This is not the same as "Find a strict definition for a Riemmanlike sum that is easier to calculate?
"Here I assume that my Riemmanlike sum is clear enough to understand. If not, try to answer this question.
Consider $ f: A to (0.1) $ Where $ A subseteq (a, b) $ and $ lambda $ is the Lebesgue measure. I want to create a simple, userfriendly average that fits my intuition. Before I get to the definition, here's why an intuitive average is so important to me $ f $.
motivation
After reading the entire previous post, remember to read "nonborder points", "first order border points", "second order", "third order", etc. The highest order border point should be "infinitely more weight" for to have $ f $Average as lower order boundary points. For several highestorder limit points with the same order, there are the "denser" lowerorder limit points one of the highest order pointsthe more "weight" this point of the highest order should have an average of $ f $ compared to other highest order points.
While I found this intuitive, though $ lambda (A) = 0 $I think when $ lambda (A) $ is greater than zero, the average should be
$$ frac {1} { lambda (A)} int P (x) dx $$
Where $ lambda $ is the Lebesgue measure.
In addition, if $ A $ is countless and has a dimension of zero, I want to "ignore" countable subsets of $ A $ since countable subsets are "infinitely smaller" than noncountable subsets. Take the general peicewise function, for example:
$ f (x) = f_i (x) $when $ x in A_i $ so that $ f_i: A_i bis (a, b) $ and $ A_1, …, A_m $ are not overlapping subsets of $ A $
We "ignore" the countable $ A_i $ (Give them a zero value) and take the average of $ f $ in terms of countless $ A_i $.
The only time I expect an average to have no "intuitive existence" is when $ A $ is countable and more than one $ A_i $ is countable and dense in $ (a, b) $. In this case, it is impossible to find a clear and intuitive average.
To define the average of $ f $ there is no such thing in this case; we need an upper and a lower sum (Riemmanlike sums) that do not converge; Before I do this, however, I have to define the following.
First definitions
Consider $ S subseteq A $
$$ M (S) = begin {cases}
frac { lambda (S)} { lambda (A)} & lambda (A)> 0 \
0 & S text {is countable and} A text {is countable, but} lambda (A) = 0 \
1 & text {else}
end {cases} $$
This prevents the average from being zero if $ lambda (A) = 0 $ and allows us to ignore countable subsets of $ A $ when $ A $ is countless and $ lambda (A) = 0 $
Ownership of $ M (S) $
$ M ( Emptyset) = text {undefined} $
(1) $ M (A) = 1 $
(2) When $ lambda (A)> 0 $,
If $ {A_i } _ {i = 1} ^ { infty} $ are disjoint and $ bigcup_ {i = 1} ^ { infty} A_i = A $ then $ M left ( bigcup_ {i = 1} ^ { infty} A_i right) = sum_ {i = 1} ^ { infty} M (A_i) = M (A_1) + … = 1 $
(3) When $ lambda (A) = 0 $,
We split up $ A $ into a union of countable subsets $ A_c $ and a union of countless subsets $ A_u $ from $ A $. If $ M (A_c) = 0 $, then $ M (A_u) = 1 $, because $ M (A_c) + M (A_u) = M (A) = 1 $. If $ M (A_c) = 1 $, then $ M (A_u) = 0 $ for the same reason. (I think the additivity is true).
Upper and lower total
Now we can create upper and lower sums.
Given $ S subseteq (0.1) $, and let $ P $ to be a partition of $ (0.1) $ (Note: a partition is a finite set of disjoint subintervals $ X $) you can define $ P & # 39; (S) = {X in P: X cap S neq Emptyset } $. And you can define $ n & # 39; =  P & # 39; (S)  $ (the cardinality of a finite set). Note each disjoint subinterval $ X $ has the same length.
For future intuition, consider the following:
$$ tilde {L} _ {f, P} = frac {1} {n ^ { prime}} sum_ {X in P ^ { prime} (S)} ( inf_ {t in X} f (t)) $$
Define the limit as $  P  to $ 0with refinements from $ P $ how so:
$$ lim _ {  P } ( tilde {L} _ {f, P}) = tilde {L} _f $$
This is the "lower average" of $ f $ on $ (0.1) $ (in relation to the partition $ P $).
Likewise the "upper Kinda average" of $ f $ on $ (0.1) $ in terms of partition $ P $ is:
$$ tilde {U} _f = lim _ {  P } ( tilde {U} _ {f, P}) $$
Where
$$ tilde {U} _ {f, P} = frac {1} {n ^ { prime}} sum_ {X in P ^ { prime} (S)} ( sup_ {t in X} f (t)) $$
and the limit lasts $  P  to $ 0.
We want these lower and upper average limits to converge to the same value.
Note that this @ WillieWong's expanded comment and chat are still not strict and successful.
Full definition
However, the last section was only for intuition. Now we make the real definitions.
We define the full "lower average" as:
$$ L_ {f, P} = frac {M (A)} {n ^ { prime}} sum_ {X in P ^ { prime} (A)} ( inf_ {t in X} f (t)) $$
and the full "upper average" as:
$$ U_ {f, P} = frac {M (A)} {n ^ { prime}} sum_ {X in P ^ { prime} (A)} ( sup_ {t in X} ) f (t)) $$
If these lower and upper average limits converge to the same value (id est: are the same), we get "my definition of the average" from $ f $ for each $ A $. If they don't converge, the average is undefined. Note that I define "upper" and "lower" averages to indicate when an average cannot exist.
Example with general piecewise function
Consider a general piecewise function, $ f (x) = f_i (x) $when $ x in A_i $ so that $ f_i: A_i bis (a, b) $ and $ A_1, …, A_m $ are not overlapping subsets of $ A $.
When $ lambda (A)> 0 $, the lower average of $ f $ is
$$ L_ {f, P} = frac {M (A_1)} {n ^ { prime}} sum_ {X in P ^ { prime} (A_1)} ( inf_ {t in X} f_1 (t)) + … + frac {M (A_m)} {n ^ { prime}} sum_ {X in P ^ { prime} (A_m)} ( inf_ {t in X .} f_m (t)) $$
and the top average of $ f $ is
$$ U_ {f, P} = frac {M (A_1)} {n ^ { prime}} sum_ {X in P ^ { prime} (A_1)} ( sup_ {t in X} f_1 (t)) + … + frac {M (A_m)} {n ^ { prime}} sum_ {X in P ^ { prime} (A_m)} ( sup_ {t in X .} f_m (t)) $$
When the upper and lower average limits converge, we have a defined average. If not, the average is undefined. That's why I create upper and lower sums. I want cases where we can't have an average.
Finally, if $ lambda (A) = 0 $countable $ A_i $ are combined in $ A_c $ and countless $ A_i $ are combined in $ A_u $, then with property $ (3) $, the lower average of $ f $ is
$$ L_ {f, P} = frac {M (A_c)} {n ^ { prime}} sum_ {X in P ^ { prime} (A_c)} ( inf_ {t in X} f (t)) + frac {M (A_u)} {n ^ { prime}} sum_ {X in P ^ { prime} (A_u)} ( inf_ {t in X} f (t ))) $$
and the top average of $ f $ is
$$ U_ {f, P} = frac {M (A_c)} {n ^ { prime}} sum_ {X in P ^ { prime} (A_c)} ( sup_ {t in X} f (t)) + frac {M (A_u)} {n ^ { prime}} sum_ {X in P ^ { prime} (A_u)} ( sup_ {t in X} f (t ))) $$
question
Are they well defined?
Could we find a definition that is easier to calculate and gives accurate values?
My first guess is that we can change the definition of the generalized Riemann integral. This could give the intuitive result $ 0 $for example in $ P (x) = x $ and $ A = left { frac {1} {2 ^ x} + frac {1} {2 ^ y} + frac {1} {2 ^ z}: x, y, z in mathbb { Z} right } $.
My second guess is that we can use translationinvariant metrics along the line from Density with Folner
Nets. Here are some papers that can help

