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 $A$.

1. Motivation

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 Hausdorff measure using Hyperbolic Cantor sets; however, when $A$ is non-fractal (e.g. countably infinite), we have $\mathbb{E}[f]$ is undefined.

The main issue though is neither 1. or 2. give a positive, finite expected value for all $f$ in a prevalent subset of $\mathbb{R}^{A}$. Infact, either definitions might instead give a positive and finite expected value for all $f$ in only a shy subset of $\mathbb{R}^{A}$

2. Question

Is there a natural extension of 1. or 2. that gives a positive and finite expected value for all $f$ in a prevalent subset of $\mathbb{R}^{A}$?

Does this paper already answer the question?