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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?

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    $\begingroup$ Similarly to Gerald's answer for analytic sets, one can show that $\mathbf{\Sigma}^1_n$ sets are closed under continuous (even Borel) preimages, so that continuous functions are $\mathcal B^n/\mathcal B^n$ measurable. $\endgroup$ Jul 18, 2022 at 7:07
  • $\begingroup$ The following result might also be of interest: for a function $f$ between Polish spaces the following are equivalent, where $\Gamma$-measurable for a pointclass $\Gamma$ means that the preimages of open sets are in $\Gamma$: 1. $f$ is $\mathbf{\Delta}^1_n$-measurable 2. $f$ is $\mathbf{\Sigma}^1_n$-measurable 3. $f$ is $\mathbf{\Pi}^1_n$-measurable 4. the graph of $f$ is $\mathbf{\Sigma}^1_n$ 5. the graph of $f$ is $\mathbf{\Delta}^1_n$. (at the $n=1$ level this says that a function is Borel iff its graph is Borel iff its graph is analytic) $\endgroup$ Jul 18, 2022 at 7:10

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"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.

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  • $\begingroup$ Thank you! I also add the link as a comment as further details. $\endgroup$
    – Giafazio
    Oct 20, 2023 at 13:17

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