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[no right to comment, so I post this as an answer]

By the strong Markov property, what you seem to imply is eaquivalent to the following: starting at time 0 from the (uniform) stationary distribution restricted on $U$, $\pi(\cdot \cap U)/\pi(U)$, you are distributed according $\pi$ at time 1. This is wrong in general (see the previous comment by Liviu): you may need more time.

(Aperiodic) example: for the 2 regular graph $\mathbb Z/n\mathbb Z$, $n$ even, if you take $U$ to be the set of odd/even numbers, then the probability to stay in U the next step after entering it is 0. On the other hand, if choice $U=\{n/2,...,n-1\}$, then the probability to stay in $U$ the next step after entering it at a strictly positive time is $1/2$, and this coincides with $\pi(U)$. This means the boundary of U matters.

[no right to comment, so I post this as an answer]

By the strong Markov property, what you seem to imply is equivalent to the following: starting at time 0 from the (uniform) stationary distribution restricted on $U$, $\pi(\cdot \cap U)/\pi(U)$, the chain is distributed according $\pi$ at time 1. This is wrong in general (see the previous comment by Liviu): you need more time to reach $\pi$.

(Aperiodic) example: for the 2 regular graph $\mathbb Z/n\mathbb Z$, $n$ even, if you take $U$ to be the set of odd/even numbers, then the probability to stay in U the next step after entering it is 0. On the other hand, if you choose $U=\{n/2,...,n-1\}$, then the probability to stay in $U$ the next step after entering it at a strictly positive time is $1/2$, and this coincides with $\pi(U)$. This means the boundary of U matters.

[no right to comment, so I post this as an answer]

By the strong Markov property, what you seem to believe is the following: starting at time 0 from the (uniform) stationary distribution restricted on $U$, $\pi(\cdot \cap U)/\pi(U)$, you are distributed according $\pi$ at time 1. This is wrong in general (see the previous comment by Liviu): you may need more time.

(Aperiodic) example: for the 2 regular graph $\mathbb Z/n\mathbb Z$, $n$ even, if you take $U$ to be the set of odd/even numbers, then your probability is 0, since you will leave the set immediately after entering it.

[no right to comment, so I post this as an answer]

By the strong Markov property, what you seem to imply is eaquivalent to the following: starting at time 0 from the (uniform) stationary distribution restricted on $U$, $\pi(\cdot \cap U)/\pi(U)$, you are distributed according $\pi$ at time 1. This is wrong in general (see the previous comment by Liviu): you may need more time.

(Aperiodic) example: for the 2 regular graph $\mathbb Z/n\mathbb Z$, $n$ even, if you take $U$ to be the set of odd/even numbers, then the probability to stay in U the next step after entering it is 0. On the other hand, if choice $U=\{n/2,...,n-1\}$, then the probability to stay in $U$ the next step after entering it at a strictly positive time is $1/2$, and this coincides with $\pi(U)$. This means the boundary of U matters.

[no right to comment, so I post this as an answer]

By the strong Markov property, what you seem to believe is the following: starting at time 0 from the (uniform) stationary distribution restricted on $U$, $\pi(\cdot \cap U)/\pi(U)$, you are distributed according $\pi$ at time 1. This is wrong in general (see the previous comment by Liviu): you may need more time.

[no right to comment, so I post this as an answer]

By the strong Markov property, what you seem to believe is the following: starting at time 0 from the (uniform) stationary distribution restricted on $U$, $\pi(\cdot \cap U)/\pi(U)$, you are distributed according $\pi$ at time 1. This is wrong in general (see the previous comment by Liviu): you may need more time.

(Aperiodic) example: for the 2 regular graph $\mathbb Z/n\mathbb Z$, $n$ even, if you take $U$ to be the set of odd/even numbers, then your probability is 0, since you will leave the set immediately after entering it.