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Consider the function $f: [0,1] \to \mathbb{R}$ where $f(x)=$

\begin{cases} \frac 1q & \text{if } x\in \mathbb{Q} \text{ and } x=\frac pq \text{ in lowest terms}\\ 0 & \text{otherwise} \end{cases}

Determine whether or not $g$ is in $\mathscr{R}$ on $[0,1]$ and prove your assertion. For this problem you may consider $0= 0/1$ to be in lowest terms.

Here's an attempt. I may have abused a bit of notation here, but the ideas are there.

Proof:
Let $M_i = \sup \limits_{x \in [x_{i-1},x_i]} f(x)$.

Notice first that the lower Riemann sums are always $0$, since every interval contains an irrational number. Thus, to prove $f \in \mathscr{R}$, it suffices to prove that, given any $\epsilon >0$, $\sum \limits_{i \in P} M_i \Delta x_i < \epsilon$ for some partition.

Let $\epsilon > 0 $ and $M > \frac{2}{\epsilon}$. We first show that there exists $\eta(x,\frac{1}{M})$ so that $|f(x) - f(y)| < \frac{1}{M}$ if $|x-y| < \eta$. Fix $x \in (\mathbb{R} \setminus \mathbb{Q}) \cap [0,1]$. Now, consider the set $$R_{M} := \{ r \in \mathbb{Q} : r = \frac{p}{n}, n \leq M, p \leq n, p \in \mathbb{N} \}.$$ Clearly this set is finite, enumerate it as $\{q_1,\ldots, q_m\}$. So, let $$\eta(x,\frac{1}{M}) = \min_{i=1,\ldots, m} |x- q_i|.$$ We see then, $|f(x) - f(y)| < \frac{1}{M}$ on this $\eta$-neighborhood.

After we choose that $\eta$ so that $x \in (\mathbb{R} \setminus \mathbb{Q}) \cap [0,1]$, is continuous in a $\eta$-neighborhood, we see $$ A:= [0,1] \setminus R_M \subset \left( \bigcup_{ x \in ( \mathbb{R} \setminus \mathbb{Q}) \cap [0,1]} B_{\eta(x)} (x) \right) \cap [0,1].$$ Since $A$ is compact, we may take finite sub-covering, and let $\delta = \min \limits_{i=1,\ldots,n} \{\eta(x_i)\}$. Take a partition $P_1$ of $A$ so that $\Delta x_i < \delta$. Since $R_M$ is non-empty, we can take a partition $P_2$ of $R_M$ so that $\Delta x_i < \frac{\epsilon}{2m}.$ Moreover, we see that, on $[0,1]$, $f$ is at most $1$. Let $P = P_1 \cup P_2$. Thus,

\begin{eqnarray*} \sum_{i \in P} M_i \Delta x_i &=& \sum_{i \in P_1} M_i \Delta x_i + \sum_{i \in P_2} M_i \Delta x_i \\ &\leq& \frac{1}{M} \sum_{i \in P_1} \Delta x_i + \sum_{i \in P_2} \Delta x_i \\ &<& \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{eqnarray*}

Comments?

EDITED I think I resolved the issue.

Starlight
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Anthony Peter
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5 Answers5

42

For an alternative approach, choose $N \in \mathbb{N}$ such that $1/(N+1) < \epsilon/2$ and let

$$B_N = \left\{1,\frac{1}{2},\frac{1}{3},\frac{2}{3}, \ldots, \frac{1}{N}, \ldots, \frac{N-1}{N} \right\}.$$

If $x \notin B_N$, then $f(x) \leqslant 1/(N+1) < \epsilon/2$. With $m = \#(B_N)$, choose a partition $P = (x_0,x_1, \ldots, x_n)$ where $n > m$ and $\|P\| < \frac{\epsilon}{4m}.$ There are at most $2m$ subintervals such that $[x_{j-1},x_j] \cap B_n \neq \varnothing.$

Let $M_j = \sup_{x \in [x_{j-1},x_j]} f(x).$

Then, the upper sum satisfies

$$U(P,f) = \sum_{j=1}^n M_j(x_j-x_{j-1}) = \\ \sum_{[x_{j-1},x_j] \cap B_N \neq \varnothing} M_j(x_j-x_{j-1}) + \sum_{[x_{j-1},x_j] \cap B_N = \varnothing} M_j(x_j-x_{j-1}) \\ \leqslant 2m \cdot 1 \cdot \frac{\epsilon}{4m} + \frac{1}{N+1}(1-0) \\ \leqslant \epsilon.$$

