Jumping in only because I find the terminology irritating.
A. Content zero vs measure zero.
Most authors use "content zero" to refer to the 19th century measure theory of Cantor-Peano-Jordan and "measure zero" to refer to the Lebesgue measure, dating from the first years of the 20th century.
You don't have to know much to connect them: a set has content zero if and only if the closure of the set has measure zero.
Nobody now or in the 19th century would have said (mainly because it is way too weak)
Bounded and set of discontinuities is content zero $\implies$ Riemann
integrable.
They knew back then exactly how to express this using the notion of content. We nowadays formulate this using "measure zero" since it is rather clumsy to state it using content.
Bounded and set of discontinuities is measure zero $\iff$ Riemann
integrable.
This is usually called the Lebesgue criterion for Riemann integrability. It is, however, just a restatement of an earlier criterion due to Hankel (1870), Volterra (1875) and Ascoli (1881). Lebesgue gets credit only for restating it using his measure.
B. Jump continuous vs regulated
The OP uses "jump continuous" to refer to a function whose only discontinuities are jump discontinuities. Yeech!
That is not an interesting class of functions really, especially since there is a larger class of functions of considerable importance. This strange usage suggests that removable discontinuities are not allowed?
Definition. A function is said to be regulated if it has one-sided limits at each point.
Properties:
- Every regulated function has only countably many discontinuities.
- All discontinuities are either removable or jump discontinuities.
- Step functions are regulated.
- A function is regulated if and only if it is the uniform limit of a sequence of step functions.
- All regulated functions are Riemann integrable.
C. Connections you ask?
A countable set of real numbers has measure zero. It need not have content zero. So regulated functions are Riemann integrable because all regulated functions are bounded and the set of discontinuities is countable and therefore has measure zero [not content zero please!].
Alternatively all regulated functions are uniform limits of step functions. But step functions are Riemann integrable and uniform limits of Riemann integrable functions are also Riemann integrable.
Here is an example:
We use an auxiliary continuous function $\chi: [0, 1] \to [0,1]$ such that $\chi(0) = \chi(1) = 0$, $\chi(1/2) = 1$, and $\chi^{-1}(\mathbb{Q})$ has full measure.
To construct $g: [0,1] \to [0,1]$, take a map $t: \mathbb{Z}_+ \to \mathbb{Q} \cap [0,1]$ such that each rational in $[0,1]$ appears in the image infinitely many times. Consider
$$g = \sum_{n = 1}^{\infty} c_n\chi_{I_n}$$
where $\chi_{I_n}$ is a scaling of $\chi$ function such that its support is $I_n =[t_n - \ell_n, t_n + \ell_n]$, and $c_n \in \mathbb{Q}^+$ with $c_n \leq 2^{-n}$ and $\ell_n \leq 2^{-n}q_n^{-3}$, $q_n$ being the denominator of $t_n$. Then by uniform convergence, $g$ is continuous. Furthermore, $g$ has the following property:
By choosing $c_n$ appropriately, $g(\mathbb{Q})$ can be disjoint from $\mathbb{Q}$. To achieve this, for each $q \in \mathbb{Q} \cap [0,1]$, let $n \in S_q$ be the set of $n$ such that $I_n$ is centered at $q$. $q$ lies in the intervals $\{I_n\}_{n \in S_q}$ plus finitely many other intervals $\{J_m\}$. Thus, it suffice to take the $c_n$ for $n \in S_q$ such that
$$\sum_{n \in S_q} c_n$$
is outside the $\mathbb{Q}$-span of $\{1, \chi_{J_m}(q)\}_m$.
$g^{-1}(\mathbb{Q})$ has measure $1$, as each number in its complement must either appear in infinitely many $I_n$, or is in $\chi_{I_n}^{-1}(\mathbb{R} \backslash \mathbb{Q})$ for some $n$, both of which have measure zero.
Now let $f(p/q) = 1/q$ for any $p/q \in \mathbb{Q}$, and let $f \equiv 0$ outside $\mathbb{Q}$. Then $f$ is continuous except on $\mathbb{Q}$, but $f \circ g$ is discontinuous on $g^{-1}(\mathbb{Q})$.
Note: $\chi$ itself can be constructed similar to the cantor set. Given a tenary representation $0.a_1a_2\cdots$ of a number in $[0,1]$, let $i$ be the first such that $a_i = 1$, then let
$$\eta(a) = \sum_{j = 1}^{i - 1} 2^{-j - 1}a_j + 2^{-i}.$$
Then $\eta(0) = 0, \eta(1) = 1$, and $\eta^{-1}(\mathbb{Q})$ is full measure. Glue two copies of $\eta$ together to get $\chi$.
Best Answer
The construction is correct. I’ll use your example as an illustration. Let $\{q_n:n\in\omega\}$ be an enumeration of $\mathbb{Q}\cap [0,1]$, and let $f(x)=\sum\limits_{q_n<x}2^{-n}$ for $x\in [0,1]$.
First consider what happens at some $q_m$: $$\lim\limits_{x\to {q_m}^-}f(x) = \sum_{q_n<q_m}2^{-n}=f(q_m),$$ because as $x$ moves up towards $q_m$, $\{q_n:q_n<x\}$ includes more and more of the rationals less than $q_m$. Thus, $f$ is continuous from the left at $q_m$, but for every $x>q_m$ we have $$f(x)=\sum_{q_n<x}2^{-n} \ge \sum_{q_n\le q_m}2^{-n} = f(q_m)+2^{-m},$$ so $f$ jumps by at least $2^{-m}$ at $q_m$. In fact $$\lim_{x\to {q_m}^+}f(x) = f(q_m)+2^{-m},$$ and the jump is exactly $2^{-m}$.
At each irrational $a \in [0,1]$, however, $f$ is easily seen to be continuous: $$\lim_{x\to a^-}f(x) = \lim_{x\to a^+}f(x) = \sum_{q_n<x}2^{-n}=f(x).$$