Let $c \in {^\omega\omega}$ be a real. We say that $c$ is a $G_\delta$ code iff $c(0) \gt 1$. And we let $A_c = \bigcap_n(\bigcup\{O_{s_i}: c(2^n3^{i+1}) = 1\})$, where the $s_i$'s are a fixed recursive enumeration of $^{\lt\omega}\omega$ and $O_{s_i} = \{ x \in {^\omega\omega} : s_i \subset x \}$. Now there is this statement in the proof of theorem $14.1$ of Kanamori's "The Higher Infinite" which I can't figure out:
If $C \subseteq {^2(^\omega\omega)}$ is given by $$C(x, c) \leftrightarrow c \text{ is a } G_\delta \text{ code for a null set } \wedge x \in A_c $$ then it is straightforward to check that $C$ is arithmetical.
My thoughts:
The "$x \in A_c$" part is arithmetical since we can check it using $\forall^0 n \exists^0 i(c(2^n3^{i+1}) = 1 \wedge x \in O_{s_i})$, which is clearly arithmetical. The "$c \text{ is a } G_\delta \text{ code}$" part is obvious. But I am struggling to put "$A_c \text{ is null}$" into arithmetical terms. One idea that popped into my mind was using the fact that if $B = \bigcap_n O_n$ is $G_\delta$ then $m_L(B) = \inf\{m_L(\bigcap^K_{n=0}O_n) : K \in \omega\}$. But if I were to use this then it would be complicated to translate the codes to the $O_n$'s and even then I would have to somehow define $\frac{1}{n}$ in arithmetic so I could use a formula like: $\forall^0 N \exists^0 i(m_L(\bigcap^i_{n=0}O_n) \lt \frac{1}{N+1})$ and I would also need to see if $m_L(X)$ is arithmetical and that would ultimately amount to translating measure theory and real analysis into second order arithmetic. I know that this must be easier than I think it is and I would be really glad if someone could help me out.