Expectation is zero; left-limit of minimum (proof of Doob-Meyer decomposition, Karatzas and Shreve)

Question

Two questions about the proof of Doob-Meyer (p. 25 and p. 27, respectively):

1. Suppose $\{B_t,\mathcal F_t,t\ge0\}$ is a martingale of bounded variation, and that for every bounded and right-continuous martingale $\{\xi_t,\mathcal F_t\}$, we have $E(\xi_tB_t)=0$. Then for any bounded random variable $\xi$, $E(\xi B_t)=0$.

2. Suppose [we have for some $a>0$ given (added; thanks to Saz for clarifing)] that $E\int_{(0,a]}\xi_s\,dA_s=E\int_{(0,a]}\xi_{s-}\,dA_s$ for every bounded, right-continuous martingale $\{\xi_t,\mathcal F_t\}$. Then $E\int_{(0,t]}\xi_s\,dA_s=E\int_{(0,t]}\xi_{s-}\,dA_s$ for all $t\in(0,a]$.

For 1, KS notes that we can select $\{\xi_t,\mathcal F_t\}$ a right-continuous modification of $\{E[\xi|\mathcal F_t],\mathcal F_t\}$. So we have $E[\xi_t B_t]=0$ for every $t$. This is fine; I don't see how to go from this observation to $E[\xi B_t]=0$. Independence of $\xi_t$ and $B_t$ would suffice, but we don't have anything like that, and I don't see how to use a conditioning argument to achieve something similar.

For 2, it would suffice to verify that $s\mapsto X_{(s\wedge t)-}\equiv \lim_{k\nearrow s} X_{k\wedge t}$, since then $$E\int_{(0,a]}\xi_{(s\wedge t)-}=E\int_{(0,t]}\xi_{s-}+E\int_{(t,a]}\xi_t,$$ $$E\int_{(0,a]}\xi_{s\wedge t}=E\int_{(0,t]}\xi_s+E\int_{(t,a]}\xi_t.$$ But I'm not sure if this is the correct definition of a left-limit of a minimum. Maybe it's $X_{(s\wedge t)-}\equiv\lim_{k\nearrow s\wedge t} X_k$.

Any suggestions on 1 and input on definitions in 2 would be greatly appreciated. Thanks in advance!

Answer 1

1. By the tower property, $$\mathbb{E}(\xi B_t) = \mathbb{E} \bigg( \mathbb{E}(\xi B_t \mid \mathcal{F}_t) \bigg).$$ As $B_t$ is $\mathcal{F}_t$-measurable, we find for $\xi_t := \mathbb{E}(\xi \mid \mathcal{F}_t)$ $$\mathbb{E}(\xi B_t) = \mathbb{E} \bigg( B_t \mathbb{E}(\xi \mid \mathcal{F}_t) \bigg) = \mathbb{E}(B_t \xi_t)=0.$$
2. Unfortunately, $\xi_{(s \wedge t)-} = \lim_{k \uparrow s} \xi_{k \wedge t}$ does, in general, not hold true for $s>t$. You can use the following idea instead: For given $(\xi_s)_{s \geq 0}$ and fixed $t \geq 0$ define $$\eta_s := \begin{cases} \xi_s, & s \leq t, \\ \xi_t & s > t. \end{cases}$$ It is not difficult to see that $(\eta_s,\mathcal{F}_s)_{s \geq 0}$ is a right-continuous bounded martingale. By assumption, $$\mathbb{E} \int_{(0,a]} \eta_s \, dA_s = \mathbb{E} \int_{(0,a]} \eta_{s-} \, dA_s.$$ Using that $$\int_{(t,a]} \eta_s \, dA_s = \xi_t (A_a-A_t) = \int_{(t,a]} \eta_{s-} \, dA_s$$ this implies $$\mathbb{E} \int_{(0,t]} \eta_s \, dA_s = \mathbb{E} \int_{(0,t]} \eta_{s-} \, dA_s.$$