This is question 19 in Chapter 3 of Spivak's Calculus on Manifolds.

Suppose $I_n=(a_n,b_n)$ is a sequence of open subintervals of $(0,1)$ such that every rational number in $(0,1)$ is contained in one of them, and such that $\sum (b_n-a_n)<1$. Let $A=\bigcup I_n$; it's not hard to show the boundary of $A$ is $[0,1]\setminus A$ and that it's not of measure zero.

Let $\chi_A$ be the characteristic function of $A$ and let $f$ be a function which equals to $\chi_A$ except on a set of measure zero.

Show $f$ is not Riemann integrable.

I found solutions here and here, but they both seem wrong to me. The latter seems to use the equation $\partial U\setminus V=\partial(U\setminus V)$ which is obviously wrong in general (and I don't see why it should apply here...), and the former seems to use the assertion that if $f_1$ differs from the function $f_2$ on a set $B$ and is discontinuous at $x_0$, then either $f_2$ is discontinuous at $x_0$ or $x_0\in B$ (which is again not true in general, and I don't see why it should apply here).

So how do you solve this question?

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    $\begingroup$ Riemann integrable or Lebesgue integrable? $\endgroup$ – Shashi Dec 24 '17 at 20:07
  • $\begingroup$ I apologise if I upset anyone; I wasn't trying to make fun of anyone or anything like that. I simply meant the solution seemed wrong to me. In any case, I edited the question: I meant Riemann integrable. $\endgroup$ – Cronus Dec 24 '17 at 20:43
  • $\begingroup$ It seems that whoever I upset deleted their comment... $\endgroup$ – Cronus Dec 24 '17 at 20:44

We start by making the clarification that we're talking about Riemann-integrability.

Second, I agree with you that the links are making somewhat convoluted claims that are not true in general, but the following claim is easy to justify:

Claim: $f$ is discontinuous at $\{x \in [0,1] : f(x) = 0\}$.

To see this, suppose $f(x) = 0$ and consider an open interval $I$ containing $x$. Then take any rational $q \in I$. By definition, $q \in I_n = (a_n, b_n)$ for some $n$. Obviosly $I \cap I_n$ has nonzero measure, hence there exists some point $w \in I \cap I_n$ such that $f(w) = \chi_A(w) = 1$, hence $f$ is discontinuous at $x$.

Now, $\{x \in [0,1] : \chi_A(x) = 0\} \setminus \{x \in [0,1] : \chi_A(x) \neq f(x)\} \subset \{x \in [0,1] : f(x) = 0\}$, hence the latter set has nonzero measure.

  • $\begingroup$ Wonderful, thank you! $\endgroup$ – Cronus Dec 24 '17 at 20:50

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