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Upper and lower sums, the two integrals every bounded function has, and the criterion that makes them agree.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to compute upper and lower sums for a partition, explain why refining narrows the gap between them, and state the upper and lower integrals as a supremum and an infimum over all partitions. You will also be able to use the integrability criterion — one partition per epsilon, with the gap bounded in terms of the number of pieces — and to show that the indicator of the rationals fails it.
Suprema and infima, from unit 1, and uniform continuity, from unit 3. The integral is built out of suprema and infima on pieces of an interval, and the theorem that continuous functions are integrable is uniform continuity applied once.
A partition of $[a,b]$ is a finite set of points $a = x_0 < x_1 < \cdots < x_n = b$. On the $i$-th piece write $M_i$ for the supremum of $f$ and $m_i$ for the infimum. The upper sum is $U(f, P) = \sum M_i \Delta x_i$ and the lower sum $L(f, P) = \sum m_i \Delta x_i$. A refinement of $P$ is a partition containing all of its points.
Let $f$ be bounded on $[a,b]$ — the whole construction needs it, since otherwise some $M_i$ is not a number.
Refining narrows the gap. Adding a cut point can only raise a lower sum and lower an upper sum, because a supremum over a smaller set is no larger. So $L(f,P) \le L(f,P') \le U(f,P') \le U(f,P)$ for any refinement $P'$.
Every lower sum is below every upper sum, even for unrelated partitions: compare both with their common refinement.
So the set of lower sums is bounded above and the set of upper sums is bounded below, and completeness gives two numbers: $$\underline{\int_a^b} f = \sup_P L(f,P), \qquad \overline{\int_a^b} f = \inf_P U(f,P),$$ the lower and upper integrals, with $\underline{\int} \le \overline{\int}$ always.
Definition. $f$ is integrable when the two are equal, and then $\int_a^b f$ is their common value.
The criterion. $f$ is integrable exactly when for every $\varepsilon > 0$ there is a partition $P$ with $U(f,P) - L(f,P) < \varepsilon$.
That is the working tool. It never mentions the value of the integral, so — like the Cauchy criterion for sequences — it establishes existence before anybody knows what the number is.
Another way: picture
Two staircases of rectangles, one drawn to the highest point of the graph on each piece and one to the lowest. The graph is caught between them. Refining the partition lowers the upper staircase and raises the lower one, and integrability is the question of whether they can be squeezed together.
Another way: steps
To show a function integrable by the criterion:
Step 2 is the only step that depends on the function. For a monotone one the gaps telescope; for a continuous one uniform continuity bounds every gap at once.
If the criterion holds: for each $\varepsilon$ there is $P$ with $U(f,P) - L(f,P) < \varepsilon$. Since $L(f,P) \le \underline{\int} \le \overline{\int} \le U(f,P)$, the two integrals differ by less than $\varepsilon$. True for every $\varepsilon$, so they are equal.
If $f$ is integrable: the common value $I$ is the supremum of the lower sums, so some $P_1$ has $L(f,P_1) > I - \varepsilon/2$; and it is the infimum of the upper sums, so some $P_2$ has $U(f,P_2) < I + \varepsilon/2$. Their common refinement $P$ does at least as well as both, so $U(f,P) - L(f,P) < \varepsilon$.
Both halves use the approximation property of a supremum, and the second uses the halving of $\varepsilon$. The whole of unit 1 is being spent here, which is a fair summary of why that unit came first.
Riemann's own definition takes sums $\sum f(t_i)\Delta x_i$ with an arbitrary tag $t_i$ in each piece, and asks that they converge as the mesh — the largest piece width — tends to zero.
Darboux's definition, used here, takes the supremum and infimum on each piece instead of a tag. The two are equivalent for bounded functions, and Darboux's is easier to work with because $U$ and $L$ are determined by the partition alone, with no tag to quantify over.
| Riemann | Darboux | |
|---|---|---|
| what varies | partition and tags | partition only |
| limit taken as | mesh to zero | supremum and infimum over partitions |
| needs boundedness | yes | yes |
| equivalent | — | yes, for bounded $f$ |
The name Riemann integral is kept for the object; the Darboux construction is how it is built.
Exhibiting one partition. The criterion asks for one per epsilon. A bound that depends on the number of pieces and tends to zero answers every epsilon at once; a single partition answers one.
Forgetting boundedness. An unbounded function has no upper sum, so the construction does not even begin. Integrals of unbounded functions are improper integrals, defined by a limit, which is a different definition.
Thinking discontinuity rules out integrability. A function with finitely many jumps is integrable, as the next lesson shows. What defeats integrability is discontinuity everywhere, or on a set too large to be covered cheaply.
Reading the two integrals as approximations. Both are exact numbers, defined for every bounded function. Integrability is the assertion that they coincide.
Let $f(x) = x$ on $[0,1]$, cut into $n$ equal pieces of width $1/n$.
A regular partition, in terms of $n$.
On each piece the rise is $1/n$, so $U - L = \sum (1/n)(1/n) = 1/n$.
Bound the gap in terms of the number of pieces.
Given $\varepsilon$, take $n > 1/\varepsilon$. So $f$ is integrable, and the sums squeeze to $1/2$.
Every epsilon answered, by one formula.
Let $f$ be $1$ on the rationals and $0$ on the irrationals, on $[0,1]$.
Bounded, so the construction applies.
Every piece of every partition contains both, so $M_i = 1$ and $m_i = 0$ always.
Density, used on every piece at once.
So every upper sum is $1$ and every lower sum is $0$: the upper integral is $1$, the lower is $0$, and no partition helps.
The gap cannot be made small.
Every lower sum is $0$, since every piece contains points where the function vanishes.
The lower integral, immediately.
For the upper sum, put a short piece of width $\delta$ around $1/2$: only that piece has a non-zero supremum, so the upper sum is at most $\delta$.
Isolate the bad point in a short piece.
So the upper integral is $0$ too, and the function is integrable with integral $0$. Changing a function at finitely many points changes neither integrability nor the value — which is the first sign that the integral cannot see very small sets.
Take $f(x) = x$ on the interval from $0$ to $3$, cut into two equal pieces. Give the lower sum, the upper sum, and their difference.
| value | |
|---|---|
| lower sum | |
| upper sum | |
| difference |
For $f(x) = x$ on the interval from $0$ to $7$, cut into $7$ equal pieces, what is the difference between the upper and the lower sum?
Answer:
Put the steps of showing a function integrable by the criterion into order.
Number the steps in order (write the number in the box):
Match each object in the construction of the integral to what it is.
| a partition | a refinement | an upper sum | the upper integral | |
|---|---|---|---|---|
| a finite list of $9$ cut points across the interval | ||||
| a partition containing all the cut points of another | ||||
| the sum of the largest value on each piece times its width | ||||
| the infimum of those sums over every partition |
Here is an argument that a bounded function is integrable. Mark the one step that does not follow.
This task has no paper form; do it on a device.
Let $f$ be $1$ at every rational and $0$ at every irrational, on the interval from $0$ to $5$. What are its upper and lower integrals?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Take $f(x) = x$ on the interval from $0$ to $6$, cut into two equal pieces. Give the lower sum, the upper sum, and their difference.
| value | |
|---|---|
| lower sum | |
| upper sum | |
| difference |
You can compute Darboux sums, say why every lower sum is below every upper sum, and use the criterion to show a function integrable. Say in your own words why one partition is never enough. Next: which functions the criterion actually admits.
10. Your turn: is the function that is $0$ except at $x = 1/2$, where it is $5$, integrable on $[0,1]$?, step 3