Let $\{\alpha_i\mid i<\kappa\}$ be some increasing sequence of elements of $\mathbb{H}$ and let $\alpha=\sup \{\alpha_i\mid i<\kappa\}$ Then for any $x,y<\alpha$ it must be that $x<\alpha_i$ and $y<\alpha_j$ for some $i,j<\kappa$ But then $x+y<\alpha_{\max\{i,j\}}<\alpha$

Consider any $\alpha$ and define a sequence by $\alpha_0=S\alpha$ and $\alpha_{n+1}=\alpha_n+\alpha_n$ Let $\alpha_\omega=\sup_{n<\omega}\alpha_n$ be the limit of this sequence. If $x,y<\alpha_\omega$ then it must be that $x<\alpha_i$ and $y<\alpha_j$ for some $i,j<\omega$ and therefore $x+y<\alpha_{\max\{i,j\}+1}$ Note that $\alpha_\omega$ is, in fact, the next element of $\mathbb{H}$ since every element in the sequence is clearly additively decomposable.

Since $0$ is not in $\mathbb{H}$ we have $f_\mathbb{H}(0)=1$

For any $\alpha+1$ we have $f_\mathbb{H}(\alpha+1)$ is the least additively indecomposable number greater than $f_\mathbb{H}(\alpha)$ Let $\alpha_0=Sf_\mathbb{H}(\alpha)$ and $\alpha_{n+1}=\alpha_n+\alpha_n=\alpha_n\cdot 2$ Then $f_\mathbb{H}(\alpha+1)=\sup_{n<\omega} \alpha_n=\sup_{n<\omega}S\alpha\cdot 2\cdots 2=f_\mathbb{H}(\alpha)\cdot\omega$ The limit case is trivial since $\mathbb{H}$ is closed and unbounded, so $f_\mathbb{H}$ is continuous.