8

I was reading the theory of Riemann integration when I cam across the following ,

If $f$ is bounded on $[a,b]$, and $P = \{x_0,x_1,x_2.......x_n\}$ is a partition of $[a,b]$, let $$M_j = \sup_{x_{j-1}\leq x\leq x_j}f(x)$$ The upper sum of f over P is $$S(P) = \sum_{j=1}^{n} M_j(x_j-x_{j-1})$$ and the upper integral of $f$ over $[a,b]$, denoted by $$\int_{a}^{b^-} f(x)dx$$ is the infimum of all upper sums.

The theorem similarly goes on to state the result for lower sums.

My doubt is : I do not understand how is $$\int_{a}^{b^-} f(x)dx$$ the infimum of all upper sums. I understand that if we refine the partition P, then the upper sum would decrease, so it may be a lower limit for all the upper sums computed on the refinements of P (but still being the lower limit does not prove that it is the infimum) and what about those partitions for which P itself is the refinement of? How do I know that it will be a lower limit for those, let alone a infimum?

Martin Sleziak
  • 53,687
Noob101
  • 962
  • Do you have a definition of the upper integral other than "the infimum of all upper sums"? From what you've written, it sounds like it's just the definition. – Eric Wofsey Dec 06 '16 at 00:41
  • It's not a theorem; it's a definition. – zhw. Dec 06 '16 at 00:42
  • @zhw , But is it right to claim that we could just define that that particular limit which is what the integral basically is equals the infimum ? – Noob101 Dec 06 '16 at 00:52
  • @EricWofsey , I wanted to know that is the upper integral defined in this way a riemann integral too , that is if the partition goes to less than delta , then will my upper sum converge to it ? – Noob101 Dec 06 '16 at 00:54
  • Nice but slightly ambiguous question. +1 – Paramanand Singh Dec 07 '16 at 13:17

2 Answers2

16

Your question does have some ambiguity. From the wording of your question and comments it appears that you want to know:

Does the limit of upper sums (with respect to partitions getting finer and finer) equal the infimum of all upper sums?

First of all note that when we are dealing with limits of things dependent on a partition of an interval then there are two ways in which the limit operation can be defined:

1) Limit via refinement of a partition: Let $P = \{x_{0}, x_{1}, x_{2},\ldots, x_{n} \}$ be a partition of $[a, b]$ where $$a =x_{0} < x_{1} < x_{2} < \cdots < x_{n} = b$$ A partition $P'$ of $[a, b]$ is said to be a refinement of $P$ (or finer than $P$) if $P \subseteq P'$.

Let $\mathcal{P}[a, b]$ denote the collection of all partitions of $[a, b]$ and let $F:\mathcal{P}[a, b] \to \mathbb{R}$ be a function. A number $L$ is said to be the limit of $F$ (via refinement) if for every $\epsilon > 0$ there is a partition $P_{\epsilon}\in \mathcal{P}[a, b]$ such that $|F(P) - L| < \epsilon$ for all $P \in \mathcal{P}[a, b]$ with $P_{\epsilon} \subseteq P$.

2) Limit as norm of parititon tends to $0$: If $P = \{a = x_{0}, x_{1}, x_{2}, \ldots, x_{n} = b\}$ is a partition of $[a, b]$ then the norm $||P||$ of partition $P$ is defined as $||P|| = \max_{i = 1}^{n}(x_{i} - x_{i - 1})$.

Let $\mathcal{P}[a, b]$ denote the collection of all partitions of $[a, b]$ and let $F: \mathcal{P}[a, b] \to \mathbb{R}$ be a function. A number $L$ is said to be limit of $F$ as norm of partition tends to $0$ if for every $\epsilon > 0$ there is a $\delta > 0$ such that $|F(P) - L| < \epsilon$ for all $P\in \mathcal{P}[a, b]$ with $||P|| < \delta$. This is written as $\lim_{||P|| \to 0}F(P) = L$.

