$ are independent), then
$$
\Pr(Z^{-1}\{(x,y)\in \Bbb R ^2:y>x\,\land\, x\in[0,t]\})=\int_{\Bbb R }\mu_C((s,\infty ))\mathbf{1}_{[0,t]}(s)\,\mathrm d \mu _T(s)\\
=\int_{\Bbb R }\mu _T([0,\min\{s,t\}]\setminus \{s\})\mathbf{1}_{(0,\infty )}(s)\,\mathrm d \mu _C(s)
$$
where the first integral comes when you integrate first respect to $\mu_C$ and after respect to $\mu_T$, and the second integral when you reverses the order of integration.
