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I'm trying to figure out the Darboux Integral definition as it states: $f$ is integrable if $inf${U($f$,$P$)} = $sup${L($f$,$P$)}, where U($f$,$P$) is the upper sum, L($f$,$P$) is the lower sum of $f$ and $P$ is a parition of $f$.

I'm not understanding what the $sup$/$inf$ of the sums mean. When calculating, they are a finite value (ie. L($f$,$P$) = $\sum_{i=1}^n m_i (x_i - x_{i-1})$)

Consider:

$$f(x) = 2x, x \in [0,1];~ P = \{0,\frac{1}{4},\frac{1}{2},1\}.$$

Since $P$ has 4 elements, n=3, thus 3 subintervals of $[0,1]$, can you definte these 3 subintervals however you want as long as they range [0,1]?

ie. $$[0,\tfrac{1}{4}] \,\cup\, [\tfrac{1}{4},\tfrac{1}{2}] \,\cup\,[\tfrac{1}{2},1] \text{ or } [0,\tfrac{1}{3}] \,\cup\, [\tfrac{1}{3},\tfrac{2}{3}] \,\cup\,[\tfrac{2}{3},1] $$

So to calculate $U(f,P)$ for the first set of subintervals:

$$U(f,P) = \sum_{i=1}^n M_i (x_i - x_{i-1})$$ $$ = f(\frac{1}{4})(\frac{1}{4} - 0)\;+ f(\frac{1}{2})(\frac{1}{2} - \frac{1}{4})\;+ f(1)(1 - \frac{1}{2})$$ $$ = \frac{1}{8} + \frac{1}{4} + 1= \frac{11}{8}$$

Calculate $L(f,P)$: $$L(f,P) = \sum_{i=1}^n m_i (x_i - x_{i-1})$$ $$ = f(0)(\frac{1}{4} - 0)\;+ f(\frac{1}{4})(\frac{1}{2} - \frac{1}{4})\;+ f(\frac{1}{2})(1 - \frac{1}{2})$$ $$ = 0 + \frac{1}{8} + \frac{1}{2}= \frac{5}{8}$$

First off, can someone confirm that my calculations for upper and lower sums are correct?

Secondly, back to the main question, what is the $inf${U($f$,$P$)} and $sup${L($f$,$P$)} in this? as my $U$($f$,$P$) = $\frac{11}{8}$ and $L$($f$,$P$) = $\frac{5}{8}$.

As I already know that $f$ is integrable, $U$($f$,$P$) = $L$($f$,$P$) only if $f$ is constant, but what is the set in which I'm supposed to take the $inf$ and $sup$ of? As per the Darboux Integral definition, $sup${$L(f,P)$} = $inf${$U(f,P)$} for this function. .

If someone could clear this up for me it would be greatly appreciated

miweo
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  • Darboux sums over a partition are piecewise constant - a lower sum is at most the value of the integral (assuming it exists), and an upper sum is at least the value of the integral. As for $\sup$/$\inf$, those are fundamental concepts in analysis you should understand well. – Math1000 May 26 '18 at 20:33
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    I think that $U(f,P)$ and $L(f,P)$ are viewed as sequences on the partition $P$ and the $inf$ and $sup$ conditions are there because they imply convergance as the partition gets more precise. Your calculations seem correct indeed. – Μάρκος Καραμέρης May 26 '18 at 20:34
  • @Math1000 , how are lower sums "at most" the value and upper sums "at least" value of the integral in the subinterval? Isn't the lower sum, by definition, the lower bound of the integral (thus the "at least" value) and upper sums are the upper bound of the integral of the subinterval? – miweo May 26 '18 at 20:53
  • Consider $f(x) = 1 - (1-x)^2$ on $[0,2]$ and the partition $\mathcal P={0,1/2,1,3/2,2}$. The lower sum is \begin{align} L_f(\mathcal P) &= f(0)\cdot(1/2-0) + f(1/2)\cdot(1-1/2)+f(3/2)\cdot(3/2-1)+f(2)(2-3/2)\ &= \frac12\left(0 + 3/4 + 3/4 + 0 \right) = 3/4, \end{align} – Math1000 May 26 '18 at 22:44
  • while the upper sum is \begin{align} U_f(\mathcal P) = \frac12\left(f(1/2)+f(1) + f(1)+f(3/2)\right) = \frac12\left( 3/4 + 1 + 1 + 3/4 \right) = 7/4. \end{align} The actual value of the integral is $$ \int_0^2 (1-(1-x)^2)\ \mathsf dx = 4/3. $$ – Math1000 May 26 '18 at 22:46

