# $\sigma$-algebra generated by analytic sets

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?

• 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. Commented Jul 18, 2022 at 7:07
• 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) Commented Jul 18, 2022 at 7:10

"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.

• Thank you! I also add the link as a comment as further details. Commented Oct 20, 2023 at 13:17