Taras Banakh, Extreme densities and dimensions on groups and $ G $spaces
and their combinatorial applications, arXiv: 1312.5078

Taras Banakh, The Solecki submeasurements and densities on groups, arXiv: 1211.0717

Taras Banakh, Igor Protasov, Sergiy Slobodianiuk, Density, partial measures and division of groups, arXiv: 1303.4612
EDIT: I heard that Choquet is integral
different possibility.
According to a comment by Reddit:
You may want to look in papers about OWA operators. If my memory is good, find one
Generalization in "MmOWA: a generalization of OWA operators" using the Choquet integral.
I'm not sure if it helps, but good luck.
Edit 2:
According to this post, we can use the Hausdorff measure, but it's not clear how it works
when $ A $ is countable infinite.
Edit 3: Has @WillyWong already solved this problem ?. Here is his comment.
Probably my last comment about it: I think you can probably reach
what you want by looking at the problem differently.
First, we construct a series of limited functions $ g_ sigma $ how
follows: Start with yours $ A $. Consider the crowd $ A_ sigma = cup_ {x in
A} (x – sigma, x + sigma) $. This is a union of open intervals and
therefore is an open set. As long as $ A $ is not empty this set is
not empty and therefore has a positive Lebesgue dimension.
Just consider $ sigma <1/2 $. To let $ chi_ sigma $ be the indicator
The function of $ A_ sigma $. Define $ g _ { sigma} (x) = frac {1} { A_ sigma }
int _ { 1/2} ^ x chi_ sigma (y) ~ dy $.
Here $  A_ sigma  $ is the Lebesgue measure of $ A_ sigma $.
Notice that $ g_ sigma $ is normalized so that it takes value in between
$ 0 $ and $ 1 $. (It's limited.)
And $ g_ sigma $ is continuous. The question is whether there is and
What is the limit $ lim _ { sigma to 0} g _ { sigma} $.
When $  A  > $ 0, then the family $ g_ sigma $ is equal to continuous, and it
is not too difficult to see $ g $ is formed as $ frac {1} { A }
int _ { 1/2} ^ x chi (y) ~ dy $ and here $ chi (y) $ is the indicator
The function of $ A $.
The main question is what happens when $  A  = $ 0. The guess is
the if $ A $ has zero dimension, but has a nontrivial perfect kernel,
then the limit $ g $ is a continuous function (like the Cantor
Function). And when $ A $ is scattered, the limitation $ g $ is a step
Function. In any case, the integral sought should be
the Stieltjes integral with weight function $ g $.