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?

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 Lebesgue density theorem with 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.

3. In the case $f$ is unbounded and fractal, we could use this paper, which applies a Henstock-Kurzweil type integral (i.e., $\mu$-HK integral) on a measure Metric Space. This coincides with unbounded functions with finite improper Riemman integrals, including the bounded functions with finite Lebesgue integrals, bounded function with finite integrals w.r.t the Hausdorff measure, or functions with finite Henstock-Kurzweil integrals.

4. In the case that $A$ is non-fractal (e.g. countably infinite) where 1., 2., and 3. does not return a unique, finite expected values, we could use conditional expectation

The main issue though is, even if set $M^{*}$ is the set of all measurable functions in $\mathbb{R}^{A}$, neither 1., 2., 3. or 4. give a positive, finite expected value for all $f$ in non-shy subset of $M^{*}$: (i.e., prevalent or neither prevelant nor shy subset of $M^{*}$). Infact, either definitions might give a positive and finite expected value instead for all $f$ in only a shy subset of $M^{*}$.

2. Question

Is there a natural extension of 1., 2., 3., and 4. that gives a positive and finite expected value for all $f$ in a non-shy subset of $M^{*}$?

Does this paper already answer the question?

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 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 prevelant subset of $\mathbb{R}^{A}$?

Does this paper already answer the question?

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?

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 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 prevelant subset of $\mathbb{R}^{A}$?

Does this paper already answer this?

In case the term "natural" is unclear, see the attempt in the answer (once posted) for what a solution can look like.

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 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 prevelant subset of $\mathbb{R}^{A}$?

Does this paper already answer the question?