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Let $\left\{r_1, r_2, \ldots \right\}$ be an enumeration of all the rationals in $[0,1]$ and define $$ g_n(x) = \begin{cases} 1 & \text{if} \ x = r_n \\ 0 & \text{otherwise} \end{cases} $$

a) Is $G(x) = \sum_{n=1}^{\infty} g_n(x)$ (Riemann) integrable on $[0,1]$?

b) Is $F(x) = \sum_{n=1}^{\infty} g_n(x) / n$ integrable on $[0,1]$?

For part a), I think not, but I'm not sure how to prove this. Is $G(x)$ even bounded on $[0,1]$? How do I know this series will converge? For each subinterval $[x_{k-1}, x_k]$, let $m_k = \inf\left\{f(x) \mid x \in [x_{k-1}, x_k] \right\}$ and likewise $M_k$ for the supremum. Can I then just say that, if $P$ is some partition of $[0,1]$, by the density of the rationals we will always have $M_k = 1$ and $m_k = 0$? And so the upper and lower Riemann sum are always different and hence $G$ is not integrable.

For part b) I took as my partition $P_n$ the one that divides $[0,1]$ in equal subintervals. So $x_k = a + k(b-a)/n$. Then $$x_k - x_{k-1} = \frac{b-a}{n}. $$ Then I think I will have $$ \lim_{n \to \infty} [U(F,P_n) - L(F,P_n)] = 0$$ and hance $F$ is Riemann integrable.

Help and/or feedback is appreciated.

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    $\begingroup$ Your answer to (a) is correct. Nothing you have written in (b) is false, but you haven't actually given an argument as to why it is true. $\endgroup$
    – Mees de Vries
    Aug 5, 2016 at 10:27
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    $\begingroup$ We know that $G(x)$ converges and is very much bounded, because it turns out to become $1_{\Bbb Q}$, that is $$G(x) = \cases{1 & if $x$ is rational\\0 & if $x$ is irrational}$$on the interval $[0,1]$. $F(x)$ is somewhat similar, except the value at each rational number decreases according to its index in the enumeration $r_i$. $\endgroup$
    – Arthur
    Aug 5, 2016 at 10:53
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    $\begingroup$ @Arthur. So what is the conclusion? Is $F(x)$ integrable then? $\endgroup$
    – Kamil
    Aug 5, 2016 at 12:24
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    $\begingroup$ Actually, yes, it is, because it's discontinuous only at the rationals and bounded. But you need to show that. $\endgroup$
    – Arthur
    Aug 5, 2016 at 13:16
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    $\begingroup$ Hint for (b): Let $s\in[0,1]\setminus \mathbb Q$ and $r_{n_k}\to s$ for $k\to\infty$. By compactness you may assume wlog $1/n_k$ converges. Can the limit be non-zero? $\endgroup$
    – user251257
    Aug 11, 2016 at 0:15

1 Answer 1

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The function $G$ takes the values $0$ and $1$ hence is bounded. Your argument is correct and proves that this function is not Riemann-integrable.

For $F$: fix a positive $\varepsilon$ and let $N$ be such that $1/N\lt\varepsilon$. Define the step function $f_1$ and $f_2$ by $f_1=0$ and $f_2=1/N$, except on $r_1,\dots,r_N$, where the value is $1$. Then $f_1\left(x\right)\leqslant F\left(x\right)\leqslant f_2\left(x\right)$ for any $x\in[0,1]$ and the Riemann integral of $f_2-f_1$ does not exceed $\varepsilon$. Remark that there is nothing specific about $1/n$; the function $x\mapsto\sum_{n=1}^{+\infty}\delta_n g_n\left(x\right)$ is Riemann-integrable on $[0,1]$ for any sequence $\left(\delta_n\right)_{n\geqslant 1}$ converging to $0$.

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