# Equivalent Formulations of Thompson Sampling

I am studying Chapter 36 Thompson Sampling of the book Bandit Algorithms by Lattimore and Szepesvari. The authors present two equivalent formulations of Thompson Sampling on page 460, and I am having trouble seeing why they are equivalent.

Let $$(\mathcal{E},\mathcal{G},P_0)$$ be a probability space, where $$\mathcal{E}$$ is a collection of bandit environments, $$\mathcal{G}$$ is a $$\sigma$$-algebra over $$\mathcal{E}$$, and $$P_0$$ is the prior distribution. The mean of the $$i$$th arm in bandit $$\nu\in\mathcal{E}$$ is denoted by $$\mu_i(\nu)$$. Let $$A_t$$ denote the action in round $$t$$, and $$X_t$$ the reward in round $$t$$. Let $$\mathcal{F}_t=\sigma(A_1,X_1,...,A_{t-1},X_{t-1})$$. The two formulations are as follows.

1. Select an arm according to the posterior probability that the arm is optimal, that is, Thompson Sampling is the policy $$\pi=(\pi_t)_{t=1}^{\infty}$$ with $$\pi_t(a\mid a_1,x_1,...,a_{t-1},x_{t-1}) = Q(V_a\mid a_1,x_1,...,a_{t-1},x_{t-1}),$$ where $$V_a=\{\nu\in\mathcal{E}:a=\text{argmax}_i\mu_i(\nu)\}$$, and $$Q$$ is the posterior distribution.
2. First sample $$\nu_t\sim Q(\cdot\mid \mathcal{F}_{t-1})$$, then let $$A_t=\text{argmax}_{i\in[k]}\mu_i(\nu_t).$$

Moreover, on page 461, the authors further note that $$A_t$$ satisfies $$\mathbb{P}(A^*=\cdot\mid \mathcal{F}_{t-1}) = \mathbb{P}(A_t=\cdot\mid \mathcal{F}_{t-1}), \tag{*}$$ where the authors define $$A^*$$ by $$A^*=\text{argmax}_{i\in[k]}\mu_i$$. (I think the authors are being sketchy here. A more rigorous definition should be as follows. For each $$\nu\in\mathcal{E}$$, define $$A(\nu)=\text{argmax}_{i\in[k]}\mu_i(\nu)$$. Let $$V$$ be a random variable with law $$P_0$$, and define $$A^*=A(V)$$.)

My question: How to show (using a measure-theoretic argument) that the two formulations are equivalent and that $$(*)$$ holds?