Why non-uniform learnability does not imply PAC learnability?
The following answer is based on chapter 6/7 of the book »Understanding Machine Learning: From Theory to Algorithms«, by Shalev-Shwartz and Ben-David (especially Example 7.1).

It states that the class $\mathcal{H}$ of all polynomial classifiers over $\mathbb{R}$ is not PAC learnable ($\mathrm{VCdim}(\mathcal{H}) = \infty$). We might rewrite $\mathcal{H}$ as $\bigcup_{n \in \mathbb{N}} \mathcal{H}_n$, where for every $n \in \mathbb{N}$, $\mathcal{H}_n$ is the class of all polynomial classifiers over $\mathbb{R}$ of degree $n$.

Each $\mathcal{H}_n$ is PAC learnable with $\mathrm{VCdim}(\mathcal{H}_n) = n + 1$, so using the fundamental theorem of statistical learning (quantified version), we know that the sample complexity of each $h \in \mathcal{H}_n$ is in $\Theta\left(\frac{(d + 1) + \log(1 / \delta)}{\epsilon^2}\right)$, so strictly increasing in $d$, but finite.

So for $0&lt; \epsilon, \delta &lt; 1$, the set $\{ n_{\mathcal{H}}(\epsilon, \delta, h) \mid h \in \mathcal{H} \}$ is not upper bounded. Yet, as a union of PAC learnable classes, $\mathcal{H}$ is nonuniform learnable.