= Solution
Put $M_k=S_k-mk$ as above. Centering the <linear interpolation> also interpolates the centered values: for $u\in[0,1]$,
$$
S_{k+u}-m(k+u)=(1-u)M_k+uM_{k+1}.
$$
The absolute value of this convex combination is at most the larger endpoint absolute value. For every fixed $A>0$, part (c) therefore gives
$$
\mathbb E\!\left[\sup_{0\leq t\leq A}|S_t^{(N)}-mt|^2\right]
\leq \frac{4\sigma^2\lceil NA\rceil}{N^2}\longrightarrow0.
$$
Hence there is <uniform convergence on compacts in probability> to the deterministic path $h(t)=mt$, by <Markov inequality>. Equip $C([0,\infty),\mathbb R)$ with its standard <compact-open topology>, metrized by
$$
d(f,g)=\sum_{j=1}^{\infty}2^{-j}\left(1\wedge\sup_{0\leq t\leq j}|f(t)-g(t)|\right).
$$
For each finite number of terms their suprema converge to zero in probability, and the remaining tail is bounded deterministically by $\sum_{j>J}2^{-j}$. Thus $d(S^{(N)},h)\to0$ in probability. <Convergence in probability> to a deterministic point implies <weak convergence of probability measures>, so the <fluid limit> is
$$
\boxed{\mu_N\Longrightarrow\delta_h,\qquad h(t)=mt.}
$$
Here $\delta_h$ is the <Dirac measure> concentrated on that continuous path. The topology is <locally uniform convergence>; no assertion of <uniform convergence> over the entire unbounded half-line is needed. The interpolation is intended for integers $k\geq0$, including the initial interval $[0,1]$. If the PDF's $\mathbb Z^+$ is interpreted as strictly positive integers, that initial piece is omitted from the displayed definition and must be supplied by the same formula using $S_0=0$.
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