= Solution
Let $V$ be the closure of the <simple predictable processes> in the <Hilbert space> $L^2(\mathcal P,\mu)$. Common refinement of deterministic partitions makes these processes a vector space. To prove \b[<density of simple predictable processes for finite measures>], suppose $g\in V^\perp$. Since $\mu$ is a <finite measure>, the <Cauchy-Schwarz inequality> gives $g\in L^1(\mu)$, so
$$
\nu(E)=\int_Eg\,d\mu
$$
defines a finite signed <measure>. The indicator of each predictable generating rectangle, including each time-zero rectangle, is a <simple predictable process>; hence $\nu$ vanishes on these rectangles.
Finite intersections of the rectangles are again rectangles or empty. Add the whole space to this <pi-system>. Its $\nu$-mass is zero because $\mathbf1_{\Omega\times[0,n]}$ is simple and converges to $1$ in $L^2(\mu)$ by the <dominated convergence theorem>. The zero sets of $\nu$ form a <lambda-system> now that its total mass is zero. The <Pi-lambda theorem> therefore gives $\nu(E)=0$ for every $E\in\mathcal P$. Taking the sets where $g>0$ and $g<0$ gives $g=0$ $\mu$-almost everywhere.
Thus $V^\perp=\{0\}$ and the <orthogonal complement> criterion for a closed subspace of a <Hilbert space> yields
$$
\boxed{\overline{\mathcal S}^{\,L^2(\mathcal P,\mu)}=L^2(\mathcal P,\mu).}
$$
Omitting the time-zero term would make this conclusion false when $\mu$ is concentrated at time zero.
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