Here's the question: Let $\left\{q_1, q_2, \ldots\right\}$ be an enumeration of all rational numbers in the interval $[0,1]$. This means that $q_k$ 's are rational numbers in $[0,1]$ and that every rational number $q$ in $[0,1]$ is equal to exactly one $q_k$. Now for any $x \in[0,1]$, let $S_x=\left\{k: q_k \leq x\right\}$. Fix a sequence $\left\{p_n\right\}_{n \geq 1}$ of strictly positive real numbers such that $\sum_n p_n=1$. Define a function $h:[0,1] \rightarrow[0,1]$ by $h(x)=\sum_{k \in S_x} p_k$. In other words, $h(x)$ is the sum of $p_k$ 's where corresponding $q_k$ 's are in $[0, x]$. (a) Show that $h$ is continuous at all irrational points and is discontinuous at all rational points. In particular, $h$ has infinitely many discontinuity points. (b) Show that $h$ is Riemann integrable. (c) Compute $\int^1 h$
I'm done with the first two parts, the first one following from a jump at rationals and second one following from monotonous nature of the function $h$. I need hints for the third part. I'm unable to come up with a proper argument in order to restrict the integral to a value. I would appreciate some help.
Best Answer
Let $P_n$ be the partition including $0,1$ and $q_1,...,q_n$. We denote this partition by $P_n=\{t_0=0,t_1,...t_n,t_{n+1}=1\}$. Moreover, let $s_i$ be the term in $\sum p_k$ corresponding to $t_i$ for $1\leq i\leq n$.
Observe that
$h(t_i)\geq \sum_{j=1}^is_j$ and
$h(t_i)+\sum_{j=i}^ns_j\leq 1$ $\implies h(t_i) \leq 1-\sum_{j=i}^ns_j$
We can now bound the upper and lower Riemann sums.
$$L(f,P_n)\geq \sum_{i=1}^nh(t_i)(t_{i+1}-t_i) $$$$\geq \sum_{i=1}^n\left((t_{i+1}-t_i)\sum_{j=1}^i s_j\right)=\sum_{i=1}^n(s_i-t_is_i)$$
This last sum is just a rearrangement of $\sum_{k=1}^np_k-p_kq_k$.
$$U(f,P_n)\leq \sum_{i=0}^nh(t_{i+1})(t_{i+1}-t_i)$$$$\leq \sum_{i=0}^n(t_{i+1}-t_i)(1-\sum_{j=i+1}^ns_j)$$$$=1- \sum_{i=0}^n\left((t_{i+1}-t_i)\sum_{j=i+1}^ns_j\right)$$$$ = 1-\sum_{i=1}^ns_it_i=1-\sum_{k=1}^np_kq_k.$$
Taking limits as $n\to\infty$, we find the desired integral to be $$\fbox{1-$\sum p_kq_k$}.$$