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For a stopping time $T\le s$, optional sampling for $N$ gives $\mathbb{E}[N_s\mid {\mathcal{F}}_T]=N_T$, henceThis is justified by truncation and the Cauchy-Schwarz inequality, as in solution. If $MN$ is a martingale, bounded optional sampling gives $\mathbb{E}[M_T{N}_T]=\mathbb{E}[M_0{N}_0]=0$, proving the required condition.

Conversely suppose the required condition holds. Every bounded stopping time $T$ lies below some deterministic $s$, so the displayed identity gives $\mathbb{E}[(MN{)}_T]=0$. The product is continuous and adapted, and its stopped values are integrable: the Doob L2 maximal inequality bounds the product of the two path suprema in $L^1$ by the Cauchy-Schwarz inequality. The characterization of a martingale by bounded continuous-time stopped expectations therefore makes $MN$ a martingale. Thus orthogonality is equivalent to the backward-stopping expectation test. In contrast to weakly orthogonal continuous martingales, this tests the product at random times, not only deterministic times.

Fix deterministic $0\le s<t$ and $A\in {\mathcal{F}}_s$. The pasting ordered stopping times argument shows that $U=s{1}_A+t{1}_{A^c}$ is a bounded stopping time. The given stopped-expectation property, applied to $U$ and to the deterministic stopping time $t$, yieldsThis holds for every $A\in {\mathcal{F}}_s$. Since $X_s$ is ${\mathcal{F}}_s$-measurable and both time values are integrable, it is exactly the defining test for conditional expectation:Hence $X$ is a martingale. This proof of the characterization of a martingale by bounded continuous-time stopped expectations uses only deterministic two-time pastings; the càdlàg assumption and the usual conditions for a filtration are stronger than needed for this implication.