Does my rigorous definiiton make sense and give what I want? How do we simplify my definiton?

Question

My previous measure in this post doesn't make sense so I made modifications. I need someone to check whether my definitions improved.

Suppose we have the following definition?

Definition

• $\ell$ is the length of an interval
• $\left(J_{k} \right)_{k=1}^{m}$, for $m\in\mathbb{N}$, are a sequence of open intervals where $\;\ell(J_1)=...=\ell(J_m)=g\in\mathbb{R}^{+}$ and the infimum is taken over all possible $J_k$

\begin{align} \mathcal{M}(g,S)=g\cdot\inf\left\{m\in\mathbb{N}: {S^{\prime}\text{ is countable}, \; \left(S\setminus S^{\prime}\right)\subseteq\bigcup\limits_{k=1}^{m}J_{k}}\right\} \end{align}

and

$$\mathcal{N}(S,A)=\lim\limits_{g\to 0}\frac{\mathcal{M}\left(g,S\right)}{\mathcal{M}\left(g,A\right)}$$

then if

\begin{align} & \mathcal{O}(g,S)=\\ & \begin{cases} \small{g\cdot\inf\left\{m\in\mathbb{N}: {S_{j}\subseteq S, \; \mathcal{N}(S_{j},A)=0,\;\bigcup\limits_{j=1}^{\infty}{S}_{j}=S^{\prime\prime},\;\left(S\setminus S^{\prime\prime}\right)\subseteq\bigcup\limits_{k=1}^{m}J_{k}}\right\}} & A \text{ is uncountable}\\ g\cdot\inf\left\{m\in\mathbb{N}: {S\subseteq\bigcup\limits_{k=1}^{m}J_{k}}\right\} & A \text{ is countable} \end{cases} \end{align}

then the outer measure is

$$\mathcal{P}^{*}(S,A)=\lim\limits_{g\to 0}\frac{\mathcal{O}(g,S)}{\mathcal{O}(g,A)}$$

And the inner measure is $\mathcal{P}_{*}(S,A)=1-\mathcal{P}^{*}(A\setminus S,A)$, meaning measure $\mathcal{P}(S,A)$ exists when $\mathcal{P}^{*}(S,A)=\mathcal{P}_{*}(S,A)$.

While reading note:

1. Does my definition make sense
2. Does it give the same value as the Lebesgue Measure when $A=\mathbb{R}$ and $S\subseteq A$?
3. Does it give what else I wish for? (See Intuitive explanation and Example)
4. How can we simplify it? (See the Example)

Intuitive Explanation

I wish to define an outer measure $\mathcal{P}^{*}$ where $m=m_1$ intervals cover $S$ and $m=m_2$ intervals cover $A$. Note all the intervals that cover $S$ and $A$ must have the same length.

To solve $\mathcal{P}^{*}$, if $S$ and $A$ are uncountable, find $S^{\prime}\subseteq S$ and $A^{\prime}\subseteq A$ that can be covered by points or already proven to be "infinitesmially small" such when covering $S\setminus S^{\prime}$ and $A\setminus{A^{\prime}}$ use box-counting dimensions to find total length of minimum $m_1$ intervals of $g$-length covering $S$ divided by total length of minimum $m_2$ intervals $g$ length covering $A$ (then find the infimum of the result). Note to get the total lengths from the minimum $m_1$ and $m_2$ intervals, they must be written as function of $g$ or $m_1(g)$ and $m_2(g)$ and each of must be multiplied by $g$ as $g\to 0$. Make sure you choose a $S^{\prime}$ and $A^{\prime}$ where we get the infimum bound of $\lim\limits_{g\to 0}\left(g \cdot m_1(g)\right)/\left({g \cdot m_2(g)}\right)$.

In my definitons, note $g \cdot n(g)$ becomes $\mathcal{M}(g,S)$ and $g \cdot m(g)$ becomes $\mathcal{M}(g,A)$.

Example

Suppose we define $A=\mathcal{G}\cup\mathcal{C}\cup\left(\mathbb{Q}\cap[0,1]\right)$ and $S=\mathcal{G}\cup\mathbb{Q}$ where $\mathcal{C}$ is the cantor set and $$G_{0}=[0,1]$$

$$G_{n+1}=\frac{G_{n}}{5}\cup\frac{G_{n}+2}{5}\cup\frac{G_{n}+4}{5}$$

$$\mathcal{G}=\bigcap\limits_{n=1}^{\infty}G_n$$

To solve a measure say $\mathcal{P}(S,A)$, I wish to cover $\mathbb{Q}$ with countably infinite points and cover remaining points of $S$, being $\mathcal{G}$, and remaining points of $A$, being $\mathcal{G}\cup\mathcal{C}$, with intervals of same length.

