I am supprised why nobody responded. Either because the questions are too difficult (which I cannot imagine, the first one was really easy). Or they are just not fun.

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First Problem:

Okay you have,

with .

That means,

Because otherwise, since there is no way to make the left hand smaller.

Similarly,

The best case senerio is when all are solutions in that case,

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Second Problem:

There are a number of things we need to know.

First, if the function is integratble then the Riemann sum iswell-definedas long as the norm sequence limit is zero.

Second, if exists and then both the integrals exists and,

Third, is irrational therefore, .

Now we can begin.

The idea is that we are going to assume that the Dirichlet function is Riemann integrable then arrive at a contradiction.

If the Dirichlet function is integrable on then we can partition the interval in any way we choose as long as the norm sequence limit is zero. We will use the right end-points. In, that case,

But the trick is to realize that,

is rational!

Thus,

Thus, (limit of zero)

.

But, since the Dirchelt integral exists (by assumption) then,

That means its value is by left-endpoints,

+

But,

for it is rational. Und for it is irrational.

Thus,

Thus, the limit is not well-defined by partitioning.

Thus, the function is not integrable.

If you are curious how I invented this problem is when I was in Calculus class we were doing integration over surfaces and I asked whether every bounded function (not continous) is integratble. My professor said it has to be discontinous at uncountably many points. So I decided to test whether what he said was true or false by using the Dirichlet function (since it is discontinous at uncountably many points) and he seemed to have been correct. So the credit of this problem should go to my professor.