You lost me with the line $g \in \{0, g_1 \in \mathbb{R}^+\}$. It seems to say that $g$ is an element of a set that consists of the real number $0$ and a truth claim about something called $g_1$.

@Michael $A$ is an arbitrary subset of $\mathbb{R}$ and $S$ is an arbitrary subsets of $A$. And when $A$ is uncountable, $g$ can be $0$ or a positive real number.

I also observe an instance of $\sum_{k=1}^m g$ and I do not know how that is different from $mg$, and I observe $M(g,S)$ being defined in terms of things that are more than just $g$ or $S$ (like $\{J_k\}$).

$\sum_{k=1}^{m}g$ is the same as $mg$ and all $J_k$ have the same length $g$ despite each $J_k$ possibly being different intervals. Hence $M(g,S)$ relies just on $g$ and the set that is being covered which in this case is $S$

If $g\in\{0,\mathbb R_+\}$, then you are allowing for the possibility of $g=\mathbb R_+$? Did you mean instead $g=\infty$? Or did you mean $g\in [0,\infty)$? Note that $[0,\infty) = \{0\}\cup \mathbb R_+ \neq \{0,\mathbb R_+\}$. Secondly, is your infimum also over the choices of the intervals $J_k$s? As written, it is an infimum only over $m$

I read $g=0, g\in \mathbb R_+$ to mean two mutually exclusive statements at the same time (i.e. my brain thinks commas mean "and"). It is much easier to write $$g \ge 0$$ to mean the same thing

@CalvinKhor I intended to write $\ell(J_1)=g$…$\ell(J_k)=g$ where $g=0$ or $g\in\mathbb{R}^{+}$ when $A$ is uncountable.

@CalvinKhor The infimum is over $m$

I still don't see the difference between that and $g\ge 0$. and then since the sequence of intervals stops at a finite number $n$, its a little odd to talk about your infimum which tries to use $J_m$ for all $m\in\mathbb N$. And, as Michael says, if you are not taking an infimum over possible collections $(J_k)_k$ then it seems to me like it should depend on $J_k$ and it may be better to reflect this in the notation e.g. $M(g,S,(J_k)_k)$

Closer to this perhaps? en.wikipedia.org/wiki/Lebesgue_measure#Definition Also I need to leave, but good luck

A final comment before I leave (might check back in a couple days) if $A=\mathbb R$ it seems that $M(g,A) = \inf \{\} = \infty$? Which might be an issue in your definition of $P^*$, or perhaps I have misread.

@Michael Stay with me. How much has my definition improved,

Is this the same stuff as in all your other questions along these lines, Arbuja? Shouldn't you link to those other questions?

@GerryMyerson I did link this question

You seem to be taking a limit as $g\rightarrow 0$, which would also affect the $\{J_k\}$ interval size, but it is not clear how the positioning of those $J_k$ intervals depends on $g$. Do they always have the same left endpoint, for all $g$? It is also not clear what happens if the set $S$ contains some points that are in none of the $J_k$ or $I_k$ sets.

@Michael For all $g$, $J_k$ doesn’t always have the same left end point. For example if $J_k$ covered $[0,1]$ the last $J_k$ to cover the end-point of $[0,1]$ would be $[1-\left\lfloor g \right\rfloor, 1]$. Note the left point may fluctuate between $1-2g$ and $1-g$ but note as $g\to0$ both values converge so this doesn’t matter. Also note $S$ must have all its points in the $I_k$ or $J_k$ sets, otherwise this breaks the requirements of my measure. If $S$ was too “over-covered” it wouldn’t give an infimum bound. I assume the Lebesgue measure would state the same.

I meant “under-covered”. If it was “undercovered” there is no reason for an infimum bound. As stated in my measure when $A$ is uncountable $S\subseteq\left(\bigcup\limits_{j=1}^{\infty}I_j\right)\cup\left(\bigcup\limits_{k=1}^{m}J_k\right)$

@Michael How much has my definition improved.

You seem to be assuming some unstated structure of the $\{J_k\}$ sets. Perhaps they are disjoint, or perhaps they only overlap on the endpoints, and/or perhaps they cover $\mathbb{R}$ or some relevant portion of $\mathbb{R}$. So, it is not clear whether or not I am allowed to use $$\{J_k\} = \{[k, k+g]\}_{k=1}^{\infty} = \{[1, 1+g], [2, 2+g], [3, 3+g], ...\}$$

@Michael The structure of $J_k$ shouldn't matter. You should get the same results. If not I will redefine $J_k$ as having open intervals. The main reason, I wanted to use closed intervals is for countable sets to have zero measure which couldn't be done by open intervals if they had the same length. As for your example, if you are trying to cover $\mathbb{R}$, then the sets you are using won't work for $g\le 1$ since we have $\left\{[0,1][1+g,2],[2+g,3],[3+g,4]\right\}$ which is not covered as $g\to 0$

I was not trying to cover $\mathbb{R}$. Again: It was never stated that $\cup_k J_k = \mathbb{R}$. Nothing that was stated ever ruled out my example of $\{[1, 1+g], [2, 2+g], ...\}$. [Or if you like, to be consistent with your new edits, make my intervals open instead of closed.]

@Michael If you are not covering $\mathbb{R}$, then your example is fine. As I previously mentioned the structure of $J_k$ shouldn't matter. Closed, open, overlapping or non-overlapping you should get the same result......I'm not sure how to be more specific.

So then you really are starting with a collection of intervals $\{J_k(g)\}_{k=1}^{\infty}$, defined for all $0<g<1$. Anyway, I do not think your formulation makes sense. If modified to make sense, its usefulness is unclear.

@Michael Could you give me an example where my definition makes no sense?