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 $\mathfrak{F}$ are Borel}\right\}$ is the set of all $f$, what is an example of a non-trivial, rigorous definition of a "measure zero" 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 $\mathcal{D}:=\mathrm{dom}(\dot{\mathbf{f}})$ and $m_{\dot{\mathbf{f}}}(\mathcal{D};n)$ (Definition 2.1) is defined and finite, then $F$ is a "measure zero" subset of $\mathrm{S}(n)$.

This means the set of all $f$ with a finite mean (Definition 2.1) is a "measure zero" subset of the set of all $f$. In other terms, "almost no" $f$ has a finite mean.

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

$\S$2. Definitions

Definition $\S$2.1 (Extended Mean of $f$ w.r.t. The Hausdorff Measure in the Hausdorff dimension of its domain)

Suppose:

• $\dim_{\text{H}}(\cdot)$ 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 the Hausdorff dimension of its domain
• $(A_r)_{r\in\mathbb{N}}$ is a sequence of bounded sets
• $A$ is a Borel subset of $\mathbb{R}^n$
• $\mathbf{B}(A)$ is the set of all sequences of bounded sets with set-theoretic limit $A$

The mean of $f:A\subseteq\mathbb{R}^{n}\to\mathbb{R}$ (i.e., $f$ is Borel), w.r.t. the Hausdorff measure in the Hausdorff dimension of its domain, is $m_{f}(A;n)$ when it exists:

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

The extended mean of $f$ extends the mean, w.r.t. the Hausodorff measure in the Hausdorff dimension of its domain $$\frac{1}{{\mathcal{H}}^{\dim_{\text{H}}(A)}(A)}\int_{A} f {} d{\mathcal{H}}^{\dim_{\text{H}}(A)}$$ for certain cases of unbounded $A$. In simpler terms, the extended mean of $f$ exists, when the mean on all sequences of bounded functions $(A_r)_{r\in\mathbb{N}}$ with set-theoretic limit $A$ are unique and finite.

$\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 shy.

If $F\subset\mathbb{R}^{A}$ is the set of all $f\in\mathbb{R}^{A}$, where $m_{f}(A;n)$ (Definition 2.1) is finite, then $F$ is a shy 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-question here and here.)

posted 17 days ago

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5d ago

bharathk98‭

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