Give examples of the following and justify

1) uncountable Lebesgue-null set $N$ of $[0,1]$ such that $1_N$ is not Riemann integrable on $[0,1]$

2) uncountable Lebesgue-null set $N$ of $[0,1]$ such that $1_N$ is Riemann integrable on $[0,1]$

I have trouble comping up with examples. Can someone help me with the problem?

  • $\begingroup$ Do you know the Cantor set? Do you know a characterisation of Riemann-integrability in terms of the set of discontinuities? $\endgroup$ – PhoemueX Nov 5 '14 at 6:05
  • $\begingroup$ I know cantor set would be an example to question (2) but I'm more concerned with question (1) which I still have no idea about $\endgroup$ – user186073 Nov 5 '14 at 6:06
  • $\begingroup$ I know $\mathbb{Q}$ would be an example for not R-integrable function but it's countable $\endgroup$ – user186073 Nov 5 '14 at 6:07
  • $\begingroup$ Do you have any idea about examples for (1)? $\endgroup$ – user186073 Nov 5 '14 at 6:08
  • $\begingroup$ What about taking the union of both examples that you mention? $\endgroup$ – PhoemueX Nov 5 '14 at 6:09

The key ingredient for this question is that a bounded function $f : [a,b] \to \Bbb{R}$ is Riemann-integrable iff the set of discontinuities of $f$ is a null-set.

For (2), you can take (as you note yourself in the comments) $N = C$, where $C$ is the usual Cantor set.

Note that $C$ is closed, so that $C^c$ is open, which implies that $1_C$ is continuous at every $x \in C^c$ (because of $1_C|_{C^c} \equiv 0$).

Hence, the set of discontinuities of $1_C$ is a subset of $C$ and hence a null-set.

For (1) choose $N = C \cup (\Bbb{Q} \cap [0,1])$ (as hinted at in the comments). We want to show that $1_N$ is discontinuous everywhere.

To see this, note that $N^c \subset [0,1]$ is dense, because $N$ is a null-set (why does this imply density of $N^c$?).

Hence, for every $x \in N$, there is a sequence $(x_n)_n$ in $N^c$ with $x_n \to x$. Hence, $1_n (x_n) = 0 \not\to 1 = 1_N (x)$. This shows that $1_N$ is discontinuous in every $x \in N$.

Now, let $x \in [0,1] \setminus N$. Then there is a sequence $(x_n)_n$ in $\Bbb{Q} \cap [0,1]$ with $x_n \to x$.

I will let you take it from here.

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