Riemann sums + integrability

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ame

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Nov 6, 2007, 11:55:23 PM11/6/07
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i've just learned that the definite integral of f is equal to the
limit of the riemann midpoint sum if f is continuous in the interval.
i'm curious -- can that limit exist for a function that's not riemann
integrable? if so, what would such a function be? if not, why not?

thanks,
ame

hagman

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Nov 7, 2007, 2:18:36 AM11/7/07
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Wlog. your interval is [0,1].
If you take the limit only for equidistant subintervals (or similarly
for any fixed sequence of subintervals), i.e. your n'th term of the
limit is
sum_{0<=k<n} f( (k + 1/2)/n )
then the counterexample is simple:
This limit is already 0 if f(q) is zero for all rationals q.
If f(x)=1 for all irrational x then f is not Riemann integrable
although the above limit exists. Also, the limit differs from the
Lebesgue integral, which would be 1.

hagman

Robert Israel

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Nov 7, 2007, 2:56:06 AM11/7/07
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ame <rae...@gmail.com> writes:

Yes it can exist. Let's say your interval is [0,1]. Your Riemann midpoint
sums all involve evaluating f at rational numbers. Consider a function
that is 0 on the rationals (so all the Riemann midpoint sums are 0), but
does something wild and crazy on the irrationals...
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

Kira Yamato

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Nov 7, 2007, 3:45:16 AM11/7/07
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On 2007-11-07 02:56:06 -0500, Robert Israel
<isr...@math.MyUniversitysInitials.ca> said:

> ame <rae...@gmail.com> writes:
>
>> i've just learned that the definite integral of f is equal to the
>> limit of the riemann midpoint sum if f is continuous in the interval.
>> i'm curious -- can that limit exist for a function that's not riemann
>> integrable? if so, what would such a function be? if not, why not?
>
> Yes it can exist. Let's say your interval is [0,1]. Your Riemann midpoint
> sums all involve evaluating f at rational numbers. Consider a function
> that is 0 on the rationals (so all the Riemann midpoint sums are 0), but
> does something wild and crazy on the irrationals...

You seem to be considering only nets of partitions with points in the
rationals only. In that case, all midpoints are rationals also.
However, what if you consider also nets of partitioners with points in
all of reals too?

For a partition with points in the irrationals, the midpoints need not
be rationals. So, the limit of riemann midpoint sum will not converge
in your example. That is, I can always find a finer partition with
irrational points such that the midpoints are also irrational.

Did I misunderstood something here?

--

-kira

Dave L. Renfro

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Nov 7, 2007, 10:41:19 AM11/7/07
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Kira Yamato wrote:

> You seem to be considering only nets of partitions
> with points in the rationals only. In that case,
> all midpoints are rationals also. However, what if
> you consider also nets of partitioners with points
> in all of reals too?
>
> For a partition with points in the irrationals, the
> midpoints need not be rationals. So, the limit of
> riemann midpoint sum will not converge in your example.
> That is, I can always find a finer partition with
> irrational points such that the midpoints are also
> irrational.

I posted some remarks about this issue, and the issue
of which point is selected in the various intervals,
in these December 10, 2002 sci.math posts:

http://groups.google.com/group/sci.math/msg/07d2e1f77af46765

http://groups.google.com/group/sci.math/msg/c8fd54351693a760

I thought I had also, at some later time, posted citations
to the papers I alluded to in the first post, but I can't
find to post now, so maybe I never made it. I'll post some
references tomorrow. (I have to dig them up when I get
home tonight.)

Dave L. Renfro

Dave L. Renfro

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Nov 7, 2007, 3:20:05 PM11/7/07
to
Dave L. Renfro wrote (in part):

> I thought I had also, at some later time, posted citations
> to the papers I alluded to in the first post, but I can't
> find to post now, so maybe I never made it. I'll post some
> references tomorrow. (I have to dig them up when I get
> home tonight.)

I just remembered that it wasn't a post, but rather it was
an e-mail to someone whose preliminary real analysis text
manuscript I was giving editing feedback on. The e-mail
is below, with some slight changes and omissions that have
to do with specifics about the person I was writing to.

----------------------------

I believe Cauchy proved the left endpoint integral exists
for continuous functions on compact intervals (modulo his
uniform continuity omission) and he also proved that all
Riemann sums for such a function converge to the same result.
I think his definition was the left endpoint version and its
agreement with Riemann's formulation was a theorem, but I'm
not entirely sure about this. In any event, the Riemann
version of the integral had been around much earlier than
Riemann lived, and I think this is fairly well known. I
believe that, even back in Newton's time, mathematicians dealt
with general "Riemann sums" when computing areas and other
things (i.e. the location of the evaluation point in
each interval was varied as needed for the problem at
hand). What Riemann mainly did was to put the focus
on the collection of functions that are integrable
according to some notion of integrability, rather than
(as Cauchy did) defining a notion of integrability only
to be able to rigorously prove certain desired integrability
properties.

