Re: Finding an upper Darboux integral

Noh wrote (in part):

> I'm an undergraduate student, studing a first course
> in analysis. I have a problem finding the upper
> Darboux integral of the following function
>
> f(x) = x ; for ratinal x
> f(x) = 0 ; for irrational x
>
> The problem asks me to find upper and lower Darboux
> integrals for f on the interval [0,b]

Notation: "\leq" is "less than or equals to"
and "\geq" is "greater than or equals to".

Lemma: Let c < d be real numbers and g,h be functions
from [c,d] into R. If g(x) \leq h(x) for each
x in [c,d], then

lower D-integral of g on [c,d]
\leq lower D-integral of h on [c,d]

and

upper D-integral of g on [c,d]
\leq upper D-integral of h on [c,d].

You've probably already proved this lemma in class
or it's in the text. If not, you'll need to prove
it yourself if you use it. I'm going to use it.

Using the lemma with c = 0, d = b, g(x) = 0 (the zero
function on [0,b]), and h(x) = f(x) (your function),
we have

0 \leq lower D-integral of f on [0,b].

This is because the lower D-integral of the zero
function on [0,b] is 0, which you should include
a proof of if you don't know how to deduce it
from results that you're allowed to use.

Since every lower Darboux sum of f on [0,b]
is equal to 0 (you should prove this), we also
have (why?)

0 \geq lower D-integral of f on [0,b].

Hence,

0 = lower D-integral of f on [0,b].

Using the lemma with c = 0, d = b, g(x) = f(x) (your
function), and h(x) = x on [0,b], we have

upper D-integral of f on [0,b] \leq (b^2)/2.

This is because the upper D-integral of h(x) = x
on [0,b] is (b^2)/2, which you should include
a proof of if you don't know how to deduce it
from results that you're allowed to use. One way
to do this is to show that every upper Darboux sum
of h(x) on [0,b] is less than or equal to at least
one right-endpoint Riemann sum of h(x) on [0,b]
that uses equal length subintervals, and then
show that every right-endpoint Riemann sum of
h(x) on [0,b] that uses equal length subintervals
is less than or equal to (b^2)/2. For this last
result, you'll actually show that each such
right-endpoint Riemann sum is _less_ than (b^2)/2.
(Look in an elementary calculus book to see how
one goes about setting up a right-endpoint Riemann
sum for h(x) = x using n subintervals of equal
length, and then how to get (b^2)/2 when taking
the limit of these sums as n --> oo, if you've
forgotten how to carry this out.)

Finally, to show that

upper D-integral of f on [0,b] \geq (b^2)/2,

from which we then get

upper D-integral of f on [0,b] = (b^2)/2,

note that right-endpoint Riemann sums of f
on [0,b] using equal length subintervals _are_
upper Darboux sums, and the supremum of all
these right-endpoint Riemann sums is (b^2)/2.
Hence, the supremum over all the upper Darboux
sums must be greater than or equal to (b^2)/2.

After taking into account what you're allowed
to use and for possible redundancies in what
I've outlined, a similar outline of your final
proof might be shorter than what I've written.
Looking over what I've written, there do appear
to be some ways that my argument can be streamlined,
but I'll leave this to you.

Dave L. Renfro

.