Let $f:[0,1] \rightarrow R$ defined by $f(x) = 2$ if $x = \frac{1}{n}$ for some $n∈ℕ$, $0$ otherwise. Determine if $f$ is Riemann integrable.

My attempt: Let $ε > 0$. Construct a partition P as follows: let $x_0 = 0, x_1 = ε$, and let $x_1, x_2, . . . , x_m = 2$ be a uniform partition of $[ε, 2]$ of norm $δ = \frac{ε^2}2$. We need to estimate the difference between the upper and lower sums of h with respect to P. The contribution to $$U(h, P) − L(h, P) (∗)$$ from the interval $[0, ε]$ is, at most $ε$. The contribution to (∗) from the interval $[ε, 2]$ is bounded above, by $δ$ times the number of points of the form $\frac1n$ in the interval $[ε, 2]$, which in turn is at most $4ε^{−1}$. It follows that (∗) is at most $\fracε2 + δ · 4ε^{−1} = ε$. By the Criterion of Integrability we conclude that $f$ is integrable.


Minor criticisms:

  1. There will usually not be a uniform partition of $[\varepsilon/2,1]$ of norm exactly $\varepsilon^2/4$, but any finer partition than that will do the job. For clarity you might simply calculate an $m$ large enough for the job.
  2. You didn't actually say where the estimates on the "near zero" subinterval and "far away from zero" subinterval came from in full detail.
  3. You forgot that $f=2$, not $1$, at the points where it is nonzero. Thus your estimates are actually all off by a factor of $2$.
  • $\begingroup$ What finer partition do you suggest? And I think I've edited the post to fix the fact that f = 2, not 1. Have I skipped anything? $\endgroup$ – Leaf Jul 21 '16 at 14:52
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    $\begingroup$ @Leaf You haven't fixed the $2$ issue at all. You need to go back through everything and insert factors of $2$ pretty much everywhere. Anyway, the norm of a uniform partition of $[\varepsilon/2,1]$ containing $m-1$ subintervals is what? How big must $m$ be for this to be less than some given positive number $\gamma$? $\endgroup$ – Ian Jul 21 '16 at 14:56
  • $\begingroup$ Is that better? @Ian $\endgroup$ – Leaf Jul 21 '16 at 15:20
  • $\begingroup$ @Leaf Nope; now you've tried to make a partition of $[0,2]$ and your estimates still don't go through quite how you say they do. (When I said "insert factors of $2$", I didn't actually mean by multiplication; in most places it actually winds up being by division. You need smaller $\delta$ to control the spikes than you thought because the spikes are bigger than you thought.) It was actually better (but not quite correct) before. $\endgroup$ – Ian Jul 21 '16 at 15:37
  • $\begingroup$ Oh, then I multiply 2 to everything but not the interval? $\endgroup$ – Leaf Jul 21 '16 at 15:42

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