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I've seen this written several different ways in class and in Bruckner's Elemenatry Real Analysis, but I am not convinced they are equivalent.

Does $\lim \limits_{n \to \infty} U(f,P_n) = \inf_{p \in \mathcal{P}}\ U(f,p)$?

Let $f : [a,b] \to \mathbb{R}$.

Where $\mathcal{P}$ is the set of all partitions $p$ of $[a, b] $. $U(f,P)$ is the upper sum, i.e. $U(f,P) = \sum_{k=1}^n \sup\{f(x)\ : x \in [x_{k-1},x_k]\}(x_k-x_{k-1})$.

$(P_n)$ is defined as a sequence of partitions of $[a,b]$ such that $\|P_n\| \to 0$

Paramanand Singh
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1 Answers1

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Assuming only that $f$ is bounded, let $|f(x)| \leqslant M$ for all $x \in [a,b]$ and let $A = \inf_P U(f,P)$.

By definition of the infimum, for any $\epsilon > 0$ there is a partition $P'$ such that

$$A \leqslant U(f,P') < A + \epsilon/2$$

Let $δ=ϵ/4mM\,$ where $m$ is the number of points in the partition $P'$, and let $P$ be any partition with $||P|| < \delta$ . Form the common refinement $Q=P∪P'$ .

You will see that the upper sums $U(f,P)$ and $U(f,Q)$ differ in at most $m$ subintervals and in each the deviation is bounded by $2M\delta$.

Thus,

$$U(f,P) - U(f,Q) \leqslant|U(P,f)-U(Q,f)| < m2M\delta =m2M\frac{\epsilon}{4mM}=\epsilon/2,$$

and, since $Q$ is a refinement of $P'$ implying $U(f,Q) \leqslant U(f,P')$, we have

$$A \leqslant U(f,P) < U(f,Q) + \epsilon/2 \leqslant U(f,P') + \epsilon/2 < A + \epsilon.$$

Choosing $N$ such that $\|P_n\| < \delta $ for all $n \geqslant N$, we have

$$A \leqslant U(f,P_n) < A + \epsilon$$

Therefore,

$$\lim_{n \to \infty} U(f,P_n) = A = \inf_P U(f,P).$$

RRL
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  • What if we let $f$ be unbounded below? Of course, if $f$ is unbounded above, then the upper sums are all $+\infty$, so the equality holds. But if $f$ is unbounded below, the upper sums might still be finite, so that's perhaps an interesting case. – MathematicsStudent1122 Aug 06 '17 at 08:59
  • I said $|f(x)| \leqslant M$. This is half of the equivalence of the definitions for the Riemann integral -- one in terms of partition refinement the other in terms of partition norm tending to $0$. Also, any lower sum is a a lower bound for every upper sum, so the infimum over upper sums is finite. – RRL Aug 06 '17 at 09:04
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    $U(f,P) \geqslant L(f,Q)$ for every partition $P$ and $Q$ since $U(f,P \geqslant U(f,R) \geqslant L(f,R) \geqslant L(f,Q)$ where $R = P \cup Q$. So $\inf_P U(f,P)$ is finite as long as there is at least one finite lower sum. This is a typical question that arises in trying to reconcile the two definitions of the Riemann integral so raising that point is most likely a distraction to OP. – RRL Aug 06 '17 at 09:08
  • I think this is better than the one I gave in https://math.stackexchange.com/a/2047959/72031 perhaps I really don't need to worry about the sign of $f$ in my proof. The inequality $U(f, P) - U(f, Q) \leq |U(f, P) - U(f, Q) |$ takes care of signs. Apart from that our answers are almost same. +1 – Paramanand Singh Aug 07 '17 at 07:32
  • @RRL, thanks for another great post. If $P'$ has $m$ points, shouldn't $U(f,P)$ and $U(f,Q)$ differ in at most $m-2$ subintervals, since that is at most the number of points we are adding to $P$? I am assuming the endpoints of $P'$ are the same as those of $P$. – psie Apr 27 '23 at 12:43
  • I suppose $m$ is a typo for $m-2$ in your sentence "You will see that the upper sums $U(f,P)$ and $U(f,Q)$ differ in at most $m$ subintervals and in each the deviation is bounded by $2M\delta$." – psie Apr 27 '23 at 17:59
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    @schn: Thanks for checking. Whether it is $m$ or $m-2$ does not really matter in the end in this proof, of course. To be precise I said that $P'$ has $m$ points. I did not show it but this means presumably points $a= x_0<x_1<\ldots< x_{m-1}=b$. Hence there are $m-1$ subintervals and $m-2$ partition points in the interior of $[a,b]$ – RRL Apr 27 '23 at 18:15
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    The question is -- as we add these $m-2$ points to the partition $P$ to form $Q$, in at most how many subintervals will the upper sums corresponding to $P$ and $Q$ differ? I believe it is m-2, occurring when none of the added points is already an interior partition point of $P$. – RRL Apr 27 '23 at 18:20