Challenge Yet another counterexample challenge

[micromass]

Yes, and it's still discontinuous everywhere.

 

[micromass]

https://en.wikipedia.org/wiki/Nowhere_continuous_function

 

[fresh_42]

Yes, and it's still discontinuous everywhere.

Damn density ... it's been such a nice zero set ...

 

[andrewkirk]

Re question 10: I thought I might be able to show that $S\equiv [0,1]\cap\mathbb Q$ is a counter-example, notwithstanding the above discussion, on the grounds that it has zero measure, but cannot be the set of discontinuity points of any function (let alone a Riemann-integrable one) because a function that is discontinuous on $\mathbb Q$ must be discontinuous on all of $\mathbb R$, because the former is dense in the latter.

But then I thought of the following counterexample to my counter-example, which I thought was interesting enough to be worth posting.

Here is a function $f$ that is discontinuous on the rationals and continuous on the irrationals!

For $x$ irrational, $f(x)=x$.
For $x$ rational, $f(x)=\frac m{n+1}$, where $\frac mn$ is the fully-reduced presentation of $x$ as a fraction, that is: $m,n\in\mathbb Z$ and $HCF(m,n)=1$.

This is discontinuous at $x$ rational, because $f(x)=\frac m{n+1}$ and $\lim_{\substack{x'\to x\\ x'\notin\mathbb Q}}f(x')=\frac mn$. So the limit (removing the constraint to the irrationals) either does not exist, or is not equal to $f(x)$. So it's not continuous at $x$.

But it is continuous at $x$ irrational because, given $\epsilon>0$, we can choose $\delta>0$ such that any $x'\in (x-\delta,x+\delta)$ must have such a large denominator that $|\frac m{n+1}-\frac mn|<\epsilon$.

I don't know if this is already a well-known function, but I certainly got a kick out of finding it.

Still wondering about a counter-example to 10 though. I think the above $f$ kills $\mathbb Q\cap[0,1]$ as a possible answer, because I'm pretty sure $f$ is Riemann-integrable.

 

[andrewkirk]

Aha. The Cantor Set is the answer to 10. It has measure zero. It's uncountable, hence any function that has discontinuities on all the points of the Cantor Set cannot be Riemann-Integrable!

Furthermore, since it's nowhere dense, we can even find functions whose set of discontinuity points are the Cantor Set (eg its indicator function). But they won't be Riemann Integrable.

 

[Samy_A]

Let $\pi:\mathbb{N}\times \mathbb{N}\rightarrow \mathbb{N}$ be a bijection and let $\sum_{n} a_n$ be an absolutely convergent series in $\mathbb{R}$. Then $\sum_n a_n = \sum_{k=0}^\infty \sum_{l=0}^\infty a_{\pi(k,l)}$.

I'm somewhat confused by this one.
For an absolutely convergent series of real numbers, any rearrangement of the series yields a convergent series with the same sum.
Is $\sum_{k=0}^\infty \sum_{l=0}^\infty a_{\pi(k,l)}$ not such a rearrangement? That would make the statement true.

 

[andrewkirk]

For an absolutely convergent series of real numbers, any rearrangement of the series yields a convergent series with the same sum.
Is $\sum_{k=0}^\infty \sum_{l=0}^\infty a_{\pi(k,l)}$ not such a rearrangement? That would make the statement true.

It's more than just a rearrangement. Two limits are taken, rather than just one, as it's

$$\lim_{r\to\infty}\left(\sum_{k=1}^r \lim_{s\to\infty}\sum_{l=1}^s a_{\pi(k,l)}\right)$$
A rearrangement is of the form
$$\sum_{n=1}^\infty a_{\sigma(n)}=\lim_{k\to\infty} \sum_{n=1}^k a_{\sigma(n)}$$
where $\sigma:\mathbb N\to \mathbb N$ is a permutation. In this case only one limit is taken.

A counterexample would require the two limits in the first case to interact in some way so as to give a different result from what happens when one only takes a single limit.

 

[mfb]

because a function that is discontinuous on $\mathbb Q$ must be discontinuous on all of $\mathbb R$, because the former is dense in the latter.

The counterexample I constructed #7 is also a counterexample for this claim.

