= Solution
{c}
Define $\mathbb Q$ on $\mathcal F_t$ by density $Z_t$. Nonnegativity and <countable additivity> of the integral make it a measure, and $\mathbb Q(\Omega)=\mathbb E_PZ_t=1$. Strict positivity of $Z_t$ also makes the two measures equivalent. The <Radon-Nikodym derivative> identity gives $\mathbb E_Q[\mathbf1_AY]=\mathbb E_P[\mathbf1_AYZ_t]$, first for simple variables and then for every Q-integrable $Y$ by approximation.
Write $U=\mathbb E_Q[Y\mid\mathcal F_s]$. The density on $\mathcal F_s$ is $Z_s=\mathbb E_P[Z_t\mid\mathcal F_s]$. For every $A\in\mathcal F_s$,
$$
\mathbb E_P[\mathbf1_AZ_sU]=\mathbb E_Q[\mathbf1_AU]
=\mathbb E_Q[\mathbf1_AY]=\mathbb E_P[\mathbf1_AYZ_t].
$$
Moreover $Z_sU$ is P-integrable because $\mathbb E_P[Z_s|U|]=\mathbb E_Q|U|<\infty$. Uniqueness in the definition of <conditional expectation> proves
$$
\boxed{Z_s\mathbb E_Q[Y\mid\mathcal F_s]=\mathbb E_P[YZ_t\mid\mathcal F_s].}
$$
This is <conditional expectation under a martingale density>.
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