Which Functions Are Integrable

The Cauchy criterion reduces integrability to a single task: for each $\epsilon >0$, exhibit a partition on which

The factor $M_i-m_i$ is the oscillation of $f$ on the $i$-th subinterval — how far its values spread. Three structural properties each force this sum small for a different reason, and one function shows what goes wrong when none of them holds.

Integrability is thus a statement about controlling oscillation on a fine enough grid. Two mechanisms make the sum small: small individual oscillations ${\mathrm{osc}}_i$, or small total width where the oscillation is large. Each theorem below uses one of these.

Boundedness is not a detail that can be dropped. If $f$ were unbounded on some subinterval, then $M_i$ or $m_i$ would be infinite and the corresponding Darboux sum undefined; the entire theory presumes $f$ is bounded, and unbounded integrands are treated only as improper integrals.

Continuous functions

A continuous function on a closed bounded interval is uniformly continuous, so a single $\delta$ controls the oscillation on every subinterval at once.¹

Monotone functions

A monotone function need not be continuous — it may have jumps — yet it is always integrable. The reason is that its oscillation caps telescope.²

The picture is exact: the caps, whatever their widths, stack to a single column of height $f(b)-f(a)$, and shrinking the mesh scales that column's contribution to $U-L$ linearly. The bound is uniform in the shape of $f$: for any increasing $f:[0,1]\to \mathbb{R}$ with $f(0)=0$ and $f(1)=1$, the uniform partition into $N$ pieces gives $U(P,f)-L(P,f)=\frac{1}{N}$, no matter how the graph rises between the endpoints. Ten thousand pieces force agreement to $10^{-4}$.

Monotonicity also permits infinitely many discontinuities. A monotone function can jump at every rational and still be integrable, because its jumps are at most countably many and their total height is bounded by $f(b)-f(a)$; only finitely many exceed any fixed threshold, so the caps that matter are few.

Piecewise-continuous and finitely-many-discontinuity functions

A few bad points are harmless as long as the function stays bounded. Trap each discontinuity inside a strip so thin that its contribution to $U-L$ is negligible, and integrate the continuous pieces elsewhere.

Where integrability fails, and where it does not

Two bounded functions with dense discontinuities separate the concepts. The Dirichlet function is not integrable; the Thomae function, discontinuous at every rational, is.

The difference is measure-theoretic. Dirichlet is discontinuous everywhere; Thomae is discontinuous only on $\mathbb{Q}$, a countable — hence measure-zero — set. That distinction marks the boundary.

Every class above is a special case: continuous functions have empty discontinuity set, monotone functions have at most countably many jumps, finitely-many-discontinuity functions have finite sets, and Thomae's set $\mathbb{Q}$ is countable — all measure zero. Dirichlet's discontinuity set has measure equal to the whole interval, so it alone fails.