2 Answers2

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Your calculations are fine and that's a good start. The idea of inf/sup of these upper and lower sums is easy to grasp in theory (but not in practice). Thus the value $\inf U(f, P) $ requires you to evaluate all the upper sums for each partition $P$ and then take infimum of all these sums. You can see that it is practically not possible to evaluate these sums for all partitions $P$ (because there are infinitely many partitions of a given interval). The practical method of finding the infimum here is based on the following deep and difficult theorem:

Theorem: Let $f:[a, b] \to\mathbb{R} $ be a bounded function and let $$A= \{U(f, P) \mid P\text{ is a partition of }[a, b] \}, I=\inf\, A$$ then $$I=\lim_{|P|\to 0}U(f,P)$$ where $|P|$ denotes the length of largest subinterval created by $P$.

Now one can take a uniform partition with $n$ subintervals of equal length and evaluate the upper sum over this partition and then take limit as $n\to\infty$ to get the desired infimum.

In your case let $P=\{x_0,x_1,\dots,x_n\}$ where $x_i=i/n$ and then we have $$U(f, P) =\sum_{i=1}^{n}M_i(x_i-x_{i-1})=\sum_{i=1}^{n}\frac{2i}{n}\cdot\frac{1}{n}=\frac{n(n+1)}{n^2}$$ and thus when we take limit as $n\to\infty$ we get the desired infimum as $1$.

Similarly $$L(f, P) =\sum_{i=1}^{n}\frac{2(i-1)}{n}\cdot\frac{1}{n}=\frac{n(n-1)}{n^2}$$ and hence on taking limit the supremum of lower sums is also $1$ and therefore the function $f(x) =2x$ is Riemann integrable on $[0,1]$ with integral $1$.

Paramanand Singh
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  • Wait, why can you just "take" a partition with subintervals of equal length? Aren't you supposed to take an arbitrary partition $P$ and then apply $||P||\to 0$? Because $\lim\limits_{||P||\to 0} U(f,P)$ doesn't mean just norm of partitions having equal length subintervals must approach $0$ but rather "all partitions with norm in neighbourhood of $0$". No? What am I missing? – William Aug 04 '22 at 17:13
  • @William : the deep and difficult theorem in my post says that the upper sums will tend to the infimum whenever the norm of partition tends to $0$. This holds for any arbitrary set of partitions. The case of partitions with subintervals of equal length is just a special case of the theorem. – Paramanand Singh Aug 05 '22 at 14:23
  • @William: just to clarify using your own words the set of all partitions with norm in neighborhood of $0$ includes the set of partitions with equal length subintervals and having norm in neighborhood of $0$. So the thing works for uniform partitions as well. – Paramanand Singh Aug 05 '22 at 16:36
  • I'm not sure I follow. But let me explain my thought process, may be you can point out where am I going wrong. $\lim\limits_{x \to 0} f(x)$ is not the same as $\lim\limits_{x \to 0, x \in S} f(x)$ where $S = { \frac{1}{n} : n \in \mathbb{N} }$. Former implies the latter but latter doesn't imply former. So if $\lim\limits_{n \to \infty} U(f,P_n)= L_1$ through a sequence of partitions $(P_n)$, it still isn't limit through every partition so there could be a case where $\lim\limits_{n \to \infty} U(f,Q_n)=L_2 ≠L_1$ through a different sequence of partition $(Q_n)$, no? – William Aug 07 '22 at 18:13
  • @William : your thought process is correct (with example of limit of $f(x) $ as $x\to 0$). The limit $\lim_{|P|\to 0}U(f,P)$ is like $\lim_{x\to 0}f(x)$ (so that it deals with every possible sequence of partitions) and it implies that all things like $U(f, P_n) $ will tend to same limit whenever norm of $P_n$ tends to $0$. – Paramanand Singh Aug 07 '22 at 22:19
  • Okay I may sound really dumb but if you know $\lim\limits_{x \to a} f(x) = L$ then you can comment $\lim\limits_{x \to a, x \in S} f(x) = L$ but it seems to me here, you're using $\lim\limits_{x \to a, x \in S}f(x)$ to calculate $\lim\limits_{x \to a} f(x) $ but in general, $\lim\limits_{x \to a, x \in S} f(x)=L \nRightarrow \lim\limits_{x \to a}f(x)=L$ or that it even exists. How so? – William Aug 08 '22 at 09:22
  • No @William: we are doing in exactly the reverse of what you propose in your last comment. Check definition for limit as norm of partition tends to $0$. The definition includes all partitions with norms less than $\delta$ and this also includes those partitions with special property that subintervals are of equal length. So the uniform partitions correspond to $x\in S$. To elaborate more we know that the limit via arbitrary partitions exists and equals the infimum of all upper sums. And hence the limit via uniform partitions also exists and equals the same infimum. Continued... – Paramanand Singh Aug 08 '22 at 15:40
  • @William: the existence of infimum of upper sums is a consequence of the fact that the function being integrated is bounded on interval of integration. The fact that limit of upper sums as norms of arbitrary partitions tend to $0$ equals the infimum of upper sums is formally stated in the deep and difficult theorem. And then we restrict ourselves to uniform partitions and reach the same limit. – Paramanand Singh Aug 08 '22 at 15:44
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Take infimum (greatest common lower bound) /supremum (lowest common upper bound) of all the arising values while $P$ ranges over all partitions.
Your calculations are correct and they show $$\frac58\ \le\ \int_0^12x\, dx=1\ \le\ \frac{11}8$$

