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Continuous and monotone functions are integrable by different arguments; finitely many jumps cost nothing, and the exact boundary is about size.
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By the end of this lesson you will be able to prove that a continuous function on a closed bounded interval is integrable, using uniform continuity to make every piece's gap small at once, and that a monotone function is integrable by telescoping its rises with no continuity at all. You will also be able to handle finitely many jumps, and to state the exact criterion in terms of the size of the set of discontinuities.
The integrability criterion — one partition per epsilon, with the gap between the sums made small — and uniform continuity from unit 3. This lesson spends both, and the second is what makes the first theorem short.
The oscillation of a function on a piece is the gap between its supremum and its infimum there, so the difference of the two Darboux sums is the oscillations weighted by the widths. A set has measure zero when it can be covered by intervals of arbitrarily small total length.
Continuous implies integrable. Let $f$ be continuous on $[a,b]$. Given $\varepsilon$, uniform continuity supplies one $\delta$ with $|f(x) - f(y)| < \frac{\varepsilon}{b-a}$ whenever $|x-y| < \delta$. Take any partition with every piece narrower than $\delta$. Then $M_i - m_i \le \frac{\varepsilon}{b-a}$ on every piece at once, so $$U - L = \sum (M_i - m_i)\Delta x_i \le \frac{\varepsilon}{b-a}\sum \Delta x_i = \varepsilon.$$ The word doing the work is uniform. Ordinary continuity gives a delta per point and no single partition.
Monotone implies integrable. No continuity needed. For a regular partition into $n$ pieces, $M_i - m_i$ is the rise across the $i$-th piece, and those rises telescope: $$U - L = \frac{b-a}{n}\sum (f(x_i) - f(x_{i-1})) = \frac{(b-a)(f(b)-f(a))}{n}.$$ Choose $n$ large. A monotone function may have infinitely many jumps, and this estimate never notices them.
Finitely many discontinuities. A bounded function continuous except at finitely many points is integrable: cover each bad point by a piece of width $\delta$, contributing at most (oscillation) times $\delta$, and handle the rest by continuity.
The exact boundary. A bounded function is integrable exactly when its set of discontinuities has measure zero — can be covered by intervals of arbitrarily small total length. That theorem needs a theory of size this course does not build, and it is where the next analysis course starts.
Another way: picture
Think of the difference of the two staircases as a strip of area drawn over each piece: as tall as the function's swing there, as wide as the piece. Continuity makes every strip short at once; monotonicity makes the heights add up to a fixed total; a jump makes one strip tall, and narrowing its piece makes it thin.
Another way: steps
To decide integrability of a bounded function:
And before all of these: is it bounded? If not, the construction does not begin.
Thomae's function. $f(p/q) = 1/q$ in lowest terms, $f(x) = 0$ for irrational $x$. It is discontinuous at every rational and continuous at every irrational. It is integrable, with integral $0$: for a given $\varepsilon$, only finitely many points have $f(x) \ge \varepsilon$, and those can be covered by pieces of small total width.
The indicator of the rationals. Discontinuous everywhere, and not integrable: every upper sum is $b-a$ and every lower sum is $0$.
A monotone function with infinitely many jumps. Enumerate the rationals in $[0,1]$ and set $f(x) = \sum_{q_k \le x} 2^{-k}$. It increases, jumps at every rational, and is integrable by the monotone theorem.
The three together say the useful thing: what matters is not whether a function is discontinuous, nor even at how many points, but how much room the bad points take up.
| Operation on integrable $f$, $g$ | Integrable? |
|---|---|
| $f + g$, $cf$ | yes |
| $fg$ | yes |
| $|f|$ | yes |
| $\max(f,g)$, $\min(f,g)$ | yes |
| $f$ restricted to a subinterval | yes |
| a composition $g \circ f$ | not in general |
The last row is the one to remember. Composing an integrable function with a continuous one is safe in the order $(\text{continuous}) \circ (\text{integrable})$, and the other order can fail: Thomae's function composed with the indicator of $\{0\}$ produces the indicator of the rationals.