More general, for every countable set, you can find a function that is discontinuous exactly at all points of this set.#3 is indeed surprising. All the inner limits are partial sums of an absolute convergent sum and have to exist. Without the absolute convergence it would be trivial.

 

[micromass]

Aha. The Cantor Set is the answer to 10. It has measure zero. It's uncountable, hence any function that has discontinuities on all the points of the Cantor Set cannot be Riemann-Integrable!

Furthermore, since it's nowhere dense, we can even find functions whose set of discontinuity points are the Cantor Set (eg its indicator function). But they won't be Riemann Integrable.

Why does the uncountability of the Cantor set matter? Why isn't the indicator function Riemann integrable?

 

[andrewkirk]

Why does the uncountability of the Cantor set matter? Why isn't the indicator function Riemann integrable?

Because then any function whose set of discontinuity points is the Cantor set has an uncountable discontinuity set, and a function with an uncountable discontinuity set cannot be Riemann integrable.

The indicator function is discontinuous on the Cantor set, and is hence not Riemann integrable.

 

[micromass]

Ba function with an uncountable discontinuity set cannot be Riemann integrable.

Why not?

 

[andrewkirk]

Why not?

It's not since its set of discontinuity points is uncountable.

(referring to fresh_42's suggestion in the preceding post that the indicator function of the rationals was Riemann integrable.)

 

[micromass]

(referring to fresh_42's suggestion in the preceding post that the indicator function of the rationals was Riemann integrable.)

Ah. Sorry, my post 59 is incorrect. What I should have said is that it's not Riemann integrable since its discontinuity set does not have measure zero. Sorry.

 

[andrewkirk]

Ah well, I shall have to think about it some more. It's late here and no doubt thinking about it will soon send me to sleep. Good night all.

 

[andrewkirk]

Good morning. I thought I'd just set out where I thought number 10 is up to, to help my own clarity of mind.

To be a counterexample, a set $S\subset[0,1]$ must have the following properties:

1. Measure of $S$ is zero. I presume here we mean Lebesgue measure (is that correct, Micromass?). If we were allowed to consider other measures then new avenues of inquiry might open up.
2. $S$ is uncountable [I'll explain why below].
   
3. One of these two must hold:
   1. There is no function from $[0,1]$ to $\mathbb R$ whose discontinuity set is $S$; or
   2. There may be such functions, but they are are not Riemann integrable.

$S=\mathbb Q$ fails on 2, and the Cantor Set fails 3.2 (as the Cantor Set's indicator function turns out to be Riemann integrable)

Here's why 2 is necessary. It is known that any function with a countable discontinuity set is R-integrable, and we can construct such a function as follows (a generalisation of my above proof for $\mathbb Q$):

For $S$ countable, with bijection $g:S\to\mathbb N$, define $f$ by $f(x)=x+\frac1{g(x)}$ for $x\in S$ and $f(x)$ otherwise. Then $f$ is discontinuous at $x\in S$ because $f(x)= x+\frac1{g(x)}\not= f(x)=\lim_{\substack{y\to x\\y\notin S}}f(y)$. But $f$ is continuous at $x\in[0,1]-S$ because for any $\epsilon>0$ we can choose $\delta>0$ such that (1) $\delta<\frac\epsilon 2$ and (2) for all $y=g^{-1}(n)\in S\cap (x-\delta,x+\delta)$ we have $n>\frac2\epsilon$, in which case we have

$$|f(y)-f(x)|\leq |f(y)-y|+|y-f(x)|=\frac1n+|y-x|< \frac\epsilon2+\frac\epsilon2=\epsilon$$

Currently, I think 2.1 may be more promising, if only because it has been less explored than 3.2 thus far in the thread.

The other possibility is that there is no counter-example. I only just noticed the clever rider in the OP that one of the ten propositions is actually true, which adds spice to the exercise. We could try to attack that by trying to prove the proposition, by making a construction of a suitable function, given a set $S$. The strong link between measure and integration would be the theme.

 

[micromass]

I presume here we mean Lebesgue measure (is that correct, Micromass?).

Correct.

Clever analysis of the situation in your post.

