Let $A,B$ be non-decreasing cadlag processes such that $A_0 = B_0 = 0$ and limits $A_\infty = \lim_{t \to \infty} A_t$ and $B_\infty = \lim_{t \to \infty} B_t$ are finite. I am trying to prove that $A_\infty B_\infty = \int_0^\infty A_sdB_s + \int_0^\infty B_{s-}dA_s$.
Do you have any suggestions or ideas how one could prove this. Thanks.
Since all terms are non-decreasing, it suffices to show
$$A_T \cdot B_T = \int_0^T A_s \, dB_s + \int_0^T B_{s-} \, dA_s \tag{1}$$
for any $T>0$.
Hint (Solution 1) Fix $T>0$ and a partition $\Pi:=\{0=t_0<\ldots<t_n=T\}$ of $[0,T]$. Then
$$\begin{align*} A_T \cdot B_T &= \sum_{j=1}^n (A_{t_j} \cdot B_{t_j}-A_{t_{j-1}} \cdot B_{t_{j-1}}) = \sum_{j=1}^n A_{t_j} \cdot (B_{t_j}-B_{t_{j-1}}) + \sum_{j=1}^n B_{t_{j-1}} \cdot (A_{t_j}-A_{t_{j-1}}) \end{align*}$$
Show that the right-hand side converges to the integrals in $(1)$ as the mesh size $$|\Pi| = \max_{j} |t_j-t_{j-1}|$$ tends to $0$.
Hint (Solution 2) By Fubini's theorem, we have
$$A_T \cdot B_T = \int_{[0,T] \times [0,T]} dA_r \, dB_s = \int_{F_1} dA_r \, dB_s + \int_{F_2} dB_s \, dA_r$$
where $F_1 := \{(r,s); 0<r \leq s \leq T\}$, $F_2 := \{(r,s); 0<s<r \leq T\}$. Calculate the integrals on the right-hand side explicitely to prove the assertion.