Notation
$S$ is a fine letter for an arbitrary set, but $\left[0,1\right]$
is important enough that it often gets the letter $I$, which I will
use below.
Question 1
Length
One important notion of size here is that of "length". (You may
have heard "(Lebesgue) measure"
as a technical term for a general concept of length/area/volume/etc.)
Length satisfies a lot of properties that should be intuitive: Any
set of real numbers you can think of (without doing something super
weird like use the axiom of choice) has nonnegative length (possibly $\infty$). The length of the empty
set is $0$. The length of the interval $I$ is $1$. If you translate
a set left or right, its length doesn't change. If you break up a
set into countably many separate pieces, you can just add the lengths
to get the length of the whole.
This makes any interval have the length you think it should: $\left[0,k\right]$
has length $\left|k\right|$, so the length of $\left[0,\frac{1}{6}\right]$
is exactly $\frac{1}{6}$ of the length of $I$, and $\left[\frac{1}{6},1\right]$
has $\frac{5}{6}$ the length, etc.
Answer
So if by "$90\%$ of $I$" you mean considering length, then the
answer to question 1 is yes.
Cardinality
Another important notion of size is that of "cardinality".
We say that two sets "have the same cardinality" if there is a
"one-to-one correspondence"
between them (often called a "bijection" or "a function that
is both one-to-one and onto").
All intervals that have more than one point in them have the same
cardinality. For the case of closed intervals, $x\mapsto\dfrac{x-a}{b-a}\left(d-c\right)+c$
is a one-to-one correspondence between $\left[a,b\right]$ and $\left[c,d\right]$.
Answer
So if by "$90\%$ of $I$" you mean considering cardinality, then
the answer to question 1 is no. All of your intervals
have the same cardinality, so with that perspective, things like "$90\%$"
don't really make sense for infinite sets.
Comment
There's not really a useful relation between cardinality and length.
For example, the Cantor set
has the same cardinality as $I$, but has length $0$. Sets of smaller
cardinality (like the set of rational numbers)
have length $0$ for sure, though. (For those who know about such
things, this claim is still true even if the Continuum Hypothesis
is false.)
Question 2A
Is the arithmetic mean of $I$ equal to $0.5$?
Median
The arithmetic mean of a finite arithmetic progression
coincides with the median (at very least if the progression has an
odd number of terms). We can try to generalize the median to an interval.
A good choice for a median in this sort of context of a set of reals
with positive length would be a point where half of the length is
to the left, and half of the length is to the right. Luckily for us,
$I$ only has one such point: $0.5$.
Answer
So we if generalize "arithmetic mean" as median in the above sense,
the answer is yes.
Expected value
Original meaning
The arithmetic mean of a finite list (arithmetic progression or not)
is also the "expected value"
if you give each entry in the list the same probability (imagine a
perfect die with the numbers on the faces). If this is new to you,
see this CrossValidated question
for a discussion.
The formula for the arithmetic mean/expected value for a finite list
like this is the sum of the values in the list divided by the number
of values. If the list has no repeats, we can think of it as a set
of numbers $T$, and write the expected value as $\dfrac{{\displaystyle \sum_{x\in T}x}}{\text{size of }T}$
or $\dfrac{{\displaystyle \sum_{x\in T}x}}{\sum_{x\in T}1}.$
A new perspective
To make the above calculation(s) make sense for an interval $\left[a,b\right]$,
we'll need to make an analogy. It's a bit off-topic to include a full-blown
introduction to integrals /Riemann
Sums here, but basically
the idea is that certain things we'd like to think about as infinite
sums, can be interpreted as areas, or at least a difference of areas.
