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So last class we saw Lebesgue's theorem on Riemann integrability (that is, a function is integrable over a bounded set in $\mathbb{R}^n$ if and only if its set of discontinuities has measure zero). Now, we also take this definition for volume of a set:

We say a bounded set has volume when the Riemann integral $\int_A 1_A$ exists, and we call that value the volume of A (where $1_A(x) = 1$ if $x \in A$ and $1_A(x)=0$ otherwise).

I know that volume zero implies measure zero but in general measure zero doesn't imply volume zero (there are sets with measure zero but no volume). So this leads us to the following "corollary", which we stated in class but didn't prove:

Let $f:A \subset \mathbb{R}^n \to \mathbb{R}$ a bounded function over a bounded set $A$ such that $f(x) \geq 0, \forall x \in A$ and $\int_A f(x)\ dx = 0$ (Riemann integral). Show that the set $\{x \in A: f(x) \neq 0\}$ has volume zero.

I've seen that corollary over and over again in several sources but with a small difference: they all replace volume by measure. So this combined with my small introduction (measure zero not always implies volume zero) makes me think the stronger version (that with volume zero) is not true (otherwise, why don't all just write volume?). My previous attempt to build a counterexample lead to naught (see this question), but I still think that weird functions and weird sets where the corollary's false exist. In particular, maybe a function with a set of discontinuities with measure zero but no volume.

So, can anyone come up with a counterexample? Thanks in advance!

Jesús Isea
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  • Is $\int_A f(x)\ dx$ the Riemann integral of $f(x)$ over the set $A$ or the Lebesgue integral of $f(x)$ over the set $A$? – Dave L. Renfro Jul 19 '21 at 16:36
  • @DaveL.Renfro Riemann integral. I'm editing! – Jesús Isea Jul 19 '21 at 16:38
  • Pretty sure the only way you can have measure 0 without volume is if the set is unbounded. Because if $A$ is bounded and measure 0, then the discontinuities of $1_A$ have to be measure 0 as well. So since in the corollary you are given $f$ is a bounded function over a bounded set, you're fine. – Alan Jul 19 '21 at 17:51
  • @Alan: if $A$ is bounded and measure $0,$ then the discontinuities of $1_A$ have to be measure $0$ as well --- Not if $A$ is dense in some non-degenerate ball and has measure $0$ (in fact, such a set can even be countable). Incidentally, I think "volume" here what is sometimes called Jordan measure (see here also), which is like Lebesgue measure except (for the outer versions) you use finite covers instead of countable covers. – Dave L. Renfro Jul 19 '21 at 18:36
  • @DaveL.Renfro Okay, I'm confused. How can a function be discontinuous on more than its domain? – Alan Jul 19 '21 at 19:09
  • @Alan: The function is $1_A:{\mathbb R}^n \rightarrow \mathbb R$ defined by $f(x) = 1$ if $x \in A$ and $f(x) = 0$ if $x \in {\mathbb R}^n - A,$ which has domain ${\mathbb R}^n.$ Now if, for instance, $A = {\mathbb Q}^n \cap [0,1]^n,$ then $A$ has Lebesgue $n$-measure zero and $1_A$ is discontinuous at each point of $[0,1]^n,$ which has Lebesgue $n$-measure $1.$ (unless I'm really overlooking something here . . .) – Dave L. Renfro Jul 19 '21 at 20:03
  • @DaveL.Renfro That set isn't bounded. – Alan Jul 19 '21 at 20:19
  • @Alan: That set isn't bounded. --- Which set? $A$ is certainly bounded. Maybe you mean we should be considering the restriction of $1_A$ to $[0,1]^n,$ so that the integrated function is (correctly) defined so as to be defined on a bounded set? If so, then yes, I overlooked that (and I also notice that I overlooked the "abuse of notation" $f:A \subset \mathbb{R}^n \to \mathbb{R}$ as implying $f$ is to be defined only on $A).$ – Dave L. Renfro Jul 20 '21 at 07:16

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Thomae's function $f \colon [0, 1] \to [0, \infty)$ is Riemann integrable with $\int_{0}^{1}f(x)\,dx = 0$. But $\{x : f(x) \neq 0\} = \mathbb{Q} \cap [0, 1]$, which is not Riemann measurable.

Mason
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