rae306
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RRL
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  • I have one question about your notation. What means $|P| < \frac{\epsilon}{4m}$ and $m = #(B_N)$? – RFZ Nov 12 '15 at 22:42
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    @RFZ: The partition norm is $|P| = \max_{1 \leqslant i \leqslant n}(x_i-x_{i-1})$ and $#(A)$ means the number of elements in the set $A$. – RRL Nov 13 '15 at 15:18
  • Thanks a lot! By the way nice proof! +1 – RFZ Nov 13 '15 at 16:48
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    Hi, @RRL! Can you explain why there are fewer than $2m$ subintervals such that $[x_{j-1},x_j] \cap B_n \neq \phi.$ ? Thanks! – idk Jul 15 '18 at 21:57
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    @idk. The set $B_N$ contains $m$ points. The partition is assumed to have more than $m$ subintervals. The $m$ points can belong to at most $2m$ subintervals and this happens if each point is at the intersection of a pair of intervals (for m distinct pairs). I could have worded this better - it should say "at most $2m$" rather than "fewer than $2m$ in the answer. – RRL Jul 15 '18 at 22:27
  • @RRL, how is the case of $2m$ many subintervalls possible? I think at most we can get $2m-1$ many subintervalls because $1$ can only be the endpoint of a subintervall whereas all the other elements of $B_N$ can be the endpoint of one subintervall and the startpoint of another one. Or am I missing something? – Philipp Mar 08 '21 at 00:00
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    @Philipp: All that matters is that the contribution from summing over the intervals containing the $m$ points in $B_N$ is bounded by the maximum possible number of intervals times $\frac{\epsilon}{4m}$ and this is less than $\frac{\epsilon}{2}$ regardless of whether this number of intervals is $2m-1$ or $2m$. You are probably correct. I just threw out the estimate $2m$ to be conservative. – RRL Mar 08 '21 at 19:07
  • @ZFR read the comments here – Marios Gretsas Sep 17 '21 at 15:54
  • I still cannot get why the number of intervals $\Delta_j=[x_{j-1},x_j]$ such that $\Delta_j\cap B_N\neq \varnothing$ is at most $2m$. And why your partition should have $>m$ intervals? Can you explain please? – RFZ Sep 17 '21 at 16:27
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    @ZFR: To prove Riemann integrability it is sufficient (given any $\epsilon > 0$) to find just a single a partition $P$ such that $U(P,f) - L(P,f) < \epsilon$. We know that every lower sum is zero so this reduces to finding a $P$ such that $U(P,f) < \epsilon$. I have taken a partition with $n$ subintervals that is arbitrary except for one condition: $|P| < \epsilon/4m$. The contribution to the upper sum from subintervals that do not meet $B_N$ was shown above to be no more than $\epsilon/2$. – RRL Sep 17 '21 at 16:58
  • Now it remains to estimate the contribution from the other subintervals that do contain the $m$ points in $B_N$. There are say $p \leqslant n$ of these. The contribution to the upper sum is $$\sum_{[x_{j-1},x_j] \cap B_N \neq \varnothing} M_j(x_j-x_{j-1}) \leqslant p \cdot 1 \cdot \frac{\epsilon}{4m}$$ How big can $p$ possibly be? – RRL Sep 17 '21 at 17:00
  • If $n < m$, then the contribution is less than $(p/m)\epsilon/4 < \epsilon/2$. If $n \geqslant m$, then $p$ is at most $2m$ which corresponds to every one of the $m$ points being an endpoint of a subinterval and the count of containing subintervals is $2m$. – RRL Sep 17 '21 at 17:03
  • @RRL, thank you so much for your reply! I really appreciate your time. 1) if $n<m$ then the contribution is $\leq \frac{\epsilon}{4m}\cdot n<\frac{\epsilon}{4m}\cdot m=\epsilon/4$, right? – RFZ Sep 17 '21 at 17:09
  • Yes -- $p \cdot \frac{\epsilon}{4m} \leqslant n \cdot \frac{\epsilon}{4m} = \frac{\epsilon}{4}$. Otherwise if $p > m$ then $p \leqslant 2m$ and ... – RRL Sep 17 '21 at 17:11
  • Let me ask you one more question please: I really did not get why $p\leq 2m$. That it my issue. is it possible to prove it in a rigorous way? – RFZ Sep 17 '21 at 17:15
  • I was wondering is it possible to prove by contradiction or by some combinatorics or using summation with some manipulations with Iverson bracket. If you can help me it would be great! – RFZ Sep 17 '21 at 17:32
  • You have $n$ non-overlapping closed subintervals. They are disjoint except for $n-1$ endpoints in $(a,b)$. Now you ask how many of these intervals can contain the $m$ distinct points in $B_N$? Consider the case where $n \geqslant m$. Some pigeonhole argument probably applies. To assign the $m$ points to the maximum number of subintervals, note that one point can belong to no more than two subintervals, and once one point is assigned to an interior endpoint (and hence belongs to two subintervals), no other point can be assigned there. – RRL Sep 17 '21 at 18:49
  • There are at most $m-1$ points belonging to $2(m-1) = 2m-2$ subintervals assigned this way. The remaining point can be assigned to an unoccupied subinterval if there is one or an already occupied one. Thus at most $2m-1$ subintervals can contain the $m$ points. – RRL Sep 17 '21 at 18:51
  • You wrote that we have $n$ non-overlapping closed subintervals. But two such adjacent intervals intersect each other in endpoint. What do you mean by "they are disjoint except for $n-1$ endpoints in $(a,b)$". The phrasing is not so clear – RFZ Sep 17 '21 at 19:04
  • @ZFR: We are really descending into minutiae here. If two closed intervals are not disjoint, then the intersection is exclusively either a singleton or a non-degenerate interval. I call the former non-overlapping and the latter overlapping. This may not be familiar terminology to you but I have seen and used it many times. I have precisely described the situation where the $n$ intervals are $[x_0,x_1],[x_1,x_2], \ldots, [x_{n-1},x_n]$ with $x_0 < x_1 < \ldots < x_n$ -- that is a partition. The points that lie in the non-empty pairwise intersections are $x_1, \ldots, x_{n-1}$. – RRL Sep 17 '21 at 19:19
  • Those $n-1$ points -- $x_1, \ldots, x_{n-1}$ -- are what I mean by the $n-1$ endpoints in $(a,b)$. They are endpoints, there are $n-1$ of them, and they are in the open interval $(a,b)$. – RRL Sep 17 '21 at 19:22
  • Sorry but can we move to the chat? – RFZ Sep 17 '21 at 19:24
  • Let us continue this discussion in chat. – RFZ Sep 17 '21 at 19:24
  • Have you ever used the fact that $n>m$? – RFZ Sep 19 '21 at 16:32
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    @ZFR: It does not matter. The proof uses the fact that a partition is taken with norm less than $\frac{\epsilon}{4m}$. The number of subintervals is $n$ which can be arbitrary as long as the norm constraint is satisfied. The contribution to the upper sum from subintervals that contain the $m$ pojnts in $B_N$ is bounded above by $1 \cdot \frac{\epsilon}{4m}$ times the number of such intervals. There can be no more than $n$ subintervals, so if $n \leqslant m$ the bound is $n \cdot 1 \cdot \frac{\epsilon}{4m} < \frac{\epsilon}{2}$ as desired. If $n > m$ then the the other argument applies. – RRL Sep 19 '21 at 17:53
  • Where did the (1-0) come from in this equation? – dan Apr 11 '23 at 23:51
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    @dan: Note that $\sum_{[x_{j-1},x_j] \cap B_N = \varnothing} M_j(x_j-x_{j-1})$ is over intervals that are disjoint from $B_N$ and $M_j \leqslant \frac{1}{N+1}$ and ,thus, that sum is less than or equal to $\frac{1}{N+1}\cdot \sum_{[x_{j-1},x_j] \cap B_N = \varnothing} (x_j-x_{j-1}) $ The sum is taken over subintervals of a partition of the interval $[0,1]$. Hence,

