$\sigma$-algebra generated by analytic sets

Question

The Borel $\sigma$-algebra $\cal B$ on real numbers has many good properties. For instance, all continuous functions are $\cal B/\cal B$-measurable. On the other side, not only $\cal B$ is not complete but, as discovered by Suslin, $\cal B$ is not closed under images of continuous functions.

It is tempting then to consider the $\sigma$-algebra, say ${\cal B}^1$, generated by analytic sets, or even, going up along the projective hierachy, to consider the $\sigma$-algebra ${\cal B}^{\infty}$ generated by projective sets.

It would by nice to know if these $\sigma$-algebra share some properties with the Borel one that make $\cal B$ useful.

Question I: Are continuous functions ${\cal B}^1/ {\cal B}^1$-measurable?

Answer 1

"Inverse image of an analytic set is analytic" would imply ${\cal B}^1/ {\cal B}^1$-measurable, so let's try that. [Check my argument.]

Let $f : \mathbb R \to \mathbb R$ be continuous. Let $B \subseteq \mathbb R$ be an analytic set. Write $A = f^{-1}(B)$. We want to show that $A$ is analytic.

Let $\pi_1, \pi_2: \mathbb R^2 \to \mathbb R$ be the projections onto the coordinates: $\pi_1(x,y) = x, \pi_2(x,y) = y$. There is a Borel set $C \subseteq \mathbb R^2$ such that $\pi_1(C) = B$.

Define $F : \mathbb R^2 \to \mathbb R^2$ by $F(x,y) = (f(x),y)$. So $F$ is continuous. Let $D = F^{-1}(C)$. So $D$ is a Borel set.

\begin{align} x \in \pi_1(D) &\Longleftrightarrow \exists y, (x,y) \in D \\&\Longleftrightarrow \exists y, (x,y) \in F^{-1}(C) \\&\Longleftrightarrow \exists y, F(x,y) \in C \\&\Longleftrightarrow \exists y, (f(x),y) \in C \\&\Longleftrightarrow \exists y, \pi_1(f(x),y) \in B \\&\Longleftrightarrow f(x) \in B \\&\Longleftrightarrow x \in f^{-1}(B) \\&\Longleftrightarrow x \in A \end{align}

That is, $A = \pi_1(D)$. So $A$ is an analytic set.