I want to show that for any $\Sigma_3^0$ set of reals (by real I mean an element of Baire space $\omega^{\omega}$) $B$ there is a $\Pi_1^0$ set of reals $A$ such that for all Turing degrees $d$, $B$ contains an element of degree $d$ if and only if $A$ contains an element of degree $d$. The suggested proof goes like this: let $R$ be a recursive set of reals such that for all $\alpha\in\omega^\omega$ we have $$\alpha\in B\iff\exists i\forall j\exists k R(i, j, \bar{\alpha}(k)),$$ and let $$A=\{\langle i, \alpha, \beta\rangle:\forall j (\beta(j) = \mu k R(i, j, \bar{\alpha}(k)))\}.$$
Doing this, I proceed as follows: for the forward direction, suppose $\alpha\in B$, with Turing degree $d$. Then we can take $i_0$ to be such that $\forall j\exists k R(i_0, j, \bar{\alpha}(k))$. Then we may define $\beta\in\omega^\omega$ as $\beta(j) = \mu k R(i_0, j, \bar{\alpha}(k))$ (we know such smallest $k$ always exists since $\alpha\in B$). Then, clearly $\langle i_0, \alpha, \beta\rangle\in A$. Moreover, this is the only element (that I can think of) we can construct in $A$ given $\alpha\in B$.
So showing that $[\alpha]\equiv_T [\langle i_0, \alpha, \beta\rangle]$ would give us the desired result. The $\le_T$ direction is obvious, but the $\ge_T$ direction does not seem to follow. Constructing $\langle i_0,\alpha, \beta\rangle$ in a recursive fashion from $\alpha$ doesn't seem possible, since finding $i_0$ requires an unbounded search as does finding $\beta$. Any help is appreciated.
To show $(i_0, \alpha, \beta) \leq_T \alpha$, it is clearly enough to show that $\beta \leq_T \alpha$ ($i_0$ is just an integer).
Given $j$, compute $\beta(j)$ as follows: For each $n < \omega$, check if $R(i_0, j, \alpha(n))$ holds. These computations can be run because we have access to $\alpha$. We know that for some $n$, $R(i_0, j, \alpha(n))$ does hold. Stop the computation at the least such $n$ and output it as $\beta(j)$.