This is also where the Riemann integral starts to show its limits. Every entry above is a small theorem needing its own proof, whereas the same list for the Lebesgue integral is shorter to state and stronger, and its limit theorems — the subject of the next unit's hardest question — are far better behaved.
Discontinuous means not integrable. A step function is integrable, and so is a monotone function with infinitely many jumps. Discontinuity is priced by how much room it takes, not by whether it occurs.
Integrable means almost continuous. Thomae's function is discontinuous at a dense set of points and is integrable.
Bounded is enough. The indicator of the rationals is bounded and not integrable. Boundedness starts the construction; it does not finish it.
Monotone needs continuity. The monotone proof uses no continuity anywhere. It is the one theorem here that is genuinely independent of the continuity results.
Let $f(x) = \sin(x^2)$ on $[0, 3]$. It is continuous on a closed bounded interval, so it is uniformly continuous there.
The hypothesis that matters is the uniform one.
Given $\varepsilon$, take $\delta$ for the target $\varepsilon/3$, and any partition with pieces narrower than $\delta$.
Rescale the target by the width of the interval.
Then every gap is at most $\varepsilon/3$ and the widths sum to $3$, so $U - L \le \varepsilon$.
One partition, every epsilon.
Let $f$ increase on $[0,1]$ from $f(0) = 0$ to $f(1) = 1$, jumping at every rational.
Discontinuous at a dense set.
For $n$ equal pieces, $U - L = \dfrac{(1-0)(1-0)}{n} = \dfrac{1}{n}$.
The rises telescope, jumps and all.
Choose $n > 1/\varepsilon$. Integrable, and the argument never mentioned a single point of discontinuity.
Monotonicity, doing all the work.
It is bounded by $1$, so the construction begins; and it is continuous everywhere except at $0$.
Check boundedness, then locate the discontinuities.
Cover $0$ by a piece of width $\delta$: it contributes at most $2\delta$, since the oscillation there is at most $2$.
Isolate the bad point in a narrow piece.
On the rest, $[\delta, 1]$, the function is continuous on a closed bounded interval, so a fine enough partition makes its contribution small too. Integrable — and the wild oscillation cost nothing, because it happens in a place that can be made narrow.
Match each function on the interval from $0$ to $4$ to the reason it is, or is not, integrable there.
| integrable: continuous on a closed bounded interval | integrable: one jump, which a piece of tiny width covers | not integrable: unbounded, so the sums do not exist | not integrable: discontinuous at every point | |
|---|---|---|---|---|
| $x^2 + 4x$ | ||||
| the function equal to $0$ below $1$ and to $4$ from $1$ on | ||||
| $1/x$, with the value $0$ at $0$ | ||||
| the function equal to $1$ at rationals and $0$ at irrationals |
An increasing function on the interval from $0$ to $8$ rises in total by $8$. For a partition into $3$ equal pieces, what is the difference between the upper and the lower sum?
Answer:
Put the steps of the proof that a continuous function on a closed bounded interval is integrable into order.
Number the steps in order (write the number in the box):
Build the proof that a monotone function on a closed bounded interval is integrable.
This task has no paper form; do it on a device.
Here is an argument that a function with one jump is not integrable. Mark the one step that does not follow.
This task has no paper form; do it on a device.
A bounded function on the interval from $0$ to $9$ is discontinuous at every rational point and continuous at every irrational one. Is it integrable?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Match each function on the interval from $0$ to $3$ to the reason it is, or is not, integrable there.
| integrable: continuous on a closed bounded interval | integrable: one jump, which a piece of tiny width covers | not integrable: unbounded, so the sums do not exist | not integrable: discontinuous at every point | |
|---|---|---|---|---|
| $x^2 + 3x$ | ||||
| the function equal to $0$ below $1$ and to $3$ from $1$ on | ||||
| $1/x$, with the value $0$ at $0$ | ||||
| the function equal to $1$ at rationals and $0$ at irrationals |
You can prove both integrability theorems, say which hypothesis each uses, and explain why a jump costs nothing. Say in your own words why the continuous proof needs uniform continuity rather than continuity. Next: what may be done with an integral once it exists.
10. Your turn: is $f(x) = \sin(1/x)$, with $f(0) = 0$, integrable on $[0,1]$?, step 3