Now suppose all intervals have the length $g$. For every $g\in\mathbb{R}^{+}$, take the total length of all the intervals covering $\mathcal{G}$ as approximately

$$g \cdot 3^{\ln(1/g)/\ln(5)}$$

And we get the total length of the intervals covering $\mathcal{G}\cup\mathcal{C}$ as approximately

$$g\cdot 3^{\ln(1/g)/\ln(5)}+g\cdot 2^{\ln(1/g)/\ln(3)}$$

Then our outer measure should be the infimum of

$$\frac{g \cdot 3^{\ln(1/g)/\ln(5)}}{g \cdot 3^{\ln(1/g)/\ln(5)}+g\cdot 2^{\ln(1/g)/\ln(3)}}$$

Which is

$$\lim\limits_{g\to 0}\frac{g \cdot 3^{\ln(1/g)/\ln(5)}}{g \cdot 3^{\ln(1/g)/\ln(5)}+g\cdot 2^{\ln(1/g)/\ln(3)}}$$

I calculated the limit is $1$.

Answer 1

(Edit since the bounty notice says "Current answers are outdated") The question is now on revision 15. The answer below was meant to respond to revision 7 which was written about 24 hours ago, and had the following definition:

Definition

If $S\subseteq A$:

• $\ell$ is the length of an interval
• $\left(I_{j}\right)_{j=1}^{\infty}$ and $\left(J_{k} \right)_{k=1}^{m}$ for $m\in\mathbb{N}\cup\{\infty\}$, are a sequence of closed intervals where $\ell(I_1)=...=\ell(I_n)=0$, $\;\ell(J_1)=...=\ell(J_m)=g\in\mathbb{R}^{+}$ and the infimum in the equation below is taken over $m$

\begin{align} & \mathcal{M}(g,S,\left(I_j\right)_{j=1}^{n},\left(J_k\right)_{k=1}^{m})=\\ &\begin{cases} \inf\limits_{m\in\mathbb{N}\cup\{\infty\}}\left\{gm: S\subseteq\left(\bigcup\limits_{k=1}^{m}J_{k}\right)\cup\left(\bigcup\limits_{j=1}^{\infty}I_{j}\right)\right\} & A \text{ is uncountable}\\ \inf\limits_{m\in\mathbb{N}\cup\{\infty\}}\left\{gm: S\subseteq\left(\bigcup\limits_{k=1}^{m}J_{k}\right)\right\} & A \text{ is countable} \end{cases} \end{align}

Then the outer-measure is

$$\mathcal{P}^{*}(S,A)=\lim\limits_{g\to 0}\frac{\mathcal{M}\left(g,S,\left(I_j\right)_{j=1}^{n},\left(J_k\right)_{k=1}^{m}\right)}{\mathcal{M}\left(g,A,\left(I_j\right)_{j=1}^{n},\left(J_k\right)_{k=1}^{m}\right)}$$

And the inner measure is $\mathcal{P}_{*}(S,A)=1-\mathcal{P}^{*}(A\setminus S,A)$, meaning measure $\mathcal{P}(S,A)

As far as I can tell from a quick scan of the current version of the question, some minor details were changed (e.g. closed is now open, the $I_k$s have disappeared), but you have now introduced two new intermediate objects $\mathcal{N,O}$. Regardless, it seems the second and third bullet points below still apply.

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Let me attempt to follow your definition. I'll try to compute with $S=[0,1]=A=[0,1]$. It looks like you have placed absolutely no restrictions on

• the sets consisting of one point, $I_j$. So let me set them all to be $\{0\}$.
• the number of intervals $J_k$, denoted $m$. So I will pick $m=1$.
• the intervals, $J_k$, except for their length. Since $m=1$, I only need to decide what $J_1$ is, so let me choose $J_1=[0,g]$.
• the number of points $I_j$ that are used to define $\mathcal M$, $n$. So I will pick $n=1$....but $n$ only appears on the LHS; on the RHS, you still use $\cup_{j=1}^\infty I_j$, so perhaps $n=\infty$ is chosen for me.

Now I reach the definition of $\mathcal M$. Unfortunately, you are now taking an infimum over the quantity $m$ that I already fixed to be $m=1$. So I guess I should release the definition of $m$ and allow $m$ to vary. Then I need to define what $J_k$ are for $k\ge1$. Let me set them all to be $[0,g]$. Then I compute for $g<1$, $$\mathcal M(g, [0,1],(I_j)_j, (J_k)_k)=\inf_m\{ mg : [0,1]\subset [0,g]\cup\{0\}\} = \inf \emptyset = \infty$$

Now I have an issue and cannot continue, as $\mathcal P^*$ involves $\infty/\infty$.

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I highly doubt this is what you wanted. Perhaps, it should be something like

$$\mathcal M(g,S) = g\inf\left\{m\in\mathbb N:\substack{\displaystyle \text{there exists a cover of $S$ by $m$ closed intervals of length $g$,}\ \\\displaystyle \text{up to a countable set} } \right\}$$ or more precisely, if you are uncomfortable with "up to a countable set", $$\mathcal M(g,S) = g\inf\left\{m \in\mathbb N:\substack{\displaystyle \text{there exists $m$ closed intervals $(J_i)_{i=1}^m$ of length $g$, }\ \\\displaystyle \text{such that $S\setminus \left(\bigcup_i J_i\right)$ is at most countable} } \right\}$$ With this, $\mathcal M(g,[0,1])=1$ whenever $g=1/k$ for some $k\in\mathbb N$, and as $g\to 0$ $\mathcal M(g,[0,1])\to 1$. Also, I'm no expert, but what I wrote is quite close to the thing you define with lengths at most $\delta$ before you define a Hausdorff measure...