Note that for a given function $F:\mathcal{P}[a, b] \to \mathbb{R}$ the limiting behavior of $F$ can be different according to these two definitions given above. In fact if $F(P) \to L$ as $||P||\to 0$ then $F(P) \to L$ via refinement but the converse may not hold in general.

Let us establish that if $F(P) \to L$ as $||P||\to 0$ then $F(P) \to L$ via refinement. Let $\epsilon>0$ be arbitrary and let $\delta>0$ be such that $|F(P) -L|<\epsilon$ whenever $||P||<\delta$. Let us now choose any specific partition $P_{\epsilon} $ with $||P_{\epsilon} ||<\delta$. If $P_{\epsilon} \subseteq P$ then $$||P||\leq ||P_{\epsilon} ||<\delta\tag{A} $$ and hence by our assumption $|F(P) - L|<\epsilon $. Therefore it follows that $F(P) \to L$ via refinement also.

Notice that the argument here crucially hinges on inequality $(\text{A}) $. Starting with an $\epsilon>0$ we first found a $\delta>0$ via the given assumption $\lim_{||P||\to 0}F(P)=L$. The process of finding a suitable partition $P_{\epsilon} $ crucially depends on the implication $$P, Q\in\mathcal{P} [a, b], P\subseteq Q\implies ||Q||\leq||P||$$ which leads to inequality $(\text{A}) $ above. If the reverse implication $$P, Q \in \mathcal{P} [a, b], ||Q||\leq||P||\implies P\subseteq Q $$ were true then one could provide a similar argument as in last paragraph to prove that if $F(P) \to L$ via refinement then $F(P) \to L$ as $||P||\to 0$. We just need to set $\delta=||P_{\epsilon} ||$ and we are done. But this is not the case.

Now let $f$ be a function defined and bounded on $[a, b]$ and let $P = \{x_{0}, x_{1}, x_{2}, \ldots x_{n}\}$ be a partition of $[a, b]$. Let $M_{k} = \sup\,\{f(x), x \in [x_{k - 1}, x_{k}]\}$ and let $\mathcal{P}[a, b]$ denote the collection of all partitions of $[a, b]$. We define the upper sum function $S:\mathcal{P}[a, b] \to \mathbb{R}$ by $$S(P) = \sum_{k = 1}^{n}M_{k}(x_{k} - x_{k - 1})$$ It is easy to prove that if $m = \inf\,\{f(x), x \in [a, b]\}$ then $S(P) \geq m(b - a)$ for all $P \in \mathcal{P}[a, b]$ and further if $P, P' \in \mathcal{P}[a, b]$ are such that $P \subseteq P'$ then $S(P') \leq S(P)$. It follows that $J = \inf\,\{S(P), P \in \mathcal{P}[a, b]\}$ exists.

Your question can now be worded more concretely into one of the following two forms:

Does $S(P) \to J$ via refinement?

or

Does $\lim_{||P|| \to 0}S(P) = J$?

The answer to the first question is obviously "yes" and you should be able to prove this using the definition of limit via refinement given above.

The answer to second question is also "yes" but it is difficult to prove. We first prove the result for a non-negative function $f$. Let $\epsilon > 0$ be given. Since $J = \inf\,\{S(P), P \in \mathcal{P}[a, b]\}$, there is a partition $P_{\epsilon} \in \mathcal{P}[a, b]$ such that $$J \leq S(P_{\epsilon}) < J + \frac{\epsilon}{2}\tag{1}$$ Let $P_{\epsilon} = \{x_{0}', x_{1}', x_{2}', \ldots, x_{N}'\}$ and let $M = \sup\,\{f(x), x \in [a, b]\} + 1$. Let $\delta = \epsilon / (2MN)$ and consider a partition $P = \{x_{0}, x_{1}, x_{2}, \ldots, x_{n}\}$ with $||P|| < \delta$.

