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Everywhere I've tried to look, the only two common examples of non-Riemann integrable functions are unbounded functions or Dirichlet function.

What are some examples of non-Riemann integrable functions besides those functions?

I assume for such non-Riemann integrable functions, the lower integral and the upper integral must exist but are unequal. (Because the lower integral and upper integral both either exist or don't, together and since I have removed unbounded functions from the picture, both integrals exist but are unequal.)

I'd really appreciate multiple examples, if that's possible, of functions that have unequal lower and upper integral.

William
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    Lebesgue's theorem for RIemann integrals states that $f:[a,b]\to\Bbb{R}$ is Riemann integrable if and only if $f$ is bounded, and the set of discontinuities has Lebesgue measure zero. So, the only way for $f$ to fail to be Riemann integrable is if $f$ is unbounded, or it is discontinuous at 'too many' points. The Dirichlet function $\mathbf{1}_{\Bbb{Q}}$ is a very simple example of a nowhere continuous function (which is obviously bounded) so it fits the bill perfectly. There's literally infinitely many non-Riemann-integrable functions. – peek-a-boo Jul 12 '22 at 13:38
  • @peek-a-boo I'm not familiar with Lebesgue integral or measure so I appreciate you simplifying it to "discontinuities at too many points". But if you can simplify it a little more and tell me how many is too many? It'd be nice. Like uncountable points? Or even countably infinite points is too many? – William Jul 12 '22 at 15:36
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    You don't need to know about Lebesgue integrals, or the full extent of Lebesgue measure. YOu just need to know what "Lebesgue measure zero" is (google it) (examples of measure zero sets include countably infinite sets). And here I intentionally put "too many" in quotes, because the precise meaning of "too many points" is a set "which is not of Lebesgue measure zero". Unfortunately there's no simpler description. Uncountability alone is not enough to say it is 'too many' (the middle thirds Cantor set is uncountable but has measure zero, so it is still 'small enough'). – peek-a-boo Jul 12 '22 at 15:50
  • @peek-a-boo I will check it out. Just to be clear that I understand you. Countable discontinuities are never a problem? Uncountable discontinuities may or may not be a problem, correct? – William Jul 12 '22 at 15:53
  • yup that's right. – peek-a-boo Jul 12 '22 at 15:54

2 Answers2

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You can take, for instance$$\begin{array}{rccc}f\colon&[0,1]&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}x&\text{ if }x\in\Bbb Q\\0&\text{ if }x\notin\Bbb Q.\end{cases}\end{array}$$It is bounded. And it is not Riemann integrable since it is discontinuous at every point of $(0,1]$, and this is is uncountable. Another possibility would be$$\begin{array}{rccc}g\colon&[0,1]&\longrightarrow&\Bbb R\\&x&\mapsto&\begin{cases}x&\text{ if }x\in\Bbb Q\\-x&\text{ if }x\notin\Bbb Q.\end{cases}\end{array}$$

José Carlos Santos
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  • Aha, I guess the safest bet would be to consider functions with uncountable discontinuities. If the set of discontinuities is countably infinite, would such a function be non-Riemann integrable still? (I know discontinuities on finite points are not a problem.) – William Jul 12 '22 at 15:40
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    If the function is bounded and the set points of discontinuity is countable, then the function is Riemann-integrable. An example of such a function is Thomae's function. – José Carlos Santos Jul 12 '22 at 15:45
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Let $A\subset [0, 1]$ be any set.

$\mathbf{1}_A:[0,1]\to\Bbb{R}$ defined by

$\mathbf{1}_A(x)=\begin{cases}1&x\in A\\ 0& x\notin A\end{cases}$

$\mathbf{1}_A$ is called indicator function of $A$.

Set of discontinuity of $\mathbf{1}_A=\partial(A)$

where $\partial(A)$ denote the set of all boundary points of $A$

Lebesgue's criteria for Riemann integrability : $f:[a,b]\to \Bbb{R}$ is Riemann integrable if and only if $f$ is bounded, and the set of discontinuities has Lebesgue measure zero.

Step $1$ : Choose $A\subset [0, 1]$ such that $m(\partial (A)) >0$

Where $m$ is the $1$-dimensional Lebesgue measure.

Step $2$: Then choose $\mathbf{1}_A$

Few examples: (Bounded functions which are not Riemann integrable)

  1. Choose the set of rationals $Q$ . Then $m(\partial({\Bbb{Q}\cap [0,1]}))=[0, 1]=1>0$

Hence $\mathbf{1}_{\Bbb{Q}} : [0, 1]\to \Bbb{R}$ defined by $\mathbf{1}_A(x)=\begin{cases}1&x\in \Bbb{Q}\\ 0& x\notin \Bbb{Q}\end{cases}$

(It is your known example, Dirichlet function)

  1. Choose the set of irrationals $A=\Bbb{R}\setminus \Bbb{Q}$.

Then $\mathbf{1}_{{A}} : [0, 1]\to \Bbb{R}$ defined by $\mathbf{1}_{A}(x)=\begin{cases}1&x\in \Bbb{Q}\\ 0& x\notin \Bbb{Q}\end{cases}$

Another bounded function which is not Riemann integrable.

  1. Choose the indicator function of $S\subset [0, 1] $ Cantor set of positive measure ( Specifically S-V-C set of measure $\frac{1}{2}$)

  2. Choose the indicator function of $A=\{\sin n :n\in \Bbb{N}\}$

  3. Choose the indicator function of $A=\Bbb{Q}\cup \{\text{ few of your favorite irrationals!}\}$

There are many more. Can you list few more?

Sourav Ghosh
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