Let $I$ be an index set and $a:I\to (0,\infty)$.
a.) Show that if $$\sum_{i\in I}a(i):= \sup_{J\subset I,J \text{finite}}\sum_{i\in J}a(i) < \infty$$ then $I$ is countable.
b.) Suppose $I = \mathbb{Q}$ and that $\sum_{q\in \mathbb{Q}}a(q) < \infty$. Show that the function $f$ defined by $$f(x):= \sum_{q\in\mathbb{Q},q\leq x}a(q)$$ is continuous at $x$ if and only if $x\notin \mathbb{Q}$.
I really do not know where to begin,any suggestions are greatly appreciated.
a) Consider a nested sequence $J_1 \subset J_2 \subset \cdots$ of subsets of $I$. (For example, let $J_1$ be a single point, and keep adding points one by one to form the $J_n$.) If it is impossible to construct such a sequence with each inclusion being strict ($\subsetneq$), then that means $I$ is finite, and we are finished. Otherwise, we may assume each inclusion is strict.
Then $\left(\sum_{i \in J_n} a(i)\right)_{n \ge 1}$ is an increasing sequence that is upper bounded by the finite number $\sup_{J \subset I, J \text{ finite}} \sum_{i \in J} a(i)$, so the monotone convergence theorem implies that $\lim_{n \to \infty} \sum_{i \in J_n} a(i) = \sup_{J \subset I, J \text{ finite}} \sum_{i\in J} a(i)$. Why does this imply $I = \bigcup_{n \ge 1} J_n$?
b) One relevant result that is useful is that absolute convergence of $\sum_{q \in \mathbb{Q}} a(q)$ (and any sub-series) implies that the series is unchanged by rearrangements; we may take the sum in any order.
Suppose $x \in \mathbb{Q}$. Then for any $\delta>0$, \begin{align} |f(x)-f(x-\delta)| &= f(x)-f(x-\delta)\\ &= \sum_{q \in \mathbb{Q}, q \le x} a(q) - \sum_{q \in \mathbb{Q}, q \le x-\delta} a(q)\\ &= a(x) + \sum_{q \in \mathbb{Q}, q < x} a(q) - \sum_{q \in \mathbb{Q}, q \le x-\delta} a(q)\\ & \ge a(x) > 0, \end{align} so $f(x-\delta) \not\to f(x)$ as $\delta \to 0$.
If $x \notin \mathbb{Q}$, then $\sum_{q \in \mathbb{Q}, q \le x} a(q) = \sum_{q \in \mathbb{Q}, q < x} a(q)$ so the above issue goes away. For a sequence $\delta_n \downarrow 0$, you can show $f(x-\delta_n)$ is increasing with supremum $\sum_{q \in \mathbb{Q}, q < x} a(q)$ and apply the monotone convergence theorem again.
Note that $I=\cup_{n\geq 1} \{i, a_i\geq \frac 1n\}$.
If $I$ is not countable, there is some $n_0$ such that $I':=\{i, a_i\geq \frac 1{n_0}\}$ is not countable.
For any finite subset $J$ of $I'$ we have $\sum_{j\in J} a_i \geq \frac{|J|}{n_0}$. Since $|J|$ can take arbitrary high values, $\sum_{i\in I}a(i)=\infty$ which is a contradiction.