Gillespie [1] proved that any bounded function on a compact
interval is "left endpoint integrable" if and only if it is
Riemann integrable.

Kristensen/Poulsen/Reich [5], apparently unaware of Gillespie's
result [1] (or of any of the other papers below) proved a
stronger version in which "left endpoint" becomes "selection
via the function G(x,y)", where the domain of G is all pairs
of real numbers x < y in the integration interval [0,1] such
that x <= G(x,y) <= y and G satisfies the intermediate value
property in each variable (i.e. for each fixed p and q,
G(p,t) and G(t,q) satisfy the intermediate value property
relative to the variable t). Note that the choice of
G(x,y) = x gives the left endpoint formulation.

I believe Giovanni [2] (also apparently unaware of Gillespie [1])
proved a result similar to what Kristensen/Poulsen/Reich [5]
proved, but I think he restricted himself to the case where
G(x,y) is continuous (I don't know if this is continuity
in both variables or the weaker notion of continuity in
each variable). However, I haven't tried asking anyone
to translate anything from this paper, so I'm not
completely sure what he does. [Fund. Math. papers are
on the internet at <http://matwbn.icm.edu.pl> if you
want to take a look at it.]

Hildebrandt [3] (p. 273) wrote: "It seems probable that
[Gillespie's] proof can be carried over to the Stieltjes
integral [with respect to] a continuous g(x) of bounded
variation." I thought I had a paper somewhere, or at least
had seen such a paper at one time, that did this. However,
I couldn't find one, so this might make for a good undergraduate
or Master's thesis if you know of anyone (student or faculty
member) who needs a real analysis topic.

Zorn [8] (Lemma 1, p. 148) makes an interesting (to me,
at least) observation. The proof of Zorn's observation
is immediate and Zorn's paper is otherwise not particularly
related to the other papers I'm talking about, but I don't
think I've seen it pointed out anywhere else. Let's call
a bounded function on on a compact interval "weakly Cauchy
integrable" (my term) if, for each subinterval in a partition,
the evaluation point is either the left endpoint or the right
endpoint. Note that requiring a uniform choice of "left"
or "right" for all the subintervals puts us back to Cauchy's
formulation. To see that weakly Cauchy integrable implies
Riemann integrable, we simply observe that f(t_n)*(x_n - x_(n-1))
is equal to f(t_n)*(x_n - t_n) + f(t_n)*(t_n - x_(n-1)).
That is, use the "Riemann selection point t_n" to divide
the interval [x_(n-1), x_n] into two intervals, [x_(n-1), t_n]
and [t_n, x_n].

Kieffer/Stanojevic [4] study a certain weakening of the
selection of an arbitrary Riemann evaluation point in each
subinterval of a partition that, in a certain probabilistic
sense, results in almost sure convergence to (every?) Lebesgue
integrable functions. I haven't looked at it very much and
my background is rather weak in what is involved.

Tong [7] considers ways in which the convergence of equal
width subintervals is sufficient for Riemann integrability.
He was apparently unaware of Sklar [6], who seems to prove
very similar results.

[1] David Clinton Gillespie, "The Cauchy definition of a definite
integral", Annals of Mathematics (2) 17 (1915-16), 61-63.
[JFM 45.0441.02]

[2] Dantoni Giovanni, "Sul confronto di alcune definizioni di
integrale definito", Fundamenta Mathematicae 19 (1932), 29-37.
[Zbl 5.20002; JFM 58.0237.01]

[3] Theophil Henry Hildebrandt, "Definitions of Stieltjes integrals
of the Riemann type", American Mathematical Monthly 45 #5
(May 1938), 265-278. [Zbl 19.05604; JFM 64.0198.01]

[4] John C. Kieffer and Caslav V. Stanojevic, "The Lebesgue integral
as the almost sure limit of random Riemann sums", Proceedings of
the American Mathematical Society 85 (1982), 389-392.
[MR 83h:26015; Zbl 497.28007]

[5] Erik Kristensen, Ebbe Thue Poulsen, and Edgar Reich,
"A characterization of Riemann-integrability", American
Mathematical Monthly 69 #6 (June/July 1962), 498-505.
[MR 25 #4074; Zbl 113.04201]

[6] Abe Sklar, "On the definition of the Riemann integral", American
Mathematical Monthly 67 #9 (November 1960), 897-900.
[MR1530955]

[7] Jingcheng Tong, "Partitions of the interval in the definition
of Riemann's integral", International Journal of Mathematical
Education in Science and Technology 32 (2001), 788-793.
[MR1862677; Zbl 1048.26503]

[8] Max August Zorn, "Approximating sums", American Mathematical
Monthly 54 #3 (March 1947), 148-151.
[MR 8,450g; Zbl 29.03101]

Dave L. Renfro

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