We can write $$S(P) = \sum_{k = 1}^{n}M_{k}(x_{k} - x_{k - 1}) = S_{1} + S_{2}\tag{2}$$ where $S_{1}$ is the sum corresponding to the index $k$ for which $[x_{k - 1}, x_{k}]$ does not contain any point of $P_{\epsilon}$ and $S_{2}$ is the sum corresponding to other values of index $k$. Clearly for $S_{1}$ the interval $[x_{k - 1}, x_{k}]$ lies wholly in one of the intervals $[x_{j - 1}', x_{j}']$ made by $P_{\epsilon}$ and hence $S_{1} \leq S(P_{\epsilon})$ (note that $f$ is non-negative). For $S_{2}$ we can see that the number of such indexes $k$ is no more than $N$ and hence $S_{2} < MN\delta = \epsilon / 2$ (note that $f$ is non-negative here). It follows that $$J \leq S(P) = S_{1} + S_{2} < S(P_{\epsilon}) + \frac{\epsilon}{2} < J + \epsilon\tag{3}$$ for all $P \in \mathcal{P}[a, b]$ with $||P|| < \delta$. It follows that $S(P) \to J$ as $||P|| \to 0$.

Extension to a general function $f$ can be achieved by writing $f(x) = g(x) + m$ where $m = \inf\,\{f(x), x \in [a, b]\}$ and noting that $g$ is non-negative.

Another interesting example showing the difference between two limit definitions is given in this answer.

Note: The limit of a Riemann sum is based on the two definitions given above but there is a slight complication. A Riemann sum depends not only on a partition but also on choice of tags corresponding to a partition. Formally one can view a Riemann sum not as a function from $\mathcal{P} [a, b] $ to $\mathbb{R} $ but rather as a relation from $\mathcal{P} [a, b] $ to $\mathbb {R} $ such that it relates every partition of $[a, b] $ to one or more real numbers.

