# Are these functions Riemann integrable on $[0,1]$?

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.

• 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. Aug 5, 2016 at 10:27
• 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 $r_i$. Aug 5, 2016 at 10:53
• @Arthur. So what is the conclusion? Is $F(x)$ integrable then? Aug 5, 2016 at 12:24
• Actually, yes, it is, because it's discontinuous only at the rationals and bounded. But you need to show that. Aug 5, 2016 at 13:16
• 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? Aug 11, 2016 at 0:15

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 $0$.