= Solution
Assume the displayed <Gaussian forward-rate covariance drift restriction>. Its diagonal instance gives
$$
\mu_{u,u}-\mu_{0,u}=\int_0^u c(w,w,u)\,dw.
$$
Using it on the two parts of the deterministic discounted exponent yields
$$
\begin{aligned}
A(s,T)-A(0,T)
&=\int_0^s\int_0^u c(w,w,u)\,dw\,du
+\int_s^T\int_0^u c(s\wedge w,w,u)\,dw\,du\\
&=\int_0^T\int_0^u c(s\wedge w,w,u)\,dw\,du\\
&=\frac12\int_0^T\int_0^T c(s\wedge u\wedge w,u,w)\,du\,dw
=\frac12v(s,T).
\end{aligned}
$$
The last equality uses <covariance> symmetry to reflect the triangular integration region across its diagonal. The <independent increments> of the <integrated Gaussian forward-rate process> and its conditional <Gaussian moment-generating function> then give, for $r\leq s\leq T$,
$$
\mathbb E[Z_{s,T}\mid\mathcal F_r]
=e^{-A(s,T)-Y(r,T)+(v(s,T)-v(r,T))/2}
=e^{-A(r,T)-Y(r,T)}=Z_{r,T}.
$$
All these prices are <integrable> because their exponents are <Gaussian random variables> with finite <variance>. Thus the mean restriction implies the discounted bond <martingale> condition, completing the equivalence of the first two descriptions.
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