Can unbounded functions be Riemann integrable?

Question

I need to prove or disprove that the next function is Riemann integrable on $[0,2]$:

$$ f(x) = \begin{cases} \dfrac{1}{x} &x > 0 \\ 0 &x = 0 \end{cases} $$

My intuition is that it's not, because $f$ is unbounded in that interval, so $U(f,Pn) = L(f,Pn) = \infty$

So I have two questions:

Am I right?
Can unbounded functions be Riemann integrable?

Well since a Riemann integrable function have to be bounded, then yes unbounded one cannot be Riemann integrable. — Azlif, Jan 07 '18 at 01:05
https://math.stackexchange.com/questions/2246901/how-to-show-that-a-riemann-integrable-function-is-bounded — Azlif, Jan 07 '18 at 01:11
in the title "If and only f" part is not true since bounded functton need not be Rieman Integrable — Azlif, Jan 07 '18 at 01:15

spaceisdarkgreen · Accepted Answer · 2018-01-07T01:30:43.240

4

It is not integrable, so your answer is right. As to whether an unbounded function can be Riemann integrable (and thus if the reasoning is as simple as that) it's important to distinguish between proper and improper Riemann integrability.

Regarding proper Riemann integrability, it is true that if a function is unbounded on $[a,b],$ it is not integrable. And we do have $U(f,P) = \infty$ for every partition of $[a,b].$ It is not the case that for $L(f,P)=\infty$ for all partitions, but it's enough that $U(f,P) = \infty$ to prove it the function is not integrable.

For improper Riemann integrability, the function need not be bounded. For instance $1/\sqrt{x}$ has an improper Riemann integral with value $2$ on $[0,1].$ However $1/x$ does not have an improper Riemann integral either on $[0,2]$ since $$ \int_a^2 \frac{1}{x}dx = \ln(2/a)\to_{a\to 0^+} \infty.$$

edited Jan 07 '18 at 01:30

answered Jan 07 '18 at 01:20

spaceisdarkgreen

67,151

Why not for $L(f,P)$? – McLovin Jan 07 '18 at 01:22
Why would $L(f,P) = \infty$? – spaceisdarkgreen Jan 07 '18 at 01:23
Same reason for $U$, no? – McLovin Jan 07 '18 at 01:23
No.... $L(f,P)$ is the lower sum. In the case of $1/x$ on $[0,2],$ it will never be the case for any partition $P$ that $L(f,P) = \infty.$ – spaceisdarkgreen Jan 07 '18 at 01:25
$L(f) = \sup_{P}L(f,P)$ may well be infinity (and is in this case). – spaceisdarkgreen Jan 07 '18 at 01:26
But what if you divide $[0,2]$ to a high enough number of segments $n$..? Why will $L(f,P)$ not be infinity as well? – McLovin Jan 07 '18 at 01:31
@Pilpel Read my last response. The limit may be infinite as the partition becomes finer and finer, but for any given partition, the lower sum is finite. – spaceisdarkgreen Jan 07 '18 at 01:33
@Pilpel Also, although it is in this case, the limit is not necessarily infinite. There are unbounded functions for which the limit $L(f)$ finite, e.g. ones that have improper integrals. – spaceisdarkgreen Jan 07 '18 at 01:39
@spaceisdarkgreen Do you mind elaborating why there is this distinction between proper and improper Riemann integrals? The definition for the Riemann integral doesn't seem to 'break' for unbounded functions. – Anon Sep 26 '20 at 12:08
@Alona It does break, in the sense that for an unbounded function on a closed interval $U(f,P)=\infty,$ so it's always undefined. – spaceisdarkgreen Sep 26 '20 at 21:38

Can unbounded functions be Riemann integrable?

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