Can we find a regulated function with infinitely many discontinuities and the content of the set of discontinuities is not zero Can we find a jump continuous function(EDIT: according to answers, this terminology should be a regulated function) with infinitely many(countable or uncountable) discontinuities and the content of the set of discontinuities is not zero?
Note that there are two kinds of "Riemann-integrability" theorems in Calculus textbooks:

*

*If a function $f$ is bounded on $[a,b]$ and the set of points in $[a,b]$ at which $f$ is discontinuous has zero content then $f$ is Riemann integrable on $[a,b]$. (This comes from Folland advanced calculus 3rd edition. Theorem 4.13)

*If a function is regulated on $[a.b]$(of course bounded), then it is Riemann integrable on $[a,b]$.  (This comes from Amann&Escher Analysis II(English Version) Theorem 3.4 and Remark 3.5)

I am trying to find the implication between these two theorems.
I know $1 \nrightarrow 2$ since $f(x) = sin(1/x)$ with $f(0)= 0 $ is a counterexample
How to show $2 \nrightarrow 1$?  That's why I am trying to find this example.
EDIT2: This example is possible since $\mathbb Q$ is not content zero and we can construct a monotone(which is regulated) function discontinuous on $\mathbb Q$. For more details, see. Also, Example 5.62 in the book Elementary Real Analysis in this website also gives a same example.
 A: 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.
A: It isn't even possible to construct a function with uncountably many jump discontinuities Can we construct a function that has uncountable many jump discontinuities?