 

[andrewkirk]

Theorem
The indicator function of a subset of [0,1] with Lebesgue measure zero is Riemann integrable.

Proof. [Which contains an error, the identification of which is left as an exercise for anybody that prefers this sort of activity to Sudoku]

Let $S\in\mathbb R$ have Lebesgue measure zero and let $f$ be its indicator function. Then for any $\epsilon>0$ there exists a sequence of open real intervals $I_1,..,I_n$ that cover $S$ and $\sum_k |I_k|<\frac\epsilon2$. We can take closures of the intervals and remove overlaps to make this a sequence of intervals $J_1,...J_m$ (with $m\leq n$) such that the upper bound of each $J_k$ is less than or equal to the lower bound of $J_{k+1}$.

Now construct the following set of intervals $D_1...D_r$. First define $h=\frac\epsilon{4(m+1)}$. For each connected component $(a,b)$ of $[0,1]-\bigcup_k J_k$ define $l=\min(b,a+h)$ and $h=\max(l,b-h)$. This will be one or two points and so will divide $[a,b]$ into one, two or three non-trivial components. Where there are three non-trivial components, call the central component 'insulated' and the other two 'buffers'. Include the closures of the non-trivial components in the set of $D_k$.

Call the set of intervals obtained from insulated components $F_1,...,F_u$ (with $u\leq m+1$). These intervals are all entirely contained within $[0,1]-S$.

Then the sets of $J_k$ and $D_k$ together determine a partition $P$ of [0,1]. We tag this partition by selecting the lower bound of each interval as the tagged point.

Since each insulated interval $F_k$ is interior to $[0,1]-S$, the value of of $f(x)$ is zero everywhere on the interval, and remains so at all tagged points therein for any refinement of $P$. So for any refinement, the Riemann Sum for $f$ cannot exceed the measure of all the $J_k$ and the buffers. There are at most $2(m+1)$ buffers and the measure of each is no larger than $h$. So the sum of the lengths of the buffers is less than or equal to $2(m+1)h=2(m+1)\frac\epsilon{4(m+1)}=\frac\epsilon2$. And the sum of the lengths of all the $J_k$ is less than or equal to that for all the $I_k$, which is less than $\frac\epsilon2$.

Hence, since the indicator function is bounded above at 1, the Riemann Sum of any refinement of $P$ is less than $1\cdot\frac\epsilon2+1\cdot\frac\epsilon2=\epsilon$.

Hence $f$ is Riemann-integrable. $\square$

What's next:

So, unless I have made a mistake, this knocks out the possibility of 3.2. So if there is a counterexample to number 10, it must be an uncountable set of measure zero that is not the discontinuity set of any function.

We can stop thinking about Riemann Integrability.
EDIT: Ha Ha, no we can't, as the next post demonstrates.

 

[micromass]

Theorem
The indicator function of a subset of [0,1] with Lebesgue measure zero is Riemann integrable.

$\mathbb{Q}\cap [0,1]$ is a counterexample.

 

[andrewkirk]

Hmm. I see that I assumed the sequence of intervals $I_k$ is finite. But of course it may be infinite, in which case the proof does not work, as $\mathbb Q$ demonstrates.

 

[mfb]

I would be surprised if 3 has counterexample.

(10):
A counterexample set has to be uncountable, as discussed before.
If the set is dense nowhere, we can use the indicator function and integrate it. I would expect this to work for all sets. Therefore, our set has to be dense somewhere.

 

[fresh_42]

By the way: I like to say thank you to @micromass for your patience and constructive discussions. In fact I like to think that a discussion on errors is at least as valuable as a perfect solution for it shows and reminds me (and hopefully one or another reader as well) how easyly pitfalls can be overlooked.

 

[andrewkirk]

I believe I may have a proof that 3 is correct, in which case there is no counter-example. Here it is. I wonder if it has any fewer errors than my last attempted proof.

Theorem

If the sequence \sum_{n=0}^\infty a_n is absolutely convergent with limit $L$, and {\pi:\mathbb N\times\mathbb N\to\mathbb N} is a bijection, then
L&#039;\equiv\sum_{k=0}^\infty\sum_{l=0}^\infty a_{\pi(k,l)}=L.

