Suppose $n\in\mathbb{N}$ and $f:A\subseteq\mathbb{R}^{n}\to\mathbb{R}$ is a function, where $A$ and $f$ are Borel.

Question: Assuming $\mathrm{S}(n)=\left\{\mathfrak{F}_1:\mathfrak{F}_1\in\mathbb{R}^\mathrm{K},\mathrm{K}\subseteq\mathbb{R}^{n} \text{ and $\mathfrak{F}_1$ are Borel}\right\}$ is the set of all $f$, what is an example of a non-trivial, rigorous definition of a "full measure" subset of $\mathrm{S}(n)$ which can be used to prove the following? For each $n\in\mathbb{N}$:

If $F\subset\mathrm{S}(n)$ is the set of all $\dot{\mathbf{f}}\in\mathrm{S}(n)$, where $\mathcal{D}:=\mathrm{dom}(\dot{\mathbf{f}})$ there exists $(f_r,A_r),(g_v,B_v)\to(\dot{\mathbf{f}},\mathcal{D})$ (Definition 2.1) such that $m_{\dot{\mathbf{f}},\mathcal{D}}((f_r),(A_r);n)\neq m_{\dot{\mathbf{f}},\mathcal{D}}((g_v),(B_v);n)$ (Definition 2.2), then $F$ is a "full measure" subset of $\mathrm{S}(n)$.

This means the set of all $f$, where two or more sequences of bounded functions converging to $f$ have non-equivelant expected values (Definition $\S$2.2), is a "full measure" subset of all $f$. In other terms, the "new mean" (Definition $\S$2.2) of "almost all" $f$ are non-unique.

Note: See the attempt in Section $\S$3.

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$\S$2. Definitions:

Definition $\S$2.1 (Sequence of Bounded Functions With Different Domains Converging to $f$)

The sequence of functions $(f_r)_{r\in\mathbb{N}}$, where $(A_r)_{r\in\mathbb{N}}$ is a sequence of bounded sets and $f_{r}:A_r\to\mathbb{R}$ is a bounded function, converges to $f:A\subseteq\mathbb{R}^{n}\to\mathbb{R}$ (i.e., $A$ and $f$ are Borel) when:

For any $(x_1,\cdots,x_n)\in A$, there exists an $n$-dimensional sequence $(s_1(r),\cdots,s_n(r))\in A_r$ such that $(s_1(r),\cdots,s_n(r))\to (x_1,\cdots,x_n)$ and $f_r(s_1(r),\cdots,s_n(r))\to f(x_1,\cdots,x_n)$.

This is equivalent to:

$${(f_r,A_r)\to(f,A)}$$

Definition $\S$2.2 (Mean of a Sequence of Bounded Functions With Different Domains Converging to $f$)

Hence, suppose:

• $(f_r,A_r)\to(f,A)$ (Definition 2.1)
• $|\cdot|$ is the absolute value
• $\dim_{\text{H}}(\cdot)$ is the Hausdorff dimension
• $\mathcal{H}^{\dim_{\text{H}}(\cdot)}(\cdot)$ is the Hausdorff measure in its dimension on the Borel $\sigma$-algebra
• the integral of $f$ is defined w.r.t the Hausdorff measure in its dimension

The mean of $(f_r)_{r\in\mathbb{N}}$ is a real number $m_{f,A}((f_r),(A_r);n)$, when the following is true:

$$\scriptsize{\begin{align}&\forall(\epsilon>0)\exists(N\in\mathbb{N})\forall(r\in\mathbb{N})\left(r\ge N\Rightarrow\left|\frac{1}{{\mathcal{H}}^{\text{dim}_{\text{H}}(A_{r})}(A_{r})}\int_{A_{r}}f_{r}\, d{\mathcal{H}}^{\text{dim}_{\text{H}}(A_{r})}-m_{f,A}((f_r),(A_r);n)\right|< \epsilon\right) \end{align}}$$

when no such $m_{f,A}((f_r),(A_r);n)$ exists, $m_{f,A}((f_r),(A_r);n)$ is infinite or undefined. (If the graph of $f$ has zero Hausdorff measure in its dimension, replace $\mathcal{H}^{\text{dim}_{\text{H}}(A_{r})}$ with the generalized Hausdorff measure ${\mathscr{H}^{\phi_{h,g}^{\mu}(q,t)}}$.)

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$\S$3. Attempt

We want a "full measure" or "measure zero" set $F\subset \mathrm{S}(n)$ to be similar to a prevalent or shy subset of $\mathrm{S}(n)$; however, $\mathrm{S}(n)$ is not a vector space. Despite this, we can use the cardinality of a set $|\cdot|$ such that:

1. If $F\subset \mathrm{S}(n)$ is a set and $|\mathrm{S}(n)\setminus F|<|\mathrm{S}(n)|$, then $F$ is a "full measure" subset of $\mathrm{S}(n)$
2. If $F\subset \mathrm{S}(n)$ is a set and $|F|<|\mathrm{S}(n)|$, then $F$ is a "measure zero" subset of $\mathrm{S}(n)$

Still, when Borel $A\subseteq\mathbb{R}$ is a fixed set (i.e., $A=\mathbb{R}$) and $\mathrm{S}(n)\mapsto \mathbb{R}^{A}$, the quote is true when $F$ is prevalent.

If $F\subset\mathbb{R}^{A}$ is the set of all $f\in\mathbb{R}^{A}$, where there exists $(f_r,A_r),(g_v,B_v)\to(f,A)$ (Definition 2.1) such that $m_{f,A}((f_r),(A_r);n)\neq m_{f,A}((g_v),(B_v);n)$ (Definition 2.2) is finite, then $F$ is a prevalent subset of $\mathbb{R}^{A}$.

On the other hand, if we used the cardinality of a set, then 1. and 2. are false in order for the quote to be true and $F$ is neither a "measure zero" nor "full measure" subset of $\mathbb{R}^{A}$.

Sub-question: Can we define a "full measure" or "measure zero" subset of $\mathrm{S}(n)$ similar to a prevalent or shy set? (The measures should be the same as the ones answering the sub-questions here and here.) ${}{}{}{}{}$