Berci
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  • How do I take the infimum and supremum "of all the arising values while P ranges over all partitions"? – miweo May 26 '18 at 20:49
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    It's supremum of the set ${L(f,P):P\text{ a partition of }[0,1]}$, it contains real numbers, and every such set has a unique (possibly infinite) supremum. In your example, $\frac58$ is element of this set, and its supremum is $1$. For any $\varepsilon>0$, one can find a partition $P$ such that $L(f,P)>1-\varepsilon$. – Berci May 26 '18 at 21:45
  • Find a sequence of partitions for which the lower and upper sums converge to a finite limit (which is the value of the integral). – Math1000 May 26 '18 at 22:49
  • Ok thanks again for you guys' help. I have been struggling with this topic for quite some time now. Let me just see if I understand it now... so $P$={0, $\frac{1}{4}$, $\frac{1}{2}$, 1}, $U(f,P)$ = $\frac{11}{8}$ and $L(f,P)$ = $\frac{5}{8}$. $L(f,P)$ $\in$ $P$ $\therefore$ $\frac{1}{2}$ < $L(f,P)$ < 1; $\therefore$ the supremum of that subset of $P$ is 1. Now, $U(f,P)$ = $\frac{11}{8}$, $U(f,P)$ $\in$ $P$ as $P$ $\subseteq$ $\mathbb{R}$, $U(f,P)$ is an supremum of $P$ (as subset of $\mathbb{R}$ has infinte supremums), thus in $P$ 1 < $U(f,P)$ \therefore infimum of $U(f,P)$ is 1. – miweo May 26 '18 at 23:32
  • Since $sup${$L(f,P)$} = $inf${$U(f,P)$} = 1 $\rightarrow$ $f$ is integrable. Is this correct procedure? – miweo May 26 '18 at 23:32
  • Your last comment is valid. To actually get close to the supremum, one usually needs infinitely many, finer and finer partitions. That's crucial. You only took one partition. You should calculate for all, and then take their supremum. – Berci May 26 '18 at 23:37
  • Well, for the example, draw the graph of $2x$, the area below it (until the $x$ axis) over $0\le x\le1$ is the area of a right triangle, $1$. A lower or upper sum can also be depicted as area of horizontal-vertical bars below or above the graph of $2x$. – Berci May 27 '18 at 00:26