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I have difficulty understanding the definition of a Riemann integrable function $f$ on $[a,b]$. I understand the definition of upper sum $U(f,P) = \sum\limits_{i=1}^{n}M_i \Delta x_i$ and lower sum $L(f,P) = \sum\limits_{i=1}^{n}m_i \Delta x_i$, where $M_i(f)=\sup \{\ f(x):x \in [x_{i-1},x_i] \}$ and $m_i(f)=\inf \{\ f(x):x \in [x_{i-1},x_i] \}$. I think my problem is with the definition of upper integral $U(f) = \inf \{\ U(f,P): P$ is a partition of $[a,b] \}$ and lower integral $L(f) = \sup \{\ L(f,P): P$ is a partition of $[a,b] \}$. Now we define, $\int_a^b f(x) dx = L(f) = U(f)$. Are we allowed to take any partition? To better illustrate my problem, here is an example:

For simplicity, consider $f(x) = x$ on $[0,10]$. We know $\int_0^{10} x dx=50$. I would like to create partitions of length $1$, so $P=\{\ 0,1,...,10 \}$. If I understand correctly, $U(f,P)=55$ and $L(f,P) = 45$. But to me, $U(f)$ seems to be $1$, the smallest rectangle in $U(f,P)$ and $L(f) = 9$, the largest rectangle we can have with $L(f,P)$ and obviously we do not have $L(f)=U(f)$.

I know I am missing a point. Please help me find the problem with my reasoning.

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    $\begingroup$ The point is not that the upper and lower sums should be equal for any partition, but, you consider the minimum upper sum over all possible partitions, and the largest possible lower sum. $\endgroup$ – Don Thousand Apr 15 at 2:03
  • $\begingroup$ Oh I see your point. But how do we know, that for example in this case, we would be having $L(f)=U(f)=50$? What guarantees that? $\endgroup$ – Rob Apr 15 at 2:06
  • $\begingroup$ Well, it's the fact that it is Riemann integrable that guarantees that. If it weren't, they wouldn't be equal. You can look up a proof for why Riemann integrable sets have this property. It's not trivial but also not terribly complicated. $\endgroup$ – Don Thousand Apr 15 at 2:15
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Let us define two sets:

$A= \{ U(f,P)$ where P is any partition of $[a,b] \}$,

$B= \{ L(f,P)$ where P is any partition of $[a,b] \}$,

where $U$ are upper sums and $L$ are lower sums. If we take some partition $P$ of a segment $[a,b]$ we get some upper and lower sums.

Set $A$ contains all possible upper sums for every partition $P$ (every partition generates one number to be upper sum).

Set $B$ contains all possible lower sums for every partition $P$ (every partition generates one number to be lower sum).

Now we define two numbers (which I will denote little differently from you so that it can be more clear):

$I_{*} = \inf(A)$,

$I^{*} = \sup(B)$.

Remark. We presuppose existence of $\inf(A)$ and $\sup(B)$ (if they do not exist then function is not R- integrabile).

Number $I_{*}$ is lowest of all possible upper sums, more precisely it is infimum of A.

Number $I^{*}$ is greatest of all possible lower sums, more precisely it is supremum of B.

Now we can define when function is R-integrabile (if lowest upper sum and greatest lower sum coincide). Function $f$ is Riemann integrable if $I_{*}=I^{*}$ and integral is equal to $I_{*}$ or $I^{*}$.

Let us take example of $f(x) = x$ on segment $[0,10]$, ie. $\int_{0}^{10}f(x)dx$. It will not suffice to take any partition $P$ of segment $[0,10]$ and then calculate upper and lower sum. We need to take all partitions of segment $[0,10]$ and then get upper and lower sum for all possible partitions. Then construct sets as previously explained and then find lowest possible upper sum and greatest lower sum (more precisely $\inf(A)$ and $\sup(B)$).

You can see not only rigorously (which I will not show here) but graphically that upper sum gets lower and lower sum gets bigger as equidistant partitions become shorter and shorter. So you can simply presuppose some equidistant partition of length $\Delta x$ and then calculate upper or lower sum and after that take limit as $\Delta x$ goes to zero. In that way you will get $\int_{0}^{10}f(x)dx=50$.

Feel free to ask follow up questions.

Edit: (To big to be a comment).

@Rob It is not nessecary that there even exists partition that gives $U(f)$ and $L(f)$. If you have function simple as $f(x) =x$ it will not exist, because for every partition upper and lower sums (they are geometrically rectangles) are at least slightly bigger or smaller. However, what you are intreseted in are limits of these upper and lower sums, and they can exist even if there is no particular partition which realizes these limits of lower and upper sums.

Think of this analogy: take an example of set $S= \{ \dfrac{1}{n} : n \in \mathbb{N}\}$. Infimum $\inf(S) = 0$ although there is not any $n$ for which $\dfrac{1}{n}$ is zero. In the same way, although there is no partition which gives you $U(f)$ and $L(f)$ still they are infimum and supremum of associated sets.

Intuitively your observation is correct, and that is why its sometimes said that when doing integration you divide your segment into infinite little pieces and then sum area of infinitely thin rectangles.

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  • $\begingroup$ Thanks for the explanation. I realized I misunderstood the definition of $L(f)$ and $U(f)$. However, now I have another problem. It seems to me that the partition that gives $L(f)$ and $U(f)$ have to consist of infinite number of points within in the interval. But to my understanding, these partitions must be finite in number. Is my observation correct? $\endgroup$ – Rob Apr 15 at 18:10
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    $\begingroup$ @Rob I have anwered in the question because my comment was to long. Sorry for bad english, if you have further question feel free to ask. $\endgroup$ – Thom Apr 15 at 19:19
  • $\begingroup$ I completely understand now. The fuzzy world of infinity causing problems again. Thanks! $\endgroup$ – Rob Apr 15 at 19:42

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