Every Upper Sum Says One, Every Lower Sum Says Zero, and the Gap Is the Measure of the Discontinuities

The function that is 1 on every rational and 0 on every irrational does not defeat Riemann's integral by being hard to compute. It defeats it by returning two numbers. Every upper sum is 1, every lower sum is 0, for every partition that will ever be written down, and the definition refuses to hand anything back until those two meet.

The repair is one change of question, and the machinery behind it is short: an outer measure built from coverings, an integral built from simple functions, and one theorem that says exactly which bounded functions Riemann can handle. That theorem is the interesting part, because it does not merely record that this function fails. It says why, in a currency you can compute.

Where the definition actually jams

Take $f$ bounded on $[a,b]$. A partition $P$ is a finite list of cut points $a = x_0 < x_1 < \cdots < x_n = b$, and on each cell you record the highest and lowest the function gets:

The two staircases those numbers build are the upper and lower sums, $U(f,P) = \sum_i M_i \Delta x_i$ and $L(f,P) = \sum_i m_i \Delta x_i$. Cutting a cell in two replaces one supremum by two smaller ones and one infimum by two larger ones, so refining a partition can only push $U$ down and $L$ up. That monotonicity is what makes an infimum and a supremum the right operations to finish with.

Now feed it $\mathbf{1}_{\mathbb{Q}}$ on $[0,1]$. Between any two distinct reals there is a rational and there is an irrational, so every cell of every partition of positive width contains one of each. Which fixes $M_i = 1$ and $m_i = 0$ in every cell, of every partition, forever:

Notice what is not happening. The gap is not shrinking slowly, or shrinking along some partitions and not others. It is the same 1 at every mesh, so there is no cleverer sequence of partitions waiting to be found. The upper integral is 1, the lower integral is 0, and the definition has nothing to say.

Covering the rationals for less than any price you name

Before the second integral there has to be a notion of the size of an arbitrary set, and the only honest construction is to buy the set a coat and then shop around:

Countably many intervals, not finitely many, and that is the whole difference. Now take $E = \mathbb{Q} \cap [0,1]$. It is countable, so its elements can be listed as $q_1, q_2, q_3, \dots$ in some order. Pick any $\varepsilon > 0$ and hand the $k$-th rational an interval centred on it of width $\varepsilon / 2^{k}$:

Every rational is inside its own interval, so the union covers $E$, and the total length is $\varepsilon$. Since $\lambda^{*}(E) \le \varepsilon$ holds for every positive $\varepsilon$ at once, $\lambda^{*}(E) = 0$.

The load-bearing fact is that $\sum 2^{-k}$ converges: infinitely many payments can still have a finite total. Halving is not special, any convergent series of positive terms would do the same job, and halving is chosen because it is the cheapest thing to write down.

Two things are worth stopping on. The intervals overlap grotesquely, because the rationals are dense and neighbouring brackets swallow each other. That is harmless: the sum of the lengths over-counts the union, and the inequality being claimed runs in that direction.

The other thing is what the union looks like. It is an open set containing every rational in $[0,1]$, hence dense, of total length at most $\varepsilon$. Its complement inside the interval is closed, contains no rational whatsoever, and has measure at least $1 - \varepsilon$. Nearly all of the interval, measured by length, can be kept clear of a set that meets every subinterval you can name.

If the argument looks like it proves too much, check it against $[0,1]$ itself, whose outer measure is 1. The hard half of that is the lower bound, and it is where compactness earns its keep: from any cover of a closed bounded interval by open intervals you can extract a finite subcover, and finitely many intervals covering $[0,1]$ have lengths adding to at least 1. Remove that step and the whole theory collapses to zero.

Simple functions first, and everything else by supremum

A simple function takes finitely many values on measurable sets, and its integral is the one thing it could possibly be:

with the convention $0 \cdot \infty = 0$, so that a height of zero over an infinite set contributes nothing. The representation is not unique, but any two of them refine to a common one by intersecting the pieces, and the sum survives the refinement, so the number in (5) is well defined rather than merely written down.

Everything else is a supremum of those. For measurable $f \ge 0$,

From below, not from above, and that asymmetry is doing work. Approximating from below is monotone in $f$ and behaves well along increasing sequences, which is where the monotone convergence theorem comes from; approximating from above would need the function to be bounded and would reintroduce exactly the squeeze that failed in (2). A general $f$ is split into $f^{+} = \max(f,0)$ and $f^{-} = \max(-f,0)$; it is integrable when $\int \lvert f \rvert \, d\lambda < \infty$, and then $\int f = \int f^{+} - \int f^{-}$.

The monster is now a one-line computation, because it never needed the supremum in the first place. It is already simple: two values, two sets.

And the picture of slicing horizontally is not a metaphor for (6), it is a formula. For any measurable $f \ge 0$,

an integral of a single decreasing function of the height. The right-hand side never mentions where in the domain the function is large, only how much room it takes up there.

The exact test, and a function that passes it

The theorem that decides Riemann integrability is stated in terms of oscillation, which is the local version of $M_i - m_i$:

A function is continuous at $x$ exactly when $\operatorname{osc}(f,x) = 0$, so the discontinuity set is $D_f = \{ x : \operatorname{osc}(f,x) > 0 \}$. And the quantity the definition of section one hinges on is built from the very same oscillation, one cell at a time:

where $M_i - m_i$ is the oscillation of $f$ across the $i$-th cell. Riemann integrability asks that this be made small, which means asking that the cells where the oscillation is not small be short. That is a statement about the length of a set, so it should not be a surprise that the answer is a measure.

