Given a perfect tree $T$ on $2^{<\omega}$ viewed as a function from $2^{<\omega}$ to $2^{<\omega}$ define the measure of a subset of $[T]$ to be the measure of it's preimage under the usual measure on $2^{\omega}$.
Suppose $X \subset 2^{\omega}$ is a $\Delta^1_1$ class (i.e. lightface Borel) and for every hyperarithmetic perfect tree $T$ $X \cap [T]$ has measure $0$ as a subset of $T$ (i.e., it's preimage has measure 0). Is $X$ necessarily countable? (this is true if we replace hyperarithmetic with computable in $\mathcal{O}$ -- see below).
More generally, for a pointclass $\Gamma \subset P(2^{\omega})$ (e.g. the $\Sigma^0_n$ classes in $2^\omega$) and a class of sets $\Lambda \subset P(\omega)$ (e.g.the $\Sigma^0_n$ subsets of $\omega$) say that $\Gamma$ has the measure 0 perfection property relative to $\Lambda$ if for all $X \in \Gamma$ if $X \cap [T]$ has measure $0$ relative to $T$ for all perfect trees $T \in \Lambda$ then $X$ is countable.
For what nice pairs $\Gamma, \Lambda$ does the measure 0 perfection property hold?
It's clearly true that the $\Pi^0_1$ pointclass has the measure 0 perfection property with respect to the $\Delta^0_2$ sets (in the cantor space every $\Pi^0_1$ class is equal to $[T]$ for some $\Delta^0_2$ $T$). Do the $\Pi^0_2$ classes have it with respect to the $\Delta^0_3$ sets? Arithmetic sets? What about $\Pi^0_\omega$ classes?
Note that this notion isn't totally trivial since the $\Sigma^1_1$ pointclass has the measure 0 perfection property with respect to the sets computable from Kleene's O since if $X$ is $\Sigma^1_1$ then $X$ contains a perfect set iff it contains $[T]$ for a perfect tree $T$ computable in $O$ and as every boldface $\Sigma^1_1$ set has the perfect set property it follows that if $X$ is uncountable then there is a perfect tree $T$ computable in O with $[T] \subset X$.
