Give example of a function $f: [0,1] \to \mathbb R $ which is integrable ( Lebesgue or Riemann , if possible , both) but whose set of discontinuity points is an uncountable set and dense in $[0,1]$ ?

  • $\begingroup$ Integrable in what sense? Lebesgue? The first example that comes to mind works in that case. $\endgroup$ – Najib Idrissi Nov 26 '14 at 15:21
  • $\begingroup$ @NajibIdrissi: Lebesgue is ok , I would really like to see your example ; but Riemann is also wanted $\endgroup$ – user123733 Nov 26 '14 at 15:22

An example of Lebesgue integrable function is easy, as shown in other answers.

It is a known fact that a function $f: \mathbb{R} \to \mathbb{R}$ is Riemann integrable if and only if $f$ is bounded and the set of discontinuities has measure zero.

Let $C$ be the usual ternary Cantor set, and let $\mathbb{Q}$ denote the set of rational numbers. The function

$$f(x) = \begin{cases} 200, &x \in C \\ 1/q, &x = p/q \text{ is rational in lowest terms and } x \notin C \\ 0, &\text{otherwise} \end{cases} $$ is discontinuous in $C \cup \mathbb{Q}$, which is uncountable and dense but has measure zero (hence $f$ is Riemann integrable).

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  • $\begingroup$ Awesome construction! $\endgroup$ – layman Nov 26 '14 at 15:45
  • $\begingroup$ @Pedro M. : Could you just please explain the measure $0$ part of $C \cap \mathbb Q$ $\endgroup$ – user123733 Nov 26 '14 at 16:02
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    $\begingroup$ @user123733 It's actually $C \cup \mathbb{Q}$ that has measure $0$. This is because the cantor set $C$ has measure $0$, and the rationals $\mathbb{Q}$ have measure $0$, so $m(C \cup \mathbb{Q} ) \leq m(C) + m(\mathbb{Q}) = 0 + 0 = 0$, which shows $m(C \cup \mathbb{Q}) = 0$. $\endgroup$ – layman Nov 26 '14 at 16:20

An easy example is this: Let $C$ be the Cantor set and denote by $\chi_C$ its characteristic function, so that for $x\in[0,1]$, we have that $\chi_C(x)=0$ if $x\notin C$, and $\chi_C(x)=1$ if $x\in C$. Clearly $\chi_C$ is bounded, $0\le\chi_C(x)\le 1$ for all $x$. This function is discontinuous at an uncountable set of points (all points in $C$, because they are limit of points not in $C$, since $C$ is nowhere dense, but they are alos limit of points in $C$, since $C$ is perfect).

Now, let $i$ be the increasing function obtained by enumerating the rationals in $[0,1]$, say $q_0,q_1,\dots$, and letting $i(x)=\sum_{q_n\le x}2^{-n}$, that is, given $x\in[0,1]$, we look at the set of rationals in $[0,x]$, we look at the set $I\subseteq\mathbb N$ of indices of these rationals (according to our enumeration), and then set $i(x)=\sum_{n\in I}2^{-n}$. Clearly $i$ is bounded, $0\le i(x)\le 2$ for all $x$. The set of discontinuities of $i$ is dense in $[0,1]$ (because $i$ is discontinuous precisely at the points in $\mathbb Q\cap[0,1]$. That these are discontinuity points is easy to see, since we explicitly added jumps there. But if $x$ is irrational, for any $N$, in a sufficiently small neighborhood of $x$ we only find rationals with indices larger than $N$, and $\sum_{j>N}2^{-j}=\tau$ can be made arbitrarily small by choosing $N$ sufficiently large. Now, if $y$ is any point in that neighborhood, then $|i(x)-i(y)|\le2\tau$, so $i$ is continuous at $x$).

Finally, $f=\chi_C+i$ is Riemann integrable, and its set of discontinuities is both dense and uncountable. One can argue directly about its integrability, or one can quote Lebesgue's characterization, that a function is Riemann integrable iff it is bounded and its set of discontinuities has measure zero. But both $C$ and $\mathbb Q\cap[0,1]$ have measure zero, and so does their union.

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If you mean Lebesgue integrable, then $f(x) = \begin{cases} 0 & x \in \mathbb{Q} \cap [0,1] \\ 1 & x \not \in [0,1] \cap \mathbb{Q} \\ \end{cases}$ is a function that is discontinuous at every irrational (hence on an uncountable, dense set), but it is integrable, and $\int \limits_{[0,1]} f(x) \,d\mu = 0m(\mathbb{Q} \cap [0,1]) + 1m(\mathbb{Q}^{c} \cap [0,1]) = 0 \cdot 0 + 1 \cdot 1 = 1$.

Note that this function is not Riemann integrable. Do you know why?

If you want a function that is Riemann integrable, but is discontinuous at an uncountable dense set, then you need to construct a function whose set of discontinuities is both uncountable and a set of measure $0$ (since a function is Riemann integrable iff its set of discontinuities has measure $0$).

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    $\begingroup$ In fact, it's discontinuous at all the rationals also. So, the set of discontinuity points is all of [0,1]. $\endgroup$ – Nate Eldredge Nov 26 '14 at 15:24
  • $\begingroup$ @NateEldredge Right, I was just pointing out the irrationals because they wanted an uncountable dense set of discontinuities, though I guess $[0,1]$ is dense in itself. $\endgroup$ – layman Nov 26 '14 at 15:26
  • $\begingroup$ It certainly is! $\endgroup$ – Nate Eldredge Nov 26 '14 at 15:28
  • $\begingroup$ @MathIsHardNoItsNot: But the Cantor set is uncountable of Lebesgue measure $0$ , so can we find such to satisfy my conditions ? $\endgroup$ – user123733 Nov 26 '14 at 15:33
  • $\begingroup$ @user123733 The cantor set is not dense in $[0,1]$, and you said you want a function whose set of discontinuities is dense. $\endgroup$ – layman Nov 26 '14 at 15:36