[step:Integrate the pointwise inequality to obtain the weak $CD(K,N)$ condition]
Since $\mu_t=(F_t)_{\#}\mu_0$ and $\mu_t=\rho_t m$, the change-of-variables formula for pushforwards gives
\begin{align*}
\int_M \rho_t(z)^{1-\frac1N}\,d m(z)=\int_M \rho_t(F_t(x))^{-\frac1N}\,d\mu_0(x).
\end{align*}
Integrating the pointwise density inequality with respect to $\mu_0$ yields
\begin{align*}
\int_M \rho_t(z)^{1-\frac1N}\,d m(z)\ge \int_M \left[\tau_{K,N}^{(1-t)}(d_g(x,T(x)))\rho_0(x)^{-\frac1N}+\tau_{K,N}^{(t)}(d_g(x,T(x)))\rho_1(T(x))^{-\frac1N}\right]\,d\mu_0(x).
\end{align*}
Because $\pi=(\operatorname{id}_M,T)_{\#}\mu_0$, the last integral is exactly
\begin{align*}
\int_{M\times M}\left[\tau_{K,N}^{(1-t)}(d_g(x,y))\rho_0(x)^{-\frac1N}+\tau_{K,N}^{(t)}(d_g(x,y))\rho_1(y)^{-\frac1N}\right]\,d\pi(x,y).
\end{align*}
Multiplying by $-1$ gives
\begin{align*}
-\int_M \rho_t(z)^{1-\frac1N}\,d m(z)\le -\int_{M\times M}\left[\tau_{K,N}^{(1-t)}(d_g(x,y))\rho_0(x)^{-\frac1N}+\tau_{K,N}^{(t)}(d_g(x,y))\rho_1(y)^{-\frac1N}\right]\,d\pi(x,y).
\end{align*}
This is precisely the weak Lott-Sturm-Villani $CD(K,N)$ inequality for the bounded-support absolutely continuous endpoint measures $\mu_0$ and $\mu_1$. Since the endpoint measures were arbitrary subject to those hypotheses, $(M,d_g,m)$ satisfies the claimed weak $CD(K,N)$ condition.
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