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Let $E = \{1/n : n \in \mathbb{N}\}$, and suppose $f: [0,2] \rightarrow \mathbb{R}$ is continuous everywhere but $E$. Prove that $f$ is Riemann integrable.

I know the Lebesgue criterion for integrability (discontinuities have measure $0 \Rightarrow f \in \mathscr{R}$), but how would you do this without? It is my understanding that partitions need to have finitely many internal endpoints.

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Sketch: we can confine all but finitely many discontinuities of $f$ into a small interval $I$. Outside $I$, we can choose a partition fine enough to make the upper and lower sums arbitrarily close. Simultaneously, since $f$ is bounded, we can choose $I$ small enough that it cannot contribute too much to the difference between the lower and upper sums.

Details:

Since $f$ is bounded above and below, we can subtract off the lower bound in order to assume without loss of generality that $f \geq 0$. Having done so, choose $M$ such that $f \leq M$.

Let $\varepsilon > 0$. Choose $N$ large enough that $M/N<\varepsilon/2$. Choose a partition $P$ of $[1/N,2]$ whose mesh is small enough that the lower and upper sums for $P$ differ by at most $\varepsilon/2$. (Here I assume we have already proven that a function with finitely many discontinuities is Riemann integrable.)

Now extend $P$ to a partition $P'$ of $[0,2]$ by adding in the points $0$ and $1/N$. The difference between the lower and upper sums on $[1/N,2]$ is still at most $\varepsilon/2$, while the difference between the lower and upper sums on $[0,1/N]$ is also at most $\varepsilon/2$, since $0 \leq f \leq M$ there. So the difference between the lower and upper sums for $P'$ is at most $\varepsilon$.

We can tweak this to look much like the proof of the Lebesgue criterion. Set up the same as above, but where $M/N<\varepsilon/3$ instead. Take $\delta>0$ small enough that $MN\delta<\varepsilon/3$. Take intervals of length at most $\delta$ around $1,1/2,\dots,1/N$ in your partition. Then refine the rest of the partition enough that the difference between the upper and lower sums is less than $\varepsilon/3$ there (which we can do because $f$ is continuous there). Then again the lower and upper sums for the overall partition is at most $\varepsilon$ ($\varepsilon/3$ from the small intervals near the "widely spaced" discontinuities, $\varepsilon/3$ from the small interval containing the "tightly packed" discontinuities, and $\varepsilon/3$ from the continuous part).

Here I add in the assumption that $f$ is bounded, because unbounded functions are never Riemann integrable (since there is always a subinterval of the partition where the upper or lower sum is infinite).

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  • $\begingroup$ Unfortunately, nothing from the OP suggests that $f$ is bounded. Furthermore, $f$ is not continuous on a compact set, so you cannot use this to conclude its image is compact (and therefore bounded). In fact, it is easy to produce examples of such $f$ if you think about it for a moment. However, what you need to fix the flaw in your proof is still true: $f$ is bounded in a neighbourhood of zero because it is continuous at zero. For this reason I downvote, but only until the flaw is fixed. $\endgroup$
    – J. Loreaux
    Commented Feb 21, 2015 at 19:38
  • $\begingroup$ @J.Loreaux The claim fails if you omit the boundedness hypothesis. For instance, a function which is $0$ on $[0,1]$, $1/(x-1)$ on $(1,2]$ is a nonnegative function, continuous except at one point, and is not even Lebesgue integrable, much less Riemann integrable. In fact, no unbounded function is (properly) Riemann integrable. So something has to give. Would you prefer I remark that some sort of boundedness hypothesis is required? $\endgroup$
    – Ian
    Commented Feb 21, 2015 at 19:50
  • $\begingroup$ Ah, of course. My apologies. I was thinking that sufficiently nice unbounded Lebesgue functions could be Riemann integrable, but that is obviously false as you pointed out. Thanks for the update. $\endgroup$
    – J. Loreaux
    Commented Feb 21, 2015 at 19:58
  • $\begingroup$ @J.Loreaux Would it be weaker than boundedness but still sufficient to show improper reimann integrability as long as both one-sided limits exist at each discontinuity? $\endgroup$
    – Neptune
    Commented Oct 15, 2021 at 19:00
  • $\begingroup$ @Neptune What type of improper integration are you even doing in this case? Is it principal value? And what order do you take the limits in? If you don't specify, then I think you ultimately need to postulate Lebesgue integrability anyway. If you do that, I think a Lebesgue integrable function with measure zero discontinuity set is improperly Riemann integrable with the result being equal to the Lebesgue integral regardless of how the limits are taken. $\endgroup$
    – Ian
    Commented Oct 15, 2021 at 21:15

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