Integration over a box is the one-variable construction with rectangles replaced by boxes and nothing else changed. The two real questions are which functions are integrable — answered exactly, by a measure-zero condition — and how to compute the thing, answered by Fubini, which reduces an $n$-dimensional integral to $n$ one-dimensional ones.
Volume, mass, probability and flux are all integrals over regions rather than intervals. The construction has to come before any of the theorems about changing variables or integrating over surfaces, and its limits are what motivated the Lebesgue theory that follows in MAT357H1.
Every numerical quadrature scheme approximates exactly these sums, and the convergence rate depends on the smoothness assumptions this unit isolates. Monte Carlo integration exists because the box-partition approach costs exponentially in $n$.
Riemann's 1854 definition was already known to be too fragile by the 1890s: pointwise limits of integrable functions need not be integrable. Lebesgue's 1902 thesis fixed it, and the criterion in this unit — integrable exactly when the discontinuities have measure zero — is the precise statement of what Riemann's theory can and cannot see.
A closed box is $Q=[a_1,b_1]\times\dots\times[a_n,b_n]$ with volume $v(Q)=\prod_i(b_i-a_i)$. A partition $P$ of $Q$ subdivides each edge, cutting $Q$ into finitely many subboxes $R$. For bounded $f:Q\to\mathbb{R}$ set
$f$ is integrable if $\sup_PL(f,P)=\inf_PU(f,P)$, and the common value is $\int_Qf$.
A bounded $f$ is integrable on $Q$ iff for every $\varepsilon>0$ there is a partition $P$ with $U(f,P)-L(f,P)<\varepsilon$.
Refining a partition raises $L$ and lowers $U$, and any two partitions have a common refinement, so every lower sum is at most every upper sum. The supremum and infimum therefore exist with $\sup L\le\inf U$, and the criterion is exactly the statement that no gap remains.
A continuous $f$ on a closed box $Q$ is integrable.
Uniform continuity is the load-bearing step, and it is the reason compactness had to come first.
$A\subseteq\mathbb{R}^n$ has measure zero if for every $\varepsilon>0$ it can be covered by countably many boxes of total volume less than $\varepsilon$. It has content zero if finitely many suffice.
A countable union of measure-zero sets has measure zero (split $\varepsilon$ as $\sum\varepsilon2^{-k}$). Any hyperplane, any graph of a continuous function on a box, and any smooth surface of dimension less than $n$ has measure zero in $\mathbb{R}^n$. A box of positive volume does not.
A bounded $f$ on a box $Q$ is Riemann integrable if and only if its set of discontinuities has measure zero.
This is the sharp statement. It immediately gives: continuous functions are integrable; functions with finitely or countably many jumps are integrable; the indicator of $\mathbb{Q}\cap[0,1]$ is not, because it is discontinuous everywhere.
For a bounded $S\subseteq\mathbb{R}^n$, put $S$ inside a box $Q$ and define $\int_Sf=\int_Qf\chi_S$, where $\chi_S$ is the indicator function. The definition works when $f\chi_S$ is integrable, and by Lebesgue's criterion that needs $\partial S$ to have measure zero. Such an $S$ is called Jordan measurable, and $v(S)=\int_S1$ is its volume.
$S=\mathbb{Q}^n\cap[0,1]^n$ has boundary all of $[0,1]^n$, which does not have measure zero. So $\chi_S$ is not integrable and $S$ has no Jordan volume, even though it is countable. Lebesgue measure assigns it $0$; this is one of the gaps Riemann's theory leaves.
Let $Q=A\times B$ with $A\subseteq\mathbb{R}^n$, $B\subseteq\mathbb{R}^m$ boxes, and let $f$ be integrable on $Q$. Then
where the inner integral is interpreted as a lower or upper integral at the (measure-zero set of) $x$ where $y\mapsto f(x,y)$ fails to be integrable. If $f$ is continuous no such care is needed and the two iterated orders agree.
The proof partitions $Q$ as a product of partitions of $A$ and $B$, and observes that the upper and lower sums factor. Everything else is bookkeeping.
Both iterated integrals can exist and differ when $f$ is not integrable on the product. The standard example on $[0,1]^2$ is $f(x,y)=\dfrac{x^2-y^2}{(x^2+y^2)^2}$, for which the two orders give $\pi/4$ and $-\pi/4$. Fubini needs integrability on the product as a hypothesis, not as a conclusion.
For $S=\{(x,y):a\le x\le b,\ \varphi(x)\le y\le\psi(x)\}$ with $\varphi$, $\psi$ continuous, the boundary is two graphs and two segments, all of measure zero, so $S$ is Jordan measurable and
This is the "type I region" recipe from first-year calculus, now with a reason attached.
• Confusing measure zero with content zero. They differ exactly for
non-compact sets, and $\mathbb{Q}\cap[0,1]$ is the example.
• Using Fubini without knowing $f$ is integrable on the product. Equality
of iterated integrals is the conclusion, not the hypothesis.
• Assuming every bounded set has a volume. It needs a measure-zero
boundary.
• Thinking countable implies content zero. The rationals are countable and
dense.
• Forgetting that changing $f$ on a measure-zero set changes neither
integrability nor the integral.
Original problems written for this guide in the style of Spivak and Munkres exercises. Write a full solution before opening the answer.
Is $\chi_{\mathbb{Q}^2}$ integrable on $[0,1]^2$?
Let $f=1$ on $[0,1]^2$ except $f=0$ on the segment $y=\tfrac12$. Is $f$ integrable, and what is the integral?
Evaluate $\int_0^1\int_x^1 e^{y^2}\,dy\,dx$.
Prove the graph of a continuous $g:[0,1]\to\mathbb{R}$ has measure zero in $\mathbb{R}^2$.
Compute the volume of $\{x_i\ge0,\ x_1+x_2+x_3\le1\}$.
Show a countable union of measure-zero sets has measure zero.
Can an integrable function be discontinuous on a dense set?
Show the closed unit ball is Jordan measurable.
Course coverage. This unit is part of the complete MAT257Y1 topic map. It should cover the following items before you move on:
Spivak's integration chapter is the reference for measure zero and Lebesgue's criterion; Munkres develops the same material with more worked detail on Jordan measurability. Verify the assigned text against the syllabus.
Compactness gives uniform continuity gives integrability of continuous functions. Lebesgue's criterion generalises that completely. Fubini turns the definition into something computable, and Unit 10 adds the last tool, the change of variables.
Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.
State exactly when a bounded function on a box is Riemann integrable.
Give a set with measure zero but not content zero.
What must be assumed before equating iterated integrals?
Which bounded sets are Jordan measurable?