Proof (attempted)

Given \epsilon&gt;0, use absolute convergence to choose N\in\mathbb N such that {\sum_{n=N+1}^\infty|a_n|&lt;\frac\epsilon3}. Then set

$$M\equiv\max_{k\in\{1,2,...,N\}}\left(\pi^{-1}(k)\right)_1$$

where the 1 subscript indicates the first component of an element of \mathbb N\times\mathbb N.

Then
\begin{align}
|L'-L|&=\left|\sum_{k=0}^\infty\sum_{l=0}^\infty a_{\pi(k,l)}-\sum_{n=0}^\infty a_n\right|\\
&=\left|\sum_{k=0}^M\sum_{l=0}^\infty a_{\pi(k,l)}+
\sum_{k=M+1}^\infty\sum_{l=0}^\infty a_{\pi(k,l)}
-\sum_{n=0}^N a_n
-\sum_{n=N+1}^\infty a_n\right|\\
&\leq
\left|\sum_{k=0}^M\sum_{l=0}^\infty a_{\pi(k,l)}-\sum_{n=0}^N a_n\right|+
\left|\sum_{k=M+1}^\infty\sum_{l=0}^\infty a_{\pi(k,l)}\right|+
\left|\sum_{n=N+1}^\infty a_n\right|\\
&\leq
\left|\sum_{k=0}^M\sum_{l=0}^\infty a_{\pi(k,l)}-\sum_{n=0}^N a_n\right|+
\sum_{k=M+1}^\infty\sum_{l=0}^\infty \left|a_{\pi(k,l)}\right|+
\sum_{n=N+1}^\infty \left|a_n\right|\\
&<
\left|\sum_{k=0}^M\sum_{l=0}^\infty a_{\pi(k,l)}-\sum_{n=0}^N a_n\right|+
\sum_{k=M+1}^\infty\sum_{l=0}^\infty \left|a_{\pi(k,l)}\right|+
\frac\epsilon3\\
&\leq
\left|\sum_{k=0}^M\sum_{l=0}^\infty a_{\pi(k,l)}-\sum_{n=0}^N a_n\right|+
\sum_{n=N+1}^\infty\left|a_n\right|+
\frac\epsilon3\\
&\leq
\left|\sum_{k=0}^M\sum_{l=0}^\infty a_{\pi(k,l)}-\sum_{n=0}^N a_n\right|+
\frac\epsilon3+
\frac\epsilon3\\
&=
\left|\sum_{k=0}^M\sum_{\substack{l=0\\\pi(k,l)>N}}^\infty a_{\pi(k,l)}\right|+
\frac{2\epsilon}3\\
&\leq
\sum_{k=0}^M\sum_{\substack{l=0\\\pi(k,l)>N}}^\infty \left|a_{\pi(k,l)}\right|+
\frac{2\epsilon}3\\
&\leq
\sum_{n=N+1}^\infty\left|a_n\right|+
\frac{2\epsilon}3\\
&<
\frac{\epsilon}3+
\frac{2\epsilon}3\\
&=\epsilon
\end{align}

So for any \epsilon&gt;0, we have |L&#039;-L|&lt;\epsilon, whence L&#039;=L. \square

 

[micromass]

How do you show

\sum_{k=M+1}^\infty\sum_{l=0}^\infty \left|a_{\pi(k,l)}\right| \leq \sum_{n=N+1}^\infty\left|a_n\right|

Don't you assume something like the theorem but for positive numbers here?

 

[andrewkirk]

I'm not sure, but what's needed in the context is to show that the LHS is less than or equal to $\frac\epsilon3$. I think we can do that as follows:

But first, we might need more room, so let's replace all the \frac\epsilon3 in the proof by \frac\epsilon{12}, except for the one you quoted. I'll edit that in shortly in the attempted proof post, if this works.

