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.

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*}


EDITED I think I resolved the issue.

  • $\begingroup$ What's $\mathscr{R}$? $\endgroup$ – Silvia Ghinassi Sep 9 '15 at 4:36
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    $\begingroup$ Sorry, it's the class of riemann integrable functions. We're supposed to prove this without using riemann lebesgue $\endgroup$ – Anthony Peter Sep 9 '15 at 4:38
  • $\begingroup$ I think there may be an issue with the statement that these neighborhoods form an open cover $\endgroup$ – Anthony Peter Sep 9 '15 at 5:23
  • $\begingroup$ See Petrovic, Advanced Calculus, Theory and Practice, (2014), p. 146. $\endgroup$ – Tony Piccolo Sep 9 '15 at 5:38
  • $\begingroup$ @TonyPiccolo Unfortunately the link seems to be unavailable $\endgroup$ – Anthony Peter Sep 9 '15 at 6:27

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

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

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    $\begingroup$ @RFZ: The partition norm is $\|P\| = \min_{1 \leqslant i \leqslant n}(x_i-x_{i-1})$ and $\#(A)$ means the number of elements in the set $A$. $\endgroup$ – RRL Nov 13 '15 at 15:18
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    $\begingroup$ @Fianra: Yes -- $\max$ instead of $\min$ in the comment which I can no longer edit. Thank you. $\endgroup$ – RRL Jan 31 '18 at 20:18
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    $\begingroup$ 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! $\endgroup$ – idk Jul 15 '18 at 21:57
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    $\begingroup$ @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. $\endgroup$ – RRL Jul 15 '18 at 22:27
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    $\begingroup$ @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. $\endgroup$ – RRL Mar 8 at 19:07

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:)


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