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Let $I$ be a Borel generated $\sigma$-ideal on the Baire space. We say that this ideal is $\Sigma^1_2$ if $$\{c \in \omega^\omega\ |\ c\text{ is a Borel code and }B_c \in I\} \in \Sigma^1_2.$$ Where $B_c$ denotes the Borel set decoded from $c$.

In Ikegami's PhD dissertation there is a sufficient condition for an ideal to be $\Sigma^1_2$, namely, proving that an ideal is $\Sigma^1_2$ on $\Sigma^1_1$ (i.e., for any analytic set $A \subseteq 2^\omega \times \omega^\omega$, the set $\{x\ |\ A_x \in I\}$ is $\Sigma^1_2$, where $A_x$ stands for vertical sections).

It seems that it can be more difficult to compute an ideal's complexity than one may think. I found some references on the complexity of completely Ramsey null ideal (see Sabok) and Silver's ideal. Are there any hints or references for computing directly (i.e., without appealing to $\Sigma^1_2$ on $\Sigma^1_1$ condition) the complexities for the ideals of Lebesgue null, meager, countable, $\sigma$-bounded and Laver null sets? I just need the upper bound to be $\Sigma^1_2$.

Martin Sleziak
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user1612
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1 Answers1

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You can directly compute the complexity of the ideals to be $\Sigma^1_2$, e.g., in the case of Lebesgue null ideal, $B_c$ is Lebesgue null if and only if $c$ is a Borel code and there is a sequence $(T_n \mid n \in \omega)$ of trees on $\omega$ such that (i) each tree $T_n$ gives us the closed set $[T_n]$ of Lebesgue measure zero, and (ii) for every real $x$, if $x$ is in $B_c$ then there is an $n \in \omega$ such that $x$ is in $[T_n]$. The assertion (i) is arithmetical in $(T_n \mid n \in \omega)$ and the assertion (ii) is $\Pi^1_1$ in $c$ and $(T_n \mid n \in \omega)$. All in all, the whole statement is $\Sigma^1_2$. Similar calculations will give you the same result to the other ideals listed in your question.

Daisuke Ikegami
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