Riemann integration and criteria

In one sentence

A bounded function is Riemann integrable exactly when its discontinuities form a set of measure zero — and almost every question here is that criterion applied to a deliberately strange function.

Why the exam asks it

Because it separates candidates who think "integrable means you can find the area" from those who know the definition. Lebesgue's criterion turns a question that looks like it needs work into a five-second decision, and the paper rewards knowing it by asking about functions nobody could integrate directly.

The idea

Lebesgue's criterion

A bounded $f$ on $[a,b]$ is Riemann integrable if and only if its set of discontinuities has measure zero.

Both halves matter. Boundedness is a hypothesis, not a consequence — an unbounded function is not Riemann integrable at all, whatever its continuity. And "measure zero" is generous: any countable set has measure zero, so a function may be discontinuous at every rational and still integrate perfectly well.

That last point is the one the exam exploits, and it produces the pair you must know.

The Dirichlet function

$1_{\mathbb{Q}}$ on $[0,1]$: one at every rational, zero at every irrational.

It is bounded. It is discontinuous at every point — near any $x$ there are both rationals and irrationals, so the values jump by $1$ in every neighbourhood. The discontinuity set is $[0,1]$, which has measure $1$, not $0$. Not Riemann integrable.

Directly: every upper sum is $1$ and every lower sum is $0$, whatever the partition, because every subinterval contains both kinds of point.

It is Lebesgue integrable, with integral $0$, since the rationals have measure zero. That contrast is the standard motivation for the Lebesgue theory, and it is examined as such.

Thomae's function

Now change one thing: $t(x) = 1/q$ if $x = p/q$ in lowest terms, and $t(x) = 0$ for irrational $x$.

This is discontinuous at every rational and continuous at every irrational. The discontinuity set is $\mathbb{Q}$ — countable, measure zero. Riemann integrable, with integral $0$.

The reason is visible in the picture: fix any height $\varepsilon$. Only finitely many rationals have $1/q > \varepsilon$, so only finitely many spikes rise above it, and finitely many spikes can be covered by intervals of arbitrarily small total width.

Dirichlet and Thomae differ only in the height of the spikes, and that difference decides integrability. Carrying both is worth more than any theorem in this topic.

Where intuition breaks

"Bounded, so integrable." Dirichlet. Boundedness is necessary and nowhere near sufficient. This is a hypothesis dropped, and it is the most common single error in the topic.

"$|f|$ is integrable, so $f$ is." Take $f = 1$ on the rationals and $-1$ elsewhere. Then $|f| \equiv 1$, which is as integrable as anything gets. But $f$ is discontinuous everywhere, so it is not.

Why it is tempting: for Lebesgue integration, $|f|$ integrable does give $f$ integrable when $f$ is measurable. The implication people half-remember is from a different theory, and Riemann integrability is not preserved by taking absolute values in reverse.

"A composition of integrable functions is integrable." This one is genuinely surprising, and it is worth memorising because it cannot be guessed.

Let $t$ be Thomae's function and $g = 1_{(0,1]}$ — that is, $g(0) = 0$ and $g(y) = 1$ for $y > 0$. Both are Riemann integrable. But

which is exactly the Dirichlet function. Two integrable functions compose to a non-integrable one.

What survives: $g \circ f$ is integrable when $g$ is continuous and $f$ is integrable. It is the integrability of the outer function alone that is not enough.

"Products and sums of integrable functions are integrable." These are true, and appear among the false options for exactly that reason. Sums, products and absolute values of Riemann integrable functions are integrable — it is composition, and only composition, that fails.

"$\int_0^1 f x^n \, dx = 0$ for all $n$, so $f \equiv 0$." True for continuous $f$, by Weierstrass approximation. Drop continuity and it fails, since changing $f$ on a measure-zero set changes no integral. Another hypothesis dropped, and a favourite because the conclusion feels forced.

The exam's angle

1. Is it bounded? If not, stop — not Riemann integrable.
2. Where is it discontinuous? Sketch the set. Finite or countable means measure zero means integrable. An interval's worth of discontinuity means not.
3. If the option involves a composition, be suspicious. That is the one operation that breaks, and it is the reason the option is there.
4. If it says "for all $n$" about moments, check whether continuity was assumed.

Most questions here are Dirichlet or Thomae wearing a different coat. Recognising which one you are looking at settles the question faster than any computation.

The night before

• Lebesgue's criterion: bounded, and discontinuities of measure zero. Both halves.
• Dirichlet $1_{\mathbb{Q}}$: bounded, discontinuous everywhere, not Riemann integrable. Lebesgue integrable with integral $0$.
• Thomae $1/q$ at $p/q$: discontinuous exactly on $\mathbb{Q}$, is Riemann integrable, integral $0$.
• $|f|$ integrable does not give $f$ integrable.
• Sums, products, $|f|$: fine. Composition: not. $g \circ t =$ Dirichlet.
• $g \circ f$ is safe when $g$ is continuous.
• All moments vanishing forces $f \equiv 0$ only for continuous $f$.