Let $\mu_1$ and $\mu_2$ be two probabilities on $\mathbb{N}^*$. We suppose that $\forall r \in \mathbb{N}^*, \mu_1(\mathbb{N}^* r) = \mu_2(\mathbb{N}^* r)$, where $\mathbb{N}^* r$ denotes the set of strictly positive multiples of $r$. We recall that $(p_i)_{i \in \mathbb{N}^*}$ denotes the sequence of prime numbers, ordered in increasing order. Show that for all $r \in \mathbb{N}^*$ and all integer $n \geqslant 1$: $$\bigcup_{i=1}^{n+1} \mathbb{N}^* r p_i = \left(\bigcup_{i=1}^{n} \mathbb{N}^* r p_i\right) \cup \left(\mathbb{N}^* r p_{n+1} \backslash \bigcup_{i=1}^{n} \mathbb{N}^* r p_{n+1} p_i\right)$$
Let $\mu_1$ and $\mu_2$ be two probabilities on $\mathbb{N}^*$. We suppose that $\forall r \in \mathbb{N}^*, \mu_1(\mathbb{N}^* r) = \mu_2(\mathbb{N}^* r)$, where $\mathbb{N}^* r$ denotes the set of strictly positive multiples of $r$. We recall that $(p_i)_{i \in \mathbb{N}^*}$ denotes the sequence of prime numbers, ordered in increasing order. Show that for all $r \in \mathbb{N}^*$ and all integer $n \geqslant 1$:
$$\bigcup_{i=1}^{n+1} \mathbb{N}^* r p_i = \left(\bigcup_{i=1}^{n} \mathbb{N}^* r p_i\right) \cup \left(\mathbb{N}^* r p_{n+1} \backslash \bigcup_{i=1}^{n} \mathbb{N}^* r p_{n+1} p_i\right)$$