Problem: Let M be a continuous and bounded martingale, and let A be a bounded increasing process. Show $E A_{\infty}M_{\infty}=E\int_0^{\infty} M_t dA_t$.
Lemma: if $f:[0,T] \rightarrow \mathbb{R}$ is a continuous function, and if $0=t_0^n<t_1^n<...<t_{o_n}^n=T$ is a sequence of subdivisions of $[0,T]$ whose mesh tends to 0, we have $\int_0^T f(s)da(s)=\lim_{n \rightarrow \infty} \sum_{i=1}^{p_n} f(t_{i-1}^n)(a(t_i^n)-a(t_{i=1}^n))$.
Since M and A are bounded, $A_t$ has finite variation on $[0,T]$ for $T>0$. Then there is a unique $\sigma$ finite measure on $\mathbb{R}_+$ whose restriction to every interval $[0,T]$ is the total variation measure of the restriction of $A_t$ to $[0,T]$. Then we write $\int_0^{\infty} M_t d A_t=\lim_{T \rightarrow \infty} \int_0^T M_td A_t \in (-\infty,\infty)<\infty$. Then taking expectation on both sides of $\int_0^{\infty} M_t d A_t=\lim_{T \rightarrow \infty} \int_0^T M_td A_t$, we have $$E A_{\infty}M_{\infty}=E\int_0^{\infty} M_t dA_t$$.
Not sure if this works for last step, thanks in advance!
Notice that if $M_0 = 0$:
\begin{align} \int_0^T M_t dA_t &= A_T M_T - A_0 M_0 - \int_0^t A_t dM \\ &= A_T M_T - \int_0^t A_t dM \end{align}
The integral on the LHS is a martingale, so it's expected value is constant, and in particular 0. Then taking expected values the result follows.