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[sand python]

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[queen shuttle]

you don't have to

sorry i've forgotten Darboux & Riemann's theory of integration so i can't give you an answer which is satisfactory for me

here's a similar problem for which i've provided an alternative (and more skillful) solution using fixed-width interval approximation for fun:
https://math.stackexchange.com/a/4471487/290189

there's no need to do so.
an intuitive (and non formal) verbal intuition is that
"the smaller the mesh, the finer smaller the partition, and the smaller the absolute error of the Riemann sum"

a logical (but unsatisfactory) answer to your question (for the particular case when the integrand is a monotone function) is

for any parition P,
upper Darboux sum U(f, P) - lower Darboux sum L(f, P) < [f(b) - f(a)] (mesh)
here
lower Darboux sum L(f, P) = ∑ (inf f(x) on subinterval)(subinterval width)
upper Darboux sum U(f, P) = ∑ (sup f(x) on subinterval)(subinterval width)

it's clear that for any Riemann sum with tagged partition P' made from the partition P
lower Darboux sum L(f, P) ≤ any Riemann sum with partition P ≤ upper Darboux sum U(f, P)

take mesh → 0 to see that your function f(x) = 1 / x ² is Riemann integrable on x > 0 and on x < 0.

whether the subinterval widths are uniform are not is not important

the other thing that matter is that

for each subinterval $[x_{i-1},x_i] \subseteq [a,b]$,

[pulsar smelt]

vin100

[queen shuttle]

the difference between the upper and lower Darboux sums in this subinterval is

,,[f(x_i) - f(x_{i-1})](x_i - x_{i-1}) \le [f(x_i) - f(x_{i-1})] \underbrace{\delta}_{\textrm{mesh}}

[pulsar smelt]

vin100

[queen shuttle]

we take the sum over $i = 1, \dots, n$ to see that

[pulsar smelt]

vin100

[queen shuttle]

the RHS is actually

,,[f(b)-f(a)]\delta

[pulsar smelt]

vin100

[queen shuttle]

to sum up:

• with a partition P fixed, a Riemann sum is somewhere in between a lower Darboux sum and an upper Darboux sum
• for any monotone function, as the mesh becomes arbitrarily small, the upper Darboux sum and the lower Darboux sum become arbitrarily close to one value (defined as the Darboux integral)
• use the Squeeze Theorem to see that the Riemann integral is "squeezed" to this value

[queen shuttle]

sadly, in general, no. the first interval's width is fixed, so the mesh is bounded lower below by this fixed width, no the mesh is not tending to zero

i added "in general" cuz in the equivalent epsilon-delta definition in wiki, for each epsilon, you only need to find a delta that satisfies ... It doesn't say directly how "small" this delta shld b. in fact, for the constant function, any partition would work, since there's no variation at all, so any Riemann sum would give the same value k (b - a), and
| any Riemann sum - k (b - a) | = 0 < epsilon

ok i've rmb the true reason for which any continuous function on [a, b] is Riemann integrable

in fact, it's a well-known theorem that any continous function on a closed interval (in fact, closed and bounded set) is uniformly conntinous

ofc one can prove this by contradiction using the sequential criterion

but imho the proof by invoking the Heine-Borel Theorem is worth studying

using this theorem, for each epsilon, you get a delta = delta(epsilon) (delta as a function (that we don't need to know its formula) of epsilon). partition [a, b] using subintervals with max width delta / 2

the graphical intuition is that for each pair of points inside the same subinterval (with width <= delta / 2), the absolute difference in their functional values is epsilon-small

here's an illustrutation for uniform continuity

take from wiki

we're using this so that in different subintervals, we have the same delta that "responds to the epsilon challenge"

again, you can consider the difference between the upper and lower Darboux sums

each subinterval's contribution to this difference is bounded above by

,,\varepsilon (x_i - x_{i-1})

[pulsar smelt]

vin100

[queen shuttle]

take the sum over i = 1, ..., n

so that the upper bound becomes

,,\varepsilon (b - a)

[pulsar smelt]

vin100

[queen shuttle]

i.e. difference between upper and lower Darboux sums can be made arbitrarily small

and any Riemann sum with the same partition points (and with whatever tag points) is sandwiched in the middle of these two Darboux sums

[queen shuttle]

in the whole process (of approximation of the number which is the limit of either the upper or the lower Darboux sum by a Riemann sum), we only care about finding an upper bound for such approximation, and to do so, we only care about the mesh (i.e. the max subinterval width). The choice of tagged points and the width of a particular subinterval have no role to play in this argument

[pulsar smelt]

vin100

[queen shuttle]

exercise: find the definite integral for the real exponent function $f(x) = x^s$ using the limit of Riemann sum.

[pulsar smelt]

vin100

[queen shuttle]

hint: ||consider partition points in a geometric progression, use geometric sum to find this limit||

[queen shuttle]

it's my turn to say sorry

to answer your question, you can just try the discrete and finite Riemann sum

provided in the question

then you'll see that the tag points are so-well chosen that the interval widths don't matter

but to answer properly this question (why for any tagged partition, when mesh → 0, the corresponding Riemann sum tends to this particular Riemann sum provided in the question), you need my arguments above the uniform continuity