Analytic sets are Lebesgue measurable

Question

Analytic subsets of $\mathbb{R}$ are projections of Borel sets in $\mathbb{R}^2$. I'm trying to understand a proof that these sets are always Lebesgue measurable.

One can first prove that analytic sets $(\Sigma_1^1)$ are equal to $\mathcal{A}(\Pi_1^0)$ where $\mathcal{A}$ denotes the Souslin operation, and $\Pi^0_1$ is the pointclass of closed sets. Then, one shows that Lebesgue measurable sets are closed under the Souslin operation. Doesn't it then follow by monotinicity that $\Sigma^1_1 = \mathcal{A}(\Pi_1^0) \subset \mathcal{A}(\{measurable\}) = \{measurable\}$?

Why do some authors mention the additional step that, in light of the idempotence of the Souslin operation, $\mathcal{A}(\Sigma_1^1)=\Sigma_1^1$ ?

See for example, corollary $13.5$ here, or the mention of $4.1.14$ in Theorem $4.3.1$ of Srivastava. "A Course on Borel Sets".

Answer 1

Let me try to summarise the comments to give an answer:

Despite how the proofs in the references are framed, you only need to make two arguments. First you show that any Borel subset of $\mathbb{R}^2$ is the continuous image of the 'universal' Polish space $\mathbb{N}^{\mathbb{N}}$. It then follows that $\Sigma_1^1=\mathcal{A}(\Pi_1^0)$. Second, you use the completeness property of Lebesgue measurable sets to prove that they are closed under the Souslin operation. The result should then follow.

The idempotence of the Souslin operation, while interesting in its own right, is not necessary here. For anyone who's interested, it can be proved by constructing some clever bijections $\mathbb{N}^{\mathbb{N}}\times (\mathbb{N}^{\mathbb{N}})^{\mathbb{N}}\to {\mathbb{N}}^{\mathbb{N}}, \ \mathbb{N}\times\mathbb{N}\to \mathbb{N} $ and $\mathbb{N}^{<\mathbb{N}} \to \mathbb{N}^{<\mathbb{N}}$.