Run it on $\mathbf{1}_{\mathbb{Q}}$. Every neighbourhood of every point contains rationals and irrationals, so the oscillation is 1 at every single point, and $D_f = [0,1]$ with measure 1. The criterion says no, and it says more than no: putting $\operatorname{osc}_i = 1$ into (10) returns $U - L = \sum_i \Delta x_i = 1$, which is the same 1 as in (2). The gap the definition could not close and the measure of the discontinuity set are not two facts that happen to agree. They are one number.

Which raises the obvious objection: is the criterion just a restatement of "discontinuous everywhere is bad"? No, and the function that shows it has teeth is Thomae's:

It is discontinuous at every rational and continuous at every irrational, so it is discontinuous on a dense set, exactly like the monster. But that dense set is $\mathbb{Q}$, of measure zero, so the criterion says it is Riemann integrable — and its integral is 0.

The direct proof is short enough to keep the criterion honest. Fix $\varepsilon > 0$. Only finitely many points have $t(x) \ge \varepsilon/2$, namely those with $q \le 2/\varepsilon$, so they can be shut inside finitely many cells of total width below $\varepsilon/2$. On those cells the supremum is at most 1; on all the others it is below $\varepsilon/2$. So $U(t,P) \le 1 \cdot \tfrac{\varepsilon}{2} + \tfrac{\varepsilon}{2} \cdot 1 = \varepsilon$, while $L(t,P) = 0$ because every cell contains an irrational.

So here are two functions, both discontinuous on a dense set, one integrable and one not. Density is not what Riemann objects to. Measure is.

Where the two integrals are the same integral

The criterion already carries the reassurance, but it is worth saying separately: a bounded Riemann integrable function is Lebesgue measurable and the two integrals return the same number. There is no function on which they disagree, so nothing you learned about areas has been repealed.

Watch it on $x^2$ over $[0,1]$, computed twice in incompatible ways. Vertically, with right endpoints and the sum of squares:

Horizontally, through (8): for $0 \le t < 1$ the set where $x^2 > t$ is the interval $(\sqrt{t}, 1]$, whose length is $1 - \sqrt{t}$, and above $t = 1$ the set is empty:

Two genuinely different objects were measured — lengths of pieces of the domain in (12), lengths of level sets in (13) — and landing on the same third is a theorem, not an accident of this particular parabola.

There is a second sense in which Lebesgue extends rather than replaces, and it is the one that made the construction spread. Riemann's definition is welded to a bounded interval of the real line. Lebesgue's asks only for a set, a σ-algebra on it and a measure, which is why the same symbol later carries an expectation over a probability space and a sum over a countable set without a word being changed.

One integral Riemann has and Lebesgue does not

None of this is a free upgrade, and the cleanest counterexample is one every calculus course already computes. On the half line, $\sin(x)/x$ extends continuously to the origin with value 1, so nothing goes wrong there, and the improper Riemann integral converges:

It converges the way an alternating series does: the lobes above and below the axis nearly cancel, and what is left shrinks. Take the cancellation away and the thing diverges, which is a three-line estimate. On $[k\pi, (k+1)\pi]$ the factor $1/x$ is at least $1/((k+1)\pi)$, and one full lobe of $\lvert \sin \rvert$ has area 2:

Lebesgue's definition splits a function into its positive and negative parts and asks both to be finite. Here both are infinite, so $\int f^{+} - \int f^{-}$ is $\infty - \infty$ and the integral does not exist. So $\sin(x)/x$ is improperly Riemann integrable on $(0,\infty)$ and not Lebesgue integrable there, and the containment that held on $[a,b]$ fails once the domain is unbounded.

The reason is worth more than the example. (14) is a limit of integrals over $[0,R]$, so its value depends on the order in which the domain is exhausted. Lebesgue's integral has no order anywhere in it: it sorts by value, and sorting destroys the arrangement that produced the cancellation. On a countable set the Lebesgue integral is literally a series sum, and a series whose sum survives every reordering is precisely an absolutely convergent one — a conditionally convergent one can be rearranged to any value you like. The half-line integral is the continuous form of the same fact, and $\pi/2$ is the value of one particular arrangement.

That single dominating function is a real hypothesis rather than decoration. The spikes $n \cdot \mathbf{1}_{(0,1/n)}$ tend to 0 at every point and each has integral 1, so the conclusion fails; the smallest function sitting above all of them behaves like $1/x$ near the origin, whose integral diverges, so there was never a $g$ and the theorem was never claiming anything.

Practically, nothing is lost in arithmetic. Every $\int_{(0,R]} \sin(x)/x \, d\lambda$ is a perfectly ordinary Lebesgue integral and the limit is still $\pi/2$. What is lost is the right to call that limit the integral of the function on the half line, and with it the right to feed the function to any theorem whose hypothesis reads $f \in L^1$.

Sources and further reading

• The upper and lower sums of (1) and the criterion built on them — Darboux integral
• The theorem of section four, stated where it lives — Riemann integral
• Sets of measure zero, and why countable is enough but not necessary — Null set
• The construction of (5), (6) and (8) — Lebesgue integration
• The function of (11), and the proof that it is continuous exactly at the irrationals — Thomae's function
• The value $\pi/2$ in (14) — Dirichlet integral
• Why an unordered sum has to be absolute, in the discrete case — Riemann series theorem
• The theorem the trade was made for — Dominated convergence theorem

Two questions were deliberately left standing. The first is which sets are allowed into (3) at all: not every subset of the line can be given a length consistently, and a set that cannot is built with the axiom of choice, which is why the σ-algebra in the definition is a restriction rather than a formality. The second is completeness — the space of Lebesgue integrable functions has no holes in it, and Riemann's does, which is the reason the apparatus was worth building and not merely worth admiring.