    $$\sum_{[x_{j-1},x_j] \cap B_N = \varnothing} (x_j-x_{j-1})\leqslant \sum_{j=1}^m (x_j-x_{j-1}) = (1-0)$$

     – RRL Apr 13 '23 at 16:38
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I just wanted to note that using Lebesgue's criterion for Riemann integrability, the proof is easy:

The criterion says that a bounded function on a compact interval $[a,b]$ is Riemann integrable if and only if it is continuous almost everywhere (the set of its points of discontinuity has measure zero).

Since the rationals have measure zero, it is enough to show $f$ is continuous in at all irrational points.

Let $x \in \mathbb{R} \setminus \mathbb{Q}$. Then $f(x)=0$. Suppose $x_n \to x$. We need to show $f(x_n) \to 0$. We can assume W.L.O.G that all the $x_n$ are rational. Since a sequence of rationals with an irrational limit have denominators going to infinity, we get the desired result.

Admittedly, Lebesgue's criterion is a rather "heavy gun" to use here, but for those who already know it, it is very nice:)

Asaf Shachar
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Yall, I think you're making this harder than it has to be.

First, $\overline\int_a^b f = \overline\int_a^b f\cdot \mathbf 1_{\{x\}^c}$. That is to say, you can "delete" any single point that you want, and the upper integral is unchanged.

