= Solution
Take $\tau=t_f-t_i>0$ and split it into $M$ intervals of width $\Delta t=\tau/M$. The <markov property> and Gaussian <convolution> give
$$
K=\int_{\mathbb R^{M-1}}\prod_{j=1}^{M-1}dx_j
\prod_{j=0}^{M-1}\frac{\exp[-(x_{j+1}-x_j)^2/(4D\Delta t)]}
{\sqrt{4\pi D\Delta t}},
\qquad x_0=x_i,\ x_M=x_f.
$$
This finite-dimensional identity is exact for every positive $M$. Define the path-integral measure by these normalized <time> slices:
$$
\int\mathcal D x\,(\cdots)
=\lim_{M\to\infty}(4\pi D\Delta t)^{-M/2}
\int\prod_{j=1}^{M-1}dx_j\,(\cdots).
$$
In the <time-sliced diffusion path integral>, the discrete exponent is $\sum_j[(x_{j+1}-x_j)/\Delta t]^2\Delta t/(4D)$. Its conventional continuum notation is therefore
$$
\boxed{K(x_f,t_f|x_i,t_i)
=\int_{x_i}^{x_f}\mathcal D x\,
\exp\!\left[-\int_{t_i}^{t_f}\frac{\dot x^2}{4D}\,dt\right].}
$$
The time-slice normalization is essential, not an arbitrary multiplicative constant. Brownian paths are almost surely nondifferentiable, so the displayed <action> is shorthand for the Gaussian limiting measure rather than an ordinary integral of the derivative of each typical path.
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