We want to prove that

$$\sum_{k=M+1}^\infty\sum_{l=0}^\infty \left|a_{\pi(k,l)}\right| \leq \frac\epsilon3$$

Since the limit of the outer sum exists, we can find K\in\mathbb N, with K\geq M, such that
$$\sum_{k=K+1}^\infty\sum_{l=0}^\infty\left|a_{\pi(k,l)}\right|
<\frac\epsilon{12}$$
For each k betweenM+1 and K inclusive we can then find T&#039;(k)\in\mathbb N such that \sum_{l=T&#039;(k)+1}^\infty \left|a_{\pi(k,l)}\right|&lt;\frac \epsilon{12(K+1)}. Set T\equiv\max_{k\in\{0,1,...,K\}}T_k.

Then
\begin{align*}
\sum_{k=M+1}^K\sum_{l=0}^\infty\left|a_{\pi(k,l)}\right|-
\sum_{k=M+1}^K\sum_{l=0}^T\left|a_{\pi(k,l)}\right|
&=\sum_{k=M+1}^K\left(\sum_{l=0}^\infty\left|a_{\pi(k,l)}\right|-
\sum_{l=0}^T\left|a_{\pi(k,l)}\right|\right)\\
&=\sum_{k=M+1}^K\left(\sum_{l=T+1}^\infty\left|a_{\pi(k,l)}\right|\right)\\
&\leq\sum_{k=M+1}^K\sum_{l=T'(k)+1}^\infty\left|a_{\pi(k,l)}\right|\\
&<\sum_{k=M+1}^K\frac \epsilon{12(K+1)}\leq\frac\epsilon{12}\\
\end{align*}So
\begin{align*}
\sum_{k=M+1}^\infty\sum_{l=0}^\infty\left|a_{\pi(k,l)}\right|&=
\sum_{k=M+1}^K\sum_{l=0}^\infty\left|a_{\pi(k,l)}\right|+
\sum_{k=K+1}^\infty\sum_{l=0}^\infty\left|a_{\pi(k,l)}\right|\\
&<\sum_{k=M+1}^K\sum_{l=0}^\infty\left|a_{\pi(k,l)}\right|+
\frac\epsilon{12}\\
&\leq\left( \sum_{k=M+1}^K\sum_{l=0}^T\left|a_{\pi(k,l)}\right|+\frac\epsilon{12}\right)
+\frac\epsilon{12}\\
&=\sum_{k=M+1}^K\sum_{l=0}^T\left|a_{\pi(k,l)}\right|+\frac\epsilon{6}\\
\end{align*}

The first term is a sum of terms that form a finite subset of a_{N+1},a_{N+2},... and hence cannot exceed \sum_{n=N+1}^\infty|a_n|=\frac\epsilon{12}.

Hence we have
$$\sum_{k=M+1}^\infty\sum_{l=0}^\infty\left|a_{\pi(k,l)}\right|<\frac\epsilon{12}+\frac\epsilon{6}=\frac\epsilon{4}<\frac\epsilon{3}$$

 

[micromass]

Neat! My only remark is that for this:

Since the limit of the outer sum exists, we can find K\in\mathbb N, with K\geq M, such that
$$\sum_{k=K+1}^\infty\sum_{l=0}^\infty\left|a_{\pi(k,l)}\right|
<\frac\epsilon{12}$$

you need a proof of convergence of the series. Can you provide that too?

 

[andrewkirk]

I think so. I've re-written the whole thing to make it a bit cleaner and integrate the last two posts. The proof of convergence of the absolute double sum is done at the start. I hope it makes sense now.

Theorem

If the sequence \sum_{n=1}^\infty a_n is absolutely convergent with limit L, and {\pi:\mathbb N\times\mathbb N\to\mathbb N} is a bijection, then L&#039;\equiv\sum_{k=1}^\infty\sum_{l=1}^\infty a_{\pi(k,l)} is a convergent sum, which is equal to $L$.

Proof

First we prove that \sum_{k=1}^\infty\sum_{l=1}^\infty|a_{\pi(k,l)}| exists.
If it does not exist then either one or more of the inner sums is divergent, or the inner sums converge but the outer sum diverges.

First assume that the inner sum for k diverges. Then for any R&gt;0 there exists G\in\mathbb N such that \sum_{l=1}^G|a_{\pi(k,l)}|&gt;R. Then \sum_{n=1}^N|a_n|&gt;R where
$$N\equiv\max_{l\in\{1,...,G\}}\pi(k,l)$$
which contradicts the absolute convergence of \sum_{n=0}^\infty a_n. So no inner sum can diverge.

