[proofplan] We prove invariance by testing the one-step transition measure against an arbitrary measurable set $A \in \mathcal E$. Reversibility is applied with the first set equal to the whole state space $E$ and the second set equal to $A$. The Markov kernel normalization $P(x,E)=1$ then reduces the resulting integral over $A$ to $\pi(A)$. [/proofplan] [step:Apply detailed balance with the whole state space as one argument] Fix an arbitrary set $A \in \mathcal E$. Since $P$ is a Markov kernel, the map \begin{align*} x \mapsto P(x,A) \end{align*} from $E$ to $[0,1]$ is $\mathcal E$-measurable, so the integral $\int_E P(x,A)\,d\pi(x)$ is well-defined. Apply reversibility with the measurable sets $E$ and $A$. Since $E,A \in \mathcal E$, the detailed balance identity gives \begin{align*} \int_E P(x,A)\, d\pi(x) = \int_A P(x,E)\, d\pi(x). \end{align*} [guided] We fix an arbitrary measurable set $A \in \mathcal E$ because invariance of $\pi$ must be proved set-by-set: the target identity is \begin{align*} \pi(A) = \int_E P(x,A)\, d\pi(x). \end{align*} The expression on the right is meaningful because $P$ is a Markov kernel. For each fixed $A \in \mathcal E$, the function \begin{align*} E &\to [0,1] \end{align*} \begin{align*} x &\mapsto P(x,A) \end{align*} is $\mathcal E$-measurable. Since it is also bounded between $0$ and $1$, it is integrable with respect to the probability measure $\pi$. Now we use the hypothesis that $P$ is reversible with respect to $\pi$. Reversibility says that for every pair of measurable sets $B,C \in \mathcal E$, \begin{align*} \int_B P(x,C)\, d\pi(x) = \int_C P(x,B)\, d\pi(x). \end{align*} We choose $B=E$ and $C=A$. Both are elements of $\mathcal E$, so the hypothesis applies and yields \begin{align*} \int_E P(x,A)\, d\pi(x) = \int_A P(x,E)\, d\pi(x). \end{align*} This is the key use of detailed balance: it turns the total incoming mass into $A$ into an integral over $A$ itself. [/guided] [/step] [step:Use the Markov kernel normalization to identify the integral with $\pi(A)$] Because $P$ is a Markov kernel, for every $x \in E$ the set function $B \mapsto P(x,B)$ is a probability measure on $(E,\mathcal E)$. Hence \begin{align*} P(x,E)=1 \end{align*} for every $x \in E$. Therefore \begin{align*} \int_A P(x,E)\, d\pi(x) = \int_A 1\, d\pi(x). \end{align*} By the defining property of integration of the constant function $1$ over a measurable set, \begin{align*} \int_A 1\, d\pi(x) = \pi(A). \end{align*} Combining this identity with the detailed balance identity from the previous step gives \begin{align*} \int_E P(x,A)\, d\pi(x) = \pi(A). \end{align*} Since $A \in \mathcal E$ was arbitrary, this is exactly the invariance of $\pi$ for $P$. [/step]