Riemann Integration

Attempt these from memory before reading.

• Define the upper and lower sums of a bounded function with respect to a partition.
• What does it mean for $f$ to be Riemann integrable?
• Is every continuous function on $[a,b]$ Riemann integrable?
• Is Dirichlet’s function Riemann integrable on $[0,1]$? Why not?
• State the two parts of the fundamental theorem of calculus.

Check your answers against the text. The Dirichlet question is the one that motivates everything in measure theory.

Everyone knows the area of a rectangle. The Riemann integral is the attempt to extend that to curved regions by slicing into rectangles and taking a limit.

The definition has to be careful about which rectangles, and it is here that the subject makes its first real demand. There are two obvious approaches: approximate from above with rectangles that contain the region, or from below with rectangles contained in it. If the two agree, that common value is the integral. If they do not, the region has no area in this sense — and finding out which functions are integrable turns out to be the interesting question.

That question is answered by the Lebesgue criterion: a bounded function on $[a,b]$ is Riemann integrable exactly when its discontinuities form a set of measure zero. The criterion is the bridge to measure theory, and it explains the counterexamples on this page in one line.

Let $f : [a,b] \to \mathbb{R}$ be bounded. A partition of $[a,b]$ is a finite set $P = \{x_0, x_1, \ldots, x_n\}$ with $a = x_0 \lt x_1 \lt \cdots \lt x_n = b$.

The upper sum and lower sum of $f$ with respect to $P$ are:

Where $M_i = \sup\{f(x) : x \in [x_{i-1}, x_i]\}$, $m_i = \inf\{f(x) : x \in [x_{i-1}, x_i]\}$ And $\Delta x_i = x_i - x_{i-1}$.

The mesh of $P$ is $\|P\| = \max_{1 \leq i \leq n} \Delta x_i$.

Definition. $f$ is Riemann integrable on $[a,b]$ if the upper and lower integrals are equal:

Where $\overline{\int_a^b} f = \inf\{U(f,P) : P \mathrm{\ is\ a\ partition}\}$ and $\underline{\int_a^b} f = \sup\{L(f,P) : P \mathrm{\ is\ a\ partition}\}$.

The common value is denoted $\int_a^b f(x)\, dx$.

Theorem 6.1 (Riemann Integrability Criterion). A bounded function $f : [a,b] \to \mathbb{R}$ is Riemann integrable if and only if for every $\varepsilon > 0$ There exists a partition $P$ such that

Theorem 6.2. Every continuous function on $[a,b]$ is Riemann integrable.

Proof. Let $f$ be continuous on $[a,b]$. By the Heine-Cantor theorem, $f$ is uniformly continuous. Given $\varepsilon > 0$ Choose $\delta > 0$ such that $|x - y| \lt \delta$ implies $|f(x) - f(y)| \lt \varepsilon/(b-a)$.

Let $P$ be any partition with $\|P\| \lt \delta$. On each subinterval $[x_{i-1}, x_i]$ By the Extreme Value Theorem, $f$ attains its maximum $M_i$ and minimum $m_