Paramanand Singh
  • 87,309
  • I guess I should start referring back to this as the definitive answer on the $|P| \to 0$ versus refinement question (+1). – RRL Dec 03 '18 at 01:02
  • @RRL: thanks for your encouraging words! – Paramanand Singh Dec 03 '18 at 01:05
  • @Paramanand Singh, really elaborate answer that helped me a lot to grasp the idea of limits in terms of refinements. I was just wondering if the following statement should be the otherway round: instead of "This is because of the fact that refinement of a partition leads to a decrease in the norm, but decreasing the norm of a partition does not necessarily lead to a refinement." It should be: "This is because of the fact that refinement of a partition does not lead necessarily to a decrease in the norm, but decreasing the norm of a partition does lead to a refinement." Or am I wrong? – Philipp Jan 20 '20 at 14:17
  • @Philipp: well you are wrong, but don't feel bad about it. Let me explain more. To refine a partition you add a point in existing partition. This process can never increase norm but may decrease it like from ${1,3,4}$ (norm $2$) to ${1,2,3,4}$ (norm $1$). On the other hand let's take a partition of norm $1/2$ say ${0, 1/2,1}$ and consider a partition ${0,1/3,2/3,1}$ of same interval $[0,1]$ with norm $1/3$. Clearly this is not a refinement of previous partition. – Paramanand Singh Jan 20 '20 at 15:23
  • 1
    @Philipp: Also the language in my answer is imprecise (will fix it). What I wanted to say is this: Let $P, Q$ be two partitions of $[a, b] $. Then $P\subseteq Q\implies ||Q||\leq ||P||$ but $||Q||\leq ||P||$ does not necessarily imply $P\subseteq Q$. – Paramanand Singh Jan 20 '20 at 15:26
  • @Philipp : I fixed my answer with more precise language. Thanks for noticing the problem. – Paramanand Singh Jan 20 '20 at 15:32
  • @ParamanandSingh, first you state: "In fact if $F(P)\to L$ as $\Vert P\Vert \to 0$ then $F(P)\to L$ via refinement but the converse may not hold in general." For me this means on the one hand: "if $F(P)\to L$ as $\Vert P\Vert \to 0 \Rightarrow F(P)\to L$ via refinement" and on the other hand: "if $F(P)\to L$ via refinement $ \nRightarrow F(P)\to L$ as $\Vert P\Vert \to 0$". This is perfectly clear to me. However, I haven't fully understood the reason you presented for that. I will provide you with an example. – Philipp Jan 20 '20 at 18:38
  • @ParamanandSingh, for me the reason why you cannot convert the statement "In fact if $F(P)\to L$ as $\Vert P\Vert \to 0$ then $F(P)\to L$ via refinement" is simply the following counter example: Let $P={1,7,8}$ a partitition and $P'={1,7,7.1,8}$ is a refinement of $P$. One can see that both norms are still equal (namely $=6$). So refinement does not nessecarily decrease the norm and that is why the converse of "In fact if $F(P)\to L$ as $\Vert P\Vert \to 0$ then $F(P)\to L$ via refinement" is not generally true. – Philipp Jan 20 '20 at 18:51
  • @ParamanandSingh, With this in mind I don't understand your reasoning "This is because of the fact that refinement of a partition leads to a decrease in the norm, but a partition with smaller norm is not necessarily a refinement of a partition with a larger norm. Sorry for the lengthy comments but maybe you can help me to put my thoughts on this in order. – Philipp Jan 20 '20 at 18:52
  • @Philipp : will add some explanation in my answer because adding too much in comments is not a good practice here. – Paramanand Singh Jan 21 '20 at 01:51
  • @Philipp : check updated answer. – Paramanand Singh Jan 21 '20 at 02:29
  • @ParamanandSingh, thank you for your patience and your extensive explanation. Your update helped me a lot on that issue :) – Philipp Jan 21 '20 at 14:11
  • In your second definition where you say "if for every $\epsilon > 0$ there is a $\delta > 0$ such that $|F(P) - L| < \epsilon$ for all $P\in \mathcal{P}[a, b]$ with $||P|| < \delta$", I'm wondering why do you say that EVERY partition with norm less than $\delta$ should satisfy. Why isn't "there exists a partition $P$ with $||P||< \delta$" enough? – William Feb 02 '22 at 22:19
  • @William: the definition is designed to focus on the norm of the partition and the behavior should be seen for all partitions with small norm. Compare it with "all $x$ with $0<|x-a|<\delta $" in definition of limit. – Paramanand Singh Feb 02 '22 at 22:27
  • Yes, that was actually my first thought. But I'm curious, would replacing for all with there exists work? I'm not able to see why it shouldn't. – William Feb 02 '22 at 23:07
