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.
Best Answer
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 $0$.