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 the following; however, when $A$ is non-fractal (e.g. countably infinite), we have that $\mathbb{E}[f]$ is undefined.

2. Solution for Bounded $f$

Thus, we want to define a sequence of sets $\left\{F_r\right\}_{r\in\mathbb{N}}$ where if:

• $\liminf\limits_{r\to\infty} F_r=\bigcup\limits_{r\ge 1}\bigcap\limits_{q\ge r}F_q$

• $\limsup\limits_{r\to\infty} F_r=\bigcap\limits_{r\ge 1}\bigcup\limits_{q\ge r}F_q$

we want:

1. $\liminf\limits_{r\to\infty} F_r=\limsup\limits_{r\to\infty} F_r=A$ which I'll desribe as $F_r\overset{r\in\mathbb{N}}{\rightrightarrows} A$ or "a sequence of sets converging to $A$"
2. For all $r\in\mathbb{N}$,
   • $0<{H}^{\dim_{\text{H}}(A)}(F_r)<+\infty$
   • $0<{H}^{h}(F_r)<+\infty$

If criteria 1. and 2. are satisfied, the new finite expected value of $f$ w.r.t $F_r$ or $\mathbb{E}^{*}\left[f,F_r\right]$ is:

\begin{align} & \forall(\epsilon>0)\exists(N\in\mathbb{N})\forall(r\in\mathbb{N})\left(r\ge N\Rightarrow\left|\frac{1}{{H}^{h}\left(F_r\right)}\int_{F_r}f\, d{H}^{h}-\mathbb{E}^{*}[f,F_r]\right|< \epsilon\right) \\ \end{align}

when $f$ is bounded.

3. Equivalent and Non-Equivalent Sequence of Sets

Note we can determine whether $\left\{F_r\right\}_{r\in\mathbb{N}}$ is equivelant to another sequence of sets converging to $A$ (e.g. $\{F_j^{\prime}\}_{j\in\mathbb{N}}$), if for all $f\in\mathbb{R}^{A}$ we have $\mathbb{E}^{*}[f,F_r]$ or $\mathbb{E}^{*}[f,F_j^{\prime}]$ exists such that:

$$\mathbb{E}^{*}[f,F_r]=\mathbb{E}^{*}[f,F_j^{\prime}]$$

If however, there exists an $f\in\mathbb{R}^{A}$ where:

$$\mathbb{E}^{*}[f,F_r]\neq\mathbb{E}^{*}[f,F_j^{\prime}]$$

then the sequence of sets is non-equivalent.

4. Solution for Unbounded $f$

The problem is when $f$ is unbounded, no matter what $F_r$ is chosen, we might have $\mathbb{E}^{*}[f,F_r]$ is undefined. For instance, if $f$ was continuous on set $A=\mathbb{R}\setminus\left\{0\right\}$ where $f(x)=1/x$, then the expected value of $f$ would still be undefined.

To solve this, consider the following:

If the image of $f$ under $A$ is $f[A]:=\left\{f(x):x\in A\right\}$ and the pre-image under $f$ of $F_{r}$ is $f^{-1}\left[F_r\right]:=\left\{x\in A: f(x)\in F_r\right\}$, we want:

$$F_{r}^{\dagger}\overset{r\in\mathbb{N}}{\rightrightarrows} A\times f[A]$$

where $\dagger$ symbolizes that a sequence of sets converge to $A\times f[A]$, e.g. $F_{j}^{\dagger \dagger}$ or $F_{k}^{\dagger \dagger \dagger}$ must converge to $A\times f[A]$

4.1. Generalized Expected Value: Thus, the generalized expected value of $f$ (with respect to $F_r$) is $\mathbb{E}^{**}[f,F_{r}^{\dagger}]$, when it exists where:

\begin{align} & \forall(\epsilon>0)\exists(N\in\mathbb{N})\forall(r\in\mathbb{N})\left(r\ge N\Rightarrow\left|\frac{1}{{H}^{h}\left(f^{-1}[F_r]\right)}\int_{f^{-1}[F_r]}f\, d{H}^{h}-\mathbb{E}^{**}[f,F_r^{\dagger}]\right|< \epsilon\right) \\ \end{align}

5. Choosing a specific sequence of sets

The problem is "most" unbounded $f$ have multiple $F_r^{\dagger}$ that give different $\mathbb{E}^{**}[f,F_{r}^{\dagger}]$. Therefore, we should choose a set of equivalent $F_{r}^{\dagger}$ using a choice function.

Here is what I considered:

5.1 Question

Does there exist a choice function which chooses a unique set (of equivalent sequences of sets) that converge to $A\times f[A]$, such that:

1. The sequence of sets converge to $A\times f[A]$ at a linear or super-linear rate compared to the rate non-equivelant sequences of sets converge to $A\times f[A]$

2. The generalized expected value of $f$ w.r.t the chosen (and equivalent) sequences of sets is positive and finite.

3. The choice function chooses a unique set of equivalent sequences of sets which satisfy 1. and 2., for all $f\in Q$, where set $Q\subseteq\mathbb{R}^{A}$ with $Q$ prevalent in $\mathbb{R}^{A}$. (This means the choice function chooses a unique and equivalent sequence of sets, which satisfy 1. and 2. for "almost all" $f\in\mathbb{R}^{A}$).

4. Out of all the choice functions which satisfy 1., 2. and 3., we choose the one with the simplest form, meaning for each choice function fully expanded, we take the one with the fewest variables/numbers (excluding those with quantifiers).

5.2 Notes on Question

If the solution is extraneous, what other criteria can we add to get a unique choice function? (Note if the solution is always extraneous, make the "equivelant sequences of sets", a sequence of sets that w.r.t $f$ that has the same generalized expected value.)

See my attempt in the answer (once posted) for what a solution can look like.