The area under the graph of $f\left(x\right)=x$ over the interval
on the $x$-axis is suitable here, but since negative terms in $T$
would decrease ${\displaystyle \sum_{x\in T}x}$, we need to use the
negative of the area above the graph for the negative pieces. For
example, if the interval were $\left[-1,2\right]$, then we should
replace ${\displaystyle \sum_{x\in T}x}$ with the difference between
the area of the big triangle (vertices $\left(0,0\right)$, $\left(2,0\right)$,
$\left(2,2\right)$) and the area of the small triangle (vertices
$\left(0,0\right)$, $\left(-1,0\right)$, and $\left(-1,-1\right)$)
in:
Or if we had $\left[1,3\right]$,
it would just be the area of a trapezoid with vertices $\left(1,0\right)$,
$\left(1,1\right)$, $\left(3,0\right)$, $\left(3,3\right)$, as
in:
Now, for the denominator, we can either think about replacing "size
of $T$" with "length of $T$", or the sum of $1$s with the
area under the graph of $f\left(x\right)=1$. Either one gives us
the same numerical answer for $\left[a,b\right]$: $b-a$.
Putting this together with the area formula for a trapezoid, we find
that the "expected value" for an interval $\left[a,b\right]$
when $b\ge a\ge0$ is $\left(\dfrac{a+b}{2}\left(b-a\right)\right)/\left(b-a\right)=\dfrac{a+b}{2}$.
And when $a\le0\le b$ we use a difference of triangle areas: $\left(\dfrac{b^{2}}{2}-\dfrac{\left(-a\right)^{2}}{2}\right)/\left(b-a\right)=\dfrac{a+b}{2}$.
And when $a\le b\le0$ we use the negative of the area of the trapezoid:
$\left(-\dfrac{\left(-a\right)+\left(-b\right)}{2}\left(b-a\right)\right)/\left(b-a\right)=\dfrac{a+b}{2}$.
In all cases, the expected value is the midpoint: $\dfrac{a+b}{2}$.
Answer
So we if generalize "arithmetic mean" as expected value in the
above sense, the answer is yes.
Center of Mass
The arithmetic mean of a finite set $T$ (arithmetic progression or
not) is also the location of the "center of mass"
of equal-mass particles at locations on a line given by the numbers
in the set. If every particle has equal mass $m$, the formula reduces
to $\dfrac{{\displaystyle \sum_{x\in T}mx}}{\sum_{x\in T}m}=\dfrac{{\displaystyle \sum_{x\in T}x}}{\sum_{x\in T}1}=\dfrac{{\displaystyle \sum_{x\in T}x}}{\text{size of }T}.$
Correspondingly, the center of mass of an interval/line segment $\left[a,b\right]$
with uniform linear density is just its midpoint: $\dfrac{a+b}{2}$.
Answer
So we if generalize "arithmetic mean" as center of mass in the
above sense, the answer is still yes.
Question 2B
Does it make sense to treat $I$ as an arithmetic progression whose
common difference is infinitesimal?
Calculus considerations
As discussed above, if we make analogies using area/Calculus, the
arithmetic mean still works out to $0.5$. If you know about Riemann
sums, you can think about estimating the area of the triangle under
$f\left(x\right)=x$ over $I$ with a bunch of rectangles, and have
the area either stay equal to $0.5$ or approach $0.5$ (depending
on how you decide the heights of the rectangles) as you make the rectangles
thinner.
So if you like this area idea, then I'd say the answer is almost.
Actual infinitesimals
There are a number of ways to have actual infinitesimal numbers, and
none of them will really work for your purpose here. Basically, arguably
inherent in the idea of an arithmetic progression is the idea of a
next number. But there is no next number after, say, $0.4$ no matter
what you do. Even if you allow something like "$\varepsilon$ is
an infinitesimal number that's positive but less than $\frac{1}{n}$
for any positive integer $n$", $0.4+\varepsilon$ wouldn't be the
next number, since $0.4+\varepsilon/2$ would be in-between.
And if you're in an exceptionally weird context where your $\varepsilon/2$
doesn't make sense (or equals $\varepsilon$) , then you probably
have $0.4+\varepsilon+\varepsilon=0.4+\varepsilon$ or something like
that so again the progression breaks down.
So my final answer would be not really.
