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We prove the extension from spatial test functions to time-dependent test functions for a diffusion martingale problem directly, without assuming a driving Brownian motion. Fix $0\le s<t$ and a deterministic partition $s=t_0<\cdots <t_n=t$. For each $j$, $g_j(x)=f(t_j,x)$ lies in $C_b^2$. The defining martingale problem givesCall the sum $S_n$. Telescoping separates the spatial and temporal changes:The sample path $X$ is uniformly continuous on $[s,t]$ and has compact range. Joint continuity of ${\partial }_{t}f$, the spatial first derivatives, and the spatial second derivatives therefore gives, as the mesh tends to zero,almost surely. For bounded $a,b$ and bounded derivatives of $f$, both integrals are uniformly bounded on this finite horizon, as are the two endpoint terms. The dominated convergence theorem permits passage to the limit in the expectation against $1_A$. Thus $M_t^f$ is integrable and $\mathbb{E}[1_A(M_t^f-M_s^f)]=0$ for every $A\in {\mathcal{F}}_s$, provingAdaptation and continuity follow from the formula, so $M^f$ is a continuous martingale. Under the more general integrability condition in part (a), the same proof uses an integrable bound proportional to $1+{\int }_s^t({\sum }_i|b_i(X_u)|+{\sum }_i{a}_{ii}(X_u))du$; positive semidefiniteness bounds every off-diagonal entry by the diagonal ones. With only local assumptions this reasoning proves a local martingale, and an integrability hypothesis is needed to upgrade that conclusion.