Next assume the outer sum diverges. Then for any R&gt;0 we can find G\in\mathbb N such that

$$\sum_{k=1}^G\sum_{l=1}^\infty|a_{\pi(k,l)}|>R+1$$

Since there is a finite number of inner sums and they all converge, we can find F\in\mathbb N such that for all l\in\{1,...,F\} we have \sum_{l=F+1}^\infty|a_{\pi(k,l)}|&lt;\frac1G.

Then we have

$$\sum_{k=1}^G\sum_{l=1}^F|a_{\pi(k,l)}|
=\sum_{k=1}^G\sum_{l=1}^\infty|a_{\pi(k,l)}|-
\sum_{k=1}^G\sum_{l=F+1}^\infty|a_{\pi(k,l)}|
>R+1+G\cdot\frac1G=R$$

Then \sum_{n=1}^N|a_n|&gt;R where
$$N\equiv\max_{\substack{k\in\{1,...,G\}\\l\in\{1,...,F\}}}\pi(k,l)$$
which contradicts the absolute convergence of \sum_{n=0}^\infty a_n. So the outer sum cannot diverge either. Hence the double sum of absolute values converges.

We see then that the sum \sum_{k=1}^\infty\sum_{l=1}^\infty a_{\pi(k,l)} must also converge to a limit L&#039; because sums of absolute values are no less than sums of signed numbers, so if a signed sum diverges then the sum formed by applying the same summation operators to the absolute values of the summands must also diverge.

Now we try to prove the result.

First, given \epsilon&gt;0, we choose N\in\mathbb N such that \sum_{n=N+1}^\infty|a_n|&lt;\frac\epsilon9, which we can do by the absolute convergence of that series.

Next we set

$$H\equiv\max_{k\in\{1,2,...,N\}}\left(\pi^{-1}(k)\right)_1$$

where the 1 subscript indicates the first component of an element of \mathbb N\times\mathbb N.

Next we choose G\in\mathbb N such that \sum_{k=G+1}^\infty\sum_{l=1}^\infty|a_{\pi(k,l)}|&lt;\frac\epsilon9, which we can do by the convergence of the double sum, proven above.

Then define M\equiv\max(G,H).

Next, for each k\in\{1,...,M\} we choose T&#039;(k) such that {\sum_{l=T&#039;(k)+1}^\infty|a_{\pi(k,l)}|&lt;\frac\epsilon{9M}}, which again we can do by convergence of the double sum above. We then set T\equiv\max_{k\in\{1,...,M\}}T&#039;(k).

Armed with these definitions, we proceed as follows:

\begin{align}
|L'-L|&\equiv\left|\sum_{k=1}^\infty\sum_{l=1}^\infty a_{\pi(k,l)}-\sum_{n=1}^\infty a_n\right|\\
&=\left|\left(\sum_{k=1}^M\sum_{l=1}^T
+\sum_{k=1}^M\sum_{l=T+1}^\infty
+\sum_{k=M+1}^\infty\sum_{l=1}^\infty
\right)a_{\pi(k,l)}
-\left(\sum_{n=1}^N+\sum_{n=N+1}^\infty\right) a_n\right|\\
&\leq\left|\sum_{k=1}^M\sum_{l=1}^T a_{\pi(k,l)}-\sum_{n=1}^N a_n\right|
+\left|\sum_{k=1}^M\sum_{l=T+1}^\infty a_{\pi(k,l)}\right|
+\left|\sum_{k=M+1}^\infty\sum_{l=1}^\infty a_{\pi(k,l)}\right|
+\left|\sum_{n=N+1}^\infty a_n\right|\\
&<\left|\sum_{k=1}^M\sum_{\substack{l=1\\\pi(k,l)>n}}^T a_{\pi(k,l)}\right|
+\sum_{k=1}^M\sum_{l=T+1}^\infty \left|a_{\pi(k,l)}\right|
+\sum_{k=M+1}^\infty\sum_{l=1}^\infty \left|a_{\pi(k,l)}\right|
+\sum_{n=N+1}^\infty \left|a_n\right|\\
&<\sum_{k=1}^M\sum_{\substack{l=1\\\pi(k,l)>N}}^T \left|a_{\pi(k,l)}\right|
+\sum_{k=1}^M\sum_{l=T'(k)+1}^\infty \left|a_{\pi(k,l)}\right|
+\sum_{k=G+1}^\infty\sum_{l=1}^\infty \left|a_{\pi(k,l)}\right|
+\frac\epsilon9\\
&<\sum_{n=N+1}^\infty|a_n|
+\sum_{k=1}^\infty\frac\epsilon{9M}+\frac\epsilon9
+\frac\epsilon9\\
&=\frac{\epsilon}9
+\frac{\epsilon}3=\frac{4\epsilon}9<\epsilon\\
\end{align}

