Let $\mu_1$ and $\mu_2$ be two probabilities on $\mathbb{N}^*$. We assume that $\forall r \in \mathbb{N}^*, \mu_1(\mathbb{N}^* r) = \mu_2(\mathbb{N}^* r)$. We recall that we denote by $(p_i)_{i \in \mathbb{N}^*}$ the sequence of prime numbers, ordered in increasing order. Show that for all $r \in \mathbb{N}^*$ and all integer $n \geqslant 1$: $$\mu_1\left(\mathbb{N}^* r \backslash \bigcup_{i=1}^{n} \mathbb{N}^* r p_i\right) = \mu_2\left(\mathbb{N}^* r \backslash \bigcup_{i=1}^{n} \mathbb{N}^* r p_i\right).$$
Let $\mu_1$ and $\mu_2$ be two probabilities on $\mathbb{N}^*$. We assume that $\forall r \in \mathbb{N}^*, \mu_1(\mathbb{N}^* r) = \mu_2(\mathbb{N}^* r)$.
We recall that we denote by $(p_i)_{i \in \mathbb{N}^*}$ the sequence of prime numbers, ordered in increasing order.
Show that for all $r \in \mathbb{N}^*$ and all integer $n \geqslant 1$:
$$\mu_1\left(\mathbb{N}^* r \backslash \bigcup_{i=1}^{n} \mathbb{N}^* r p_i\right) = \mu_2\left(\mathbb{N}^* r \backslash \bigcup_{i=1}^{n} \mathbb{N}^* r p_i\right).$$