  • @William: It won't work as we can then then find a set of partitions for which the integral would converge to some value and another set of partitions for which the integral would converge to another value. – Paramanand Singh Feb 03 '22 at 02:22
  • Yes, that should be the logical conclusion but I've been at it for a while and that's what I can't produce, a counterexample that $\forall \epsilon >0, \exists \delta >0, \exists P$ such that $||P||< \delta \Rightarrow |U(P,f)-L|< \epsilon$ would fail. (Taking $U(P,f)$ for example). The tricky part is that, I can always find partition for every $\epsilon$ – William Feb 03 '22 at 08:28
  • @ParamanandSingh can u give a bit hint on how to show J is lower bound by having detail that S(P) converges to J via refinement? – MEET PATEL Jul 27 '22 at 16:23
  • @MEETPATEL: that's just using the definition. Since $J$ is the greatest lower bound of sums $S(P) $, given any $\epsilon >0$ there is a partition $P_{\epsilon} $ such that $J\leq S(P_{\epsilon}) <J+\epsilon $. If $P$ is finer than $P_{\epsilon} $ then we have $S(P) \leq S(P_{\epsilon}) $ and we are done using $J\leq S(P) <J+\epsilon $. – Paramanand Singh Jul 28 '22 at 02:26
  • @ParamanandSingh i think u make mistake; plz read my question again; i want to prove J is lower bound; and as a Hypothesis i have S(P) converges to J; via refinement – MEET PATEL Jul 28 '22 at 02:30
  • @MEETPATEL: note that convergence to a limit is unique and we have proved that the sums converge to the lower bound $J$. So putting it another way whatever limit the sums converge to is the lower bound. I don't think there is anything more to be proved here. – Paramanand Singh Jul 28 '22 at 02:35
  • @MEETPATEL: If you wish you can give the argument from different direction. Let $S(P) \to L$ via refinement and let $J$ be greatest lower bound of sums like $S(P)$. Then for any $\epsilon >0$ we have a partition $P_\epsilon $ such that $L-\epsilon <S(P) <L+\epsilon $ for all $P$ with $P\supseteq P_\epsilon $. Now the lower bound $J\leq S(P) $ and hence $J<L+\epsilon $ and thus $J\leq L$. Derive a contradiction when $J<L$ – Paramanand Singh Jul 28 '22 at 02:51
  • @ParamanandSingh I got your proof and tried but did not get contradiction,i know the fact that J is largest bound will use, and gap between J and L will also,.but not getting exact contradiction.plz give me hint how can i get? – MEET PATEL Jul 28 '22 at 12:09
  • @MEETPATEL: let $\epsilon =(L-J) /2>0$ and there is a partition $P'$ such that $J\leq S(P') <J+\epsilon $. And further we have a partition $P_\epsilon $ such that $L-\epsilon <S(P) <L+\epsilon$ for all $P\supseteq P_\epsilon $. The partition $P''=P' \cup P_\epsilon $ is finer than both $P'$ and $P_\epsilon $ and hence we have $J\leq S(P'') \leq S(P') <J+\epsilon$ and $L-\epsilon<S(P'') $. Next observe the contradiction because $J+\epsilon=L-\epsilon$. – Paramanand Singh Jul 29 '22 at 02:16
  • @ParamanandSingh thank you ! – MEET PATEL Jul 29 '22 at 12:33
  • @ParamanandSingh, thank you for a nice post. There are numerous posts on the site where authors use a sequence of partitions $P_n$. Would you put it this way then that a number $L$ is said to be the limit of $F$ (via refinement) if for every $\epsilon>0$, there is an $n>N\in\mathbb{N}$ such that $|F(P_n)-L)|<\epsilon$? Would you write this as $\lim_{n\to\infty}F(P_n)=L$? – psie Apr 26 '23 at 12:45
  • The context in which this question arose is the following statement: $\lim_{|P|\to 0} L(f,P)=\sup_P L(f,P)$. Clearly the left-hand side is a limit as norm of partition tends to 0, but I am wondering if it can be rewritten in terms of limit via refinement, i.e. $\lim_{n\to \infty} L(f,P_n)=\sup_{P} L(f,P)$?? – psie Apr 26 '23 at 14:00
  • @schn: well you should ensure that the mesh/norm $||P_n||\to 0$. Any random sequence of partitions without that condition won't give us the desired result. – Paramanand Singh Apr 27 '23 at 03:29
4

You are having a fundamental misunderstanding on this topic for some reason. We have a bounded function. We define the upper integral. No question that it exists. We define the lower integral. Again, no question that it exists. We then define what it means for a bounded function to be Riemann integrable (RI): The uppper integral equals the lower integral. Plenty of questions about when this happens. The theory of the Riemann integral is all about when we are lucky enough to have $f$ RI, and about the value of the integral when it exists. For example, there is the theorem that if $f$ is continuous on $[a,b],$ then $f$ is RI on $[a,b].$ There is the FTC. A beautiful result of Lebesgue gives a necessary and sufficient condition: $f$ is RI iff $f$ is continuous a.e. All of these results go back to the definition.

zhw.
  • 105,693