Show that if $\mathcal{H}$ is PAC learnable in the standard one-oracle model, then $\mathcal{H}$ is PAC learnable in the two-oracle model This is a question $9.1$ from Understanding Machine Learning Chapter 3. It goes like this:

Consider a variant of the PAC model in which there are two example oracles: one that generates positive examples and one that generates negative examples, both according to the underlying distribution $\mathcal{D}$ on $\mathcal{X}$. Formally, given a target function $f : \mathcal{X} \to {0,1}$, let $\mathcal{D}^+$ be the distribution over $\mathcal{X}^+ = \{x \in \mathcal{X}: f(x) = 1\}$ defined by $\mathcal{D}^+(A) = \frac{\mathcal{D}(A)}{\mathcal{D}(X^+)}$, for every $A \subset \mathcal{X}$. Similary $\mathcal{D}^-$ is the distribution over $\mathcal{X}^{-}$ induced by $\mathcal{D}$.


The definition of PAC learnability in the two-oracle model is the same as the
standard definition of PAC learnability except that here the learner has access
to $m^{+}_{\mathcal{H}}(\epsilon, \delta)$ i.i.d. examples from $\mathcal{D}^+$ and $m^{-}_{\mathcal{H}}(\epsilon, \delta)$ i.i.d. examples from $\mathcal{D}^{-}$. The learner’s goal is to output $h$ s.t. with probability at least $1-\delta$ (over the choice of the two training sets, and possibly over the nondeterministic decisions made by the learning algorithm), both $L_{(\mathcal{D}^+,f)}(h) \leq \epsilon$ and $L_{(\mathcal{D}^−,f)}(h) \leq \epsilon$

I am trying to prove that if $\mathcal{H}$ is PAC learnable in the standard one-oracle model, then $\mathcal{H}$ is PAC learnable in the two-oracle model. My attempt so far:
Note that $$L_{(D,f)}(h) = \mathcal{D}(\mathcal{X}^+)L_{(\mathcal{D}^+,f)}(h) + \mathcal{D}(\mathcal{X^{-}})L_{(\mathcal{D}^-,f)}(h).$$
Let $d = min \{ \mathcal{D^+}, \mathcal{D^-}\}$, then if $m\geq m_\mathcal{H}(\epsilon d, \delta)$, then it is clear that:
$$\mathbb{P}[L_{(D,f)}(h)\leq \epsilon d] \geq 1-\delta \implies \mathbb{P}[L_{(D^+,f)}(h)\leq \epsilon] \geq 1-\delta$$
And,
$$\mathbb{P}[L_{(D,f)}(h)\leq \epsilon d] \geq 1-\delta \implies \mathbb{P}[L_{(D^-,f)}(h)\leq \epsilon] \geq 1-\delta$$
So we know that if we have $m\geq m_{\mathcal{H}}(\epsilon d, \delta)$ samples drawn iid from $\mathcal{D}$, then we can guarantee $\mathbb{P}[L_{(D^+,f)}(h)\leq \epsilon] \geq 1-\delta$ and   $\mathbb{P}[L_{(D^-,f)}(h)\leq \epsilon] \geq 1-\delta$.
How do I choose $m_{\mathcal{H}}^+(\epsilon, \delta)$ and $m_{\mathcal{H}}^-(\epsilon, \delta)$ such that if we have $m^+ \geq m_{\mathcal{H}}^-(\epsilon, \delta)$ samples iid according to $\mathcal{D}^+$ and $m^- \geq m_{\mathcal{H}}^-(\epsilon, \delta)$ drawn iid according to $\mathcal{D}^{-}$, then we can guarantee $\mathbb{P}[L_{(D^+,f)}(h)\leq \epsilon] \geq 1-\delta$ and   $\mathbb{P}[L_{(D^-,f)}(h)\leq \epsilon] \geq 1-\delta$?
When is drawing $m^+$ samples according to $\mathcal{D}^+$ and $m^{-}$ samples according to $\mathcal{D}^-$ the same as drawing
$(m^+ + m^-)$ samples according to $\mathcal{D}$?
Recall the following lemma. Let $X,Y$ be two random variables satisfying $\Pr[X\le\epsilon]\ge 1-\delta$ and $\Pr[Y\le\epsilon]\ge 1-\delta$ then $\Pr[X\le \epsilon \land Y\le \epsilon]\ge 1-2\delta$.
Suppose $\mathcal{H}$ is PAC learnable (realizable and proper). Let $m(\epsilon,\delta)$ be the guaranteed PAC sample complexity. Given $m(\epsilon,\delta/2)$ samples from $D^+$ and $m(\epsilon,\delta/2)$ samples from $D^-$, find $h_{ERM}$ that is consistent with all samples (both positive and negative). Since $\mathcal{H}$ is PAC learnable, we know that
$\Pr[L_{D^+}(h_{ERM})\le\epsilon], \hspace{1mm} \Pr[L_{D^-}(h_{ERM})\le\epsilon]\ge 1-\delta/2$. Note that we strongly used the fact that the PAC bound is achieved by any sample consistent hypothesis. Now Use the above lemma to show that the overall empirical risk minimizer is a 2-oracle PAC learner for $\mathcal{H}$.