1
$\begingroup$

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:

  1. 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)
  2. 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.

$\endgroup$
5
  • 3
    $\begingroup$ I have heard of “jump discontinuity” but I have never heard the phrase “jump continuous.” $\endgroup$
    – Michael
    Mar 5, 2022 at 3:28
  • $\begingroup$ What's your definition of "zero content"? If it means Lebesgue measure zero, then 1 is an if and only if statement. $\endgroup$
    – PhoemueX
    Mar 5, 2022 at 5:19
  • $\begingroup$ with infinitely many(countable or uncountable) discontinuities and the content of the set of discontinuities is not zero --- If you want the content to be nonzero, there's no need to clutter your requirements with "with infinitely many". Your list of requirements is like saying you want an even integer that is positive and greater than $500.$ $\endgroup$ Mar 5, 2022 at 7:07
  • $\begingroup$ @PhoemueX content zero is a 19th century Cantor-Peano-Jordan measure theory. See: B. S. Thomson's answer. $\endgroup$
    – Hamilton
    Mar 5, 2022 at 20:18
  • 1
    $\begingroup$ By the way (for those interested) Folland in his Theorem 4.13 does indeed prove a simple theorem about the Riemann integral and a set of discontinuity points having zero content. But he says "... the reader should not attach undue importance to it. ...It does however point the way toward a necessary and sufficient condition for a function to be integrable. " My own feeling is that "zero content" is an important concept for historians, but not students of analysis generally. $\endgroup$ Mar 6, 2022 at 1:18

2 Answers 2

2
$\begingroup$

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:

  1. Every regulated function has only countably many discontinuities.
  2. All discontinuities are either removable or jump discontinuities.
  3. Step functions are regulated.
  4. A function is regulated if and only if it is the uniform limit of a sequence of step functions.
  5. 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.

$\endgroup$
3
  • $\begingroup$ Sorry for irritating you so that you jump into this question. And thank you so much for your detailed explanation, which cleared my perplexity. However, in the "connections" part, is it possible to construct a regulated function such that the set of discontinuities is not content zero(In other words, the closure of this measure zero set is not measure zero). The difficult part is that the function has to be regulated. $\endgroup$
    – Hamilton
    Mar 5, 2022 at 19:56
  • 1
    $\begingroup$ @Beginner It is easy to construct a monotonic function, discontinuous at each rational number. The set of rationals is measure zero, but certainly not content zero. In fact given any finite or countable set you can construct a regulated function with jump discontinuities at precisely each of those points. All monotonic functions are regulated and they are the easiest to use for construction of examples. $\endgroup$ Mar 5, 2022 at 21:36
  • 1
    $\begingroup$ @Beginner If you have never seen how to construct a monotonic function with a specified set of jumps see Example 5.62 in this textbook: classicalrealanalysis.info/documents/… $\endgroup$ Mar 5, 2022 at 22:51
1
$\begingroup$

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?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .