Arbuja

@[email protected] Check once more.

@[email protected] I made edits.

@K.Power If you can add an answer changing my definition, I will give you the bounty.

@K.Power How would you change the defintion to get what I want?

I don't know how to define it explcitly. All I know is when the angle between a non-vertical line and vertical line approaches zero, the most number of times a near-vertical line can intersect with a hyper-discontinuous function is infinite. The least number of times a near-veritcal line can intersect with a hyper-discontinuous functions is once.

I stated the angle between the non-vertical and the vertical line approaches zero.

@K.Power The most a vertical line in $[a,b]$ can interesect with $\left. f\right|_{[a,b]}$ is once. The least is also one. Therefore, $((1-1)+(1-1))/2=0$

@K.Power I don't understand. My calculations give me zero for $f(x)=x$. Are you sure you are correct?

@Eric Could you give me an example?

Try and salvage whatever you can. Ignore the rest.

Should I then say $\left.f\right|_{\mathbb{Q}\cap[a,b]}$. I think it still make sense.

I feel like I had to remention $f:X\to Y$ from my post to make the answer clear. It's something I learned from a former professor.

Hyper-discontinuous functions cannot be defined on the interval $[a,b]$ but we can take an "almost vertical" $\ell$ in $[a,b]$ to interesect with the hyperdiscontinuous function. Note $[a,b]\cap\mathbb{Q}\cap[0,1]$ does not have to be empty.

$\ell^{\prime}$ is a non-vertical line that interects with $\ell$, and rotates around the interesection with $\ell$ to approach the line $\ell$.

I forgot to note $\left|\ell\cap[a,b]\right|=\mathfrak{c}$. I can't make more edits or I could risk being suspended.

My answer is complete.

I posted a new answer.

@AlexRavsky You're right. I don't think my answer can be salvaged.

I made one more edit. See if this measure is any better.

Oh well...thanks for responding.

If the measure D doesn't make sense, is there any way to change the measure so it gives everything I want in my answer...

What I meant to say is when the counting measure $|\ell^{\prime}\cap[a,b]|$ is positive infinity, $\ell^{\prime}$ interesects twice with the indicator function for the rationals restricted to [a,b].

Exactly

$\ell^{\prime}$ is a straight line

Try and salvage whatever you can. Ignore the rest.

Should I then say $\left.f\right|_{\mathbb{Q}\cap[a,b]}$. I think it still make sense.

I feel like I had to remention $f:X\to Y$ from my post to make the answer clear. It's something I learned from a former professor.

Hyper-discontinuous functions cannot be defined on the interval $[a,b]$ but we can take an "almost vertical" $\ell$ in $[a,b]$ to interesect with the hyperdiscontinuous function. Note $[a,b]\cap\mathbb{Q}\cap[0,1]$ does not have to be empty.

$\ell^{\prime}$ is a non-vertical line that interects with $\ell$, and rotates around the interesection with $\ell$ to approach the line $\ell$.

I forgot to note $\left|\ell\cap[a,b]\right|=\mathfrak{c}$. I can't make more edits or I could risk being suspended.

@whuber I meant $\mathbb{Q}^2\cap[0,1]$.I also edited my post. Hopefully, this time its more clear.

@SexticusEmpiricus I made edits. Hopefully this is better.

@SexticusEmpricus Since $S_s$ is finite we can take the arithmetic mean of $f(S_s)$ for every $s\in\mathbb{N}$ as $s\to\infty$. I also made edits to my post.

@Carl Apologies, $S_n$ isn’t a partial sum. Think of it as a set; and we are including new elements instead of adding them.

For example if the $S_s=\left\{\frac{1}{n}:n\in\mathbb{N},1\le n\le 2s\right\}$. Note that every time $s$ increases by one we add $2$ elements from $A=\left\{\frac{1}{m}:m\in\mathbb{N}\right\}$ to $S_s$ and as $s\to\infty$ we end up covering all elements of $A$

@SextusEmpricus Sorry for the delay, I wanted to make sure everything in my post was clear.

What about for this question I know I stated this correctly.

@whuber I added examples to clarify my statements. Is this better?

@XanderHenderson Should my question math.stackexchange.com/questions/4251598/… be closed. I tried hard and I got two upvotes but my post also got one downvote. Is my post still unclear.

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

@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 $g\to 0$

@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.

@Michael How much has my definition improved.

@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 $S\subseteq\left(\bigcup\limits_{j=1}^{\infty}I_j\right)\cup\left(\bigcup\limits_{k=1}^{m}J_k\right)$

@GerryMyerson I did link this question

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

@CalvinKhor I am close to explaining. Be patient with me. If I took the infimum over $J_k$ how would that change my definition. And how would I defined it.

@CalvinKhor The infimum is over $m$

@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.