Since |L&#039;-L|&lt;\epsilon for all positive \epsilon, it must be zero.

Phew!

 

[micromass]

I think that's ok. So only 10 remains.

I'll give another proof though that some people might find insightful. The idea is that if $(\Omega,\mathcal{B},\mu)$ is a measure space and $\Omega=\bigcup_{n} S_n$ is a partition of measurable sets, then it is not difficult to prove that for an integrable function $f:\Omega\rightarrow \mathbb{R}$ holds
\int_\Omega fd\mu = \sum_n \int_{S_n} fd\mu
Now, the result follows easily if we take $\mu$ to be the counting measure on $\mathbb{N}$. In this case, $f$ becomes a sequence which is integrable iff the series converges absolutely and whose integral is a sum.

 

[andrewkirk]

For 10, just wanted to check something else. I presume Vitali sets are ruled out, since they are not Lebesgue measurable. A given Vitali set must have a Lebesgue Outer Measure (since any subset of $\mathbb R$ does) and, for all I know, it could be zero. And I imagine it may be straightforward to show that any function whose discontinuity set is a Vitali set is not Riemann integrable. But a Vitali set is not Lebesgue measurable.

I mention this because Vitali sets are the only uncountable sets I know of, other than the Cantor set, that are not known (by me) to have positive Lebesgue Outer Measure. They also have the advantage of being dense and, as @mfb pointed, out a counterexample to 10 has to be dense somewhere. In fact, I think it will have to be dense at an infinite number of points (ie have an infinite set of limit points), and maybe even at an uncountably infinite number of points.

I think the solution to this problem, when it turns up, will be a most intriguing set.

 

[micromass]

For 10, just wanted to check something else. I presume Vitali sets are ruled out, since they are not Lebesgue measurable. A given Vitali set must have a Lebesgue Outer Measure (since any subset of $\mathbb R$ does) and, for all I know, it could be zero. And I imagine it may be straightforward to show that any function whose discontinuity set is a Vitali set is not Riemann integrable. But a Vitali set is not Lebesgue measurable.

I mention this because Vitali sets are the only uncountable sets I know of, other than the Cantor set, that are not known (by me) to have positive Lebesgue Outer Measure. They also have the advantage of being dense and, as @mfb pointed, out a counterexample to 10 has to be dense somewhere. In fact, I think it will have to be dense at an infinite number of points (ie have an infinite set of limit points), and maybe even at an uncountably infinite number of points.

I think the solution to this problem, when it turns up, will be a most intriguing set.

Right, the set in question must have an actual measure. The Vitali set does not have a measure.

Also, the Vitali sets have positive outer measure. If you construct a Vitali set inside $[0,1]$, then you can construct one which has outer measure a number inside $(0,1]$. Note that if you show something has outer measure $0$, then it actually has measure zero.

 

[micromass]

OK, let me perhaps give a series of hints to prove 10:

1) Prove that the set of discontinuity points of any function is an $F_\sigma$ set. This means that it is the countable union of closed sets.
2) Find a set of measure zero that is not an $F_\sigma$ set. This can be done in two ways (each yielding their own counterexample), by either noticing that $F_\sigma$ sets are always Borel sets, or by proving that $F_\sigma$ sets of measure zero are necessarily of first category. https://en.wikipedia.org/wiki/Meagre_set
3) So it suffices to find a measurable set that is not Borel. Or it suffices to find a set of measure zero that is not of first category.