= Solution
Assume (iii). Fix $n\geq0$ and $A\in\mathcal F_n$, and define
$$
T=n\mathbf1_A+(n+1)\mathbf1_{A^c}.
$$
This is a <stopping time>: $\{T\leq n\}=A\in\mathcal F_n$. Applying (iii) at time $n+1$ first to $T$ and then to the deterministic stopping time $n+1$ gives
$$
\mathbb E[\mathbf1_AM_n+\mathbf1_{A^c}M_{n+1}]
=\mathbb E[M_0]
=\mathbb E[M_{n+1}].
$$
Subtracting yields
$$
\mathbb E[\mathbf1_A(M_{n+1}-M_n)]=0
\qquad(A\in\mathcal F_n).
$$
By the defining characterization of <conditional expectation>, this says $\mathbb E[M_{n+1}\mid\mathcal F_n]=M_n$. Thus $M$ is a martingale, proving (i). Together the three implications establish the <characterization of a martingale by stopped expectations>.
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