## How can we show that the total variation distance of $X_s$ and $Y_s$ is bounded by the distance of $(X_t)_{t\ge s}$ and $(Y_t)_{t\ge s}$?

Let $$(X_t)_{t\ge0}$$ and $$(Y_t)_{t\ge0}$$ be real-valued time-homogeneous Markov processes with a common transition semigroup $$(\kappa_t)_{t\ge0}$$. Let $$\mathcal L(Z)$$ denote the distribution of a random variable $$Z$$ and $$\left|\mu-\nu\right|:=\sup_{B\in\mathcal E}\left|\mu(B)-\nu(B)\right|$$ denote the total variation distance of probability measures $$\mu,\nu$$ defined on a common $$\sigma$$-algebra $$\mathcal E$$.

How can we show that $$\left|\mathcal L(X_s)-\mathcal (Y_s)\right|\le\left|\mathcal L\left(\left(X_{s+t}\right)_{t\ge0}\right)-\mathcal L\left(\left(Y_{s+t}\right)_{t\ge0}\right)\right|\tag1$$ for all $$s\ge0$$?

We know that $$\mathcal L\left(X_{t_0},\ldots,X_{t_n}\right)=\mathcal L\left(X_{t_0}\right)\otimes\bigotimes_{i=1}^n\kappa_{t_i-t_{i-1}}\tag2$$ for all $$n\in\mathbb N_0$$ and $$0\le t_0\le\cdots\le t_n$$, where the right-hand side of $$(2)$$ denote the product of transition kernels. Thus, $$\mathcal L(X_t)=\mathcal L(X_0)\kappa_t\;\;\;\text{for all }t\ge0\tag3,$$ where the right-hand side of $$(3)$$ denotes the composition of transition kernels. Moreover, $$\mathcal L\left(\left(X_{s+t}\right)_{t\ge0}\right)$$ is uniquely determined of the finite-dimensional distributions $$(2)$$. So, it intuitively seems to be reasonable that $$(1)$$ holds. But how can we prove it?