We define invariant distribution of (general) Markov process with homogenne time $\left\{X_t\right\}_{t\in T}$ as $$\forall t\in T, \forall A\in B(S), \hspace{0.2cm} \pi(A)=\int\limits_{S}P(X_t\in A|X_0=x)\pi(dx),$$ where B(S) is Borel $\sigma\text{-algebra}$ defined on state space $S$.
I should show that for invariant distribution $\pi$ of Markov process with discrete time is sufficient $$ \forall A\in B(S), \hspace{0.2cm} \pi(A)=\int\limits_{S}P(X_1\in A|X_0=x)\pi(dx).$$
I can show this for Markov chains with discrete time from matrix form of invarinat distribution and I also can do it when it is in summation ( $\sum$ ) form, when I have to use Chapman-Kolmogorov equation.
I have been trying to use Chapman-Kolmogorov equation for Markov process with discrete time, but I have not be successful, yet.
Chapman-Kolmogorv equation is $$\forall t,s\in T,\forall x\in S,\forall A \in B(S): P(X_{s+t}\in A|X_0=x)=\int\limits_{S}P(X_t\in A|X_0=y)P(X_s\in dy|X_0=x).$$
I have been trying to multiply both side of $$\pi(A)=\int\limits_{S}P(X_1\in A|X_0=x)\pi(dx)$$ by some expression of the form $\int\limits_{S}\dots$ and then use CH-K equation. My idea which I had was to have something like that $$\int\limits_{S}P(X_1\in A|X_0=x)\pi(dx)=\int\limits_{S}P(X_2\in A|X_0=x)\pi(dx),$$ what is $$\pi(A)=\int\limits_{S}P(X_2\in A|X_0=x)\pi(dx)$$
And then I will continue. I will multiply both side with $\int\limits_S\dots$ and I will have $$\pi(A)=\int\limits_{S}P(X_3\in A|X_0=x)\pi(dx),$$ etc.
Any help will be appreciated. Thank you very much for any idea.