Now obviously by the ML bound, $\overline\int_0^1 f\le 1$.

Delete the point at $x=1$ and call $f_1 = f\cdot \mathbf 1_{\{1\}^c}$. Then $\overline \int_0^1 f = \overline\int_0^1 f_1 \le \frac 1 2$.

Delete the point at $x=1/2$ and the integral is shown to be at most $\frac 1 3$.

Continue likewise, or more formally, let $\varepsilon\in\Bbb R^+$ and show that $\overline\int_0^1 f < \varepsilon$.

Addem
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The lowest sums are easily seen to be zero. We are going to prove that with a suitable selection of the partitions the upper sums tend to zero. For this, take $p$ a prime number and let $k=\max \{n \mid 1+\frac{n(n+1)}{2}\leq p\}$. So $p=1+\frac{k(k+1)}{2}+q$, where $0\leq q< k+1$. We consider now the partition $P=\{0<1/p<2/p<...<(p-1)/p<1\}$ and let $m_i=\sup\{f(x)\mid x\in [(i-1)/p,i/p]$; $i=0,1,...p-1\}$. Now we reorder the sequence $M=\{m_1,m_2,...,m_p\}$ in decreasing order (not strictly) to obtain the finite sequence of $p$ terms $R=\{1,....1/p\}$. $1$ appear as the supreme in the subinterval $[(p-1)/p,1]$ and $1/p$ as the supreme in the subinterval $[0,1/p]$. Then we have

A-) If $a<b<p$ and $a/b\in [(l-1)/p,l/p]$ then $a/b\in ((l-1)/p,l/p)$. In other words, with the above conditions $a/b$ can not be of the form $n/p, n<p$ because if so, $p\mid bn \Rightarrow p\mid b$ or $p\mid n$ which is absurd since $b,n<p$. This prevent $1/b$ from being the supreme of two consecutive subintervals. For example if $p=4$ then $1/2$ is the supreme in the subintervals $[1/4,1/2]$ and $[1/2,3/4]$ and we don't want this to happen. That's the reason for which we choose p to be a prime number.

B-) As a consequence of A-) we have then that $1/b$ could appear at most $b-1$ times in the sequence $M$ (or $R$). Therefore, the first $1+k(k+1)/2$ terms of $R$ can be upper bounded (not strictly) term-to-term by the sequence $\{1,1/2,1/3,1/3,1/4,1/4,1/4,...,1/(k+1),...,1/(k+1)\}$ (note that in fact there are $1+1+2+...+k=1+k(k+1)/2$) terms). For the remaining $q$ terms we can take $1/(k+2)$ as an upper bound. Then we have

C-) $$U(f;P)\leq \frac{1}{p}[1+1/2+2/3+...+k/(k+1)+q/(k+2)]=\frac{1}{p}[1+1-1/2+1-1/3+...+1-1/(k+1)+q/(k+2)]=\frac{1}{p}[k+1-S+q/(k+2)]$$ where $S=1/2+...+1/(k+1)$. Since $S>0$, $p>\frac{k(k+1)}{2}$ and $q<k+1$,then $$U(f;P)<\frac{1}{p}[k+1+q/(k+2)]<\frac{2}{k(k+1)}[k+1+q/(k+2)]=\frac{2}{k}+\frac{q}{k(k+1)(k+2)}<\frac{2}{k}+\frac{1}{k(k+2)}$$

If $p\rightarrow \infty \Rightarrow k\rightarrow \infty \Rightarrow \frac{2}{k}+\frac{1}{k(k+2)}\rightarrow 0$. So we have proved that if $p\rightarrow \infty \Rightarrow U(f;P)\rightarrow 0 $

RAUL PEREZ
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Here is an approach which is not that lengthy. We first introduce the sandwich principle: $f:[a,b]\to \mathbb R$ is integrable if for any $\epsilon>0$, there exist integrable $g,h$ such that $g\leq f\leq h$ and $\int_a^b h-g<\epsilon$.

Then for Thomae's function $T$ and a given $\epsilon>0$, $0\leq T\leq \epsilon \chi_{[0,1]}+t(x)$, where $$t(x)=\begin{cases}\frac{1}{q},& x=\frac{p}{q}\in \mathbb Q\cap[0,1] \text{ and }\frac{1}{q}\geq \epsilon;\\0,&\text{otherwise.}\end{cases}$$ Note that $t(x)$ has only finite discontinuity (which makes $t(x)$ integrable), and $\int \epsilon\chi_{[0.1]}+t(x)dx=\epsilon$. Thus, by sandwich principle, $T$ is integrable with $\int T=0$.

SuperSupao
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