This is a follow up to this post, where I wish to verify whether one of the statements (in the post) is true but first let's recap the definitions:

Let $(X,d)$ be a metric space. If set $A\subseteq X$, let $H^{\alpha}$ be the $\alpha$-dimensional Hausdorff measure on $A$, where $\alpha\in[0,+\infty)$ and $\text{dim}_{\text{H}}(A)$ is the Hausdorff dimension of set $A$.

If $n\in\mathbb{N}$, where set $A\subseteq\mathbb{R}^{n}$ and the expected value of $f:A\to\mathbb{R}$ is

$$\mathbb{E}[f]=\frac{1}{{H}^{\text{dim}_{\text{H}}(A)}(A)}\int_{A}f \, dH^{\text{dim}_{\text{H}}(A)}$$

we can see there are cases where $\mathbb{E}[f]$ is undefined or infinite (e.g. ${H}^{\text{dim}_{\text{H}}(A)}(A)$ is zero, $+\infty$ or $f$ is unbounded).

One solution to getting a finite expected value is

1. Defining a dimension function; i.e., $h:[0,+\infty)\to[0,+\infty]$, that's monotonically increasing, strictly positive and right continuous such that, when $R$ denotes the radius of a ball in a covering for the definition of the Hausdorff Measure, we replace $R^{\text{dim}_{\text{H}}(A)}$ with $h(R)$, so $H^{h}(A)$ is positive and finite.

Note, however, not all $A$ has a dimension function which leads to:

2. If $A$ is fractal but has no gauge function, we could use this paper, which is an extension of the Lebesgue density theorem, and this paper, which is an extension of the Hausdorff measure using Hyperbolic Cantor sets; however, when $A$ is non-fractal (e.g. countably infinite), we have $\mathbb{E}[f]$ is undefined.

Question:

If the set $M^{*}$ is the set of all measurable functions in $\smash{\mathbb{R}^{A}}$, are extensions 1. and 2. finite for all $f$ in only a shy subset of $M^{*}$?

Note, I will allow partial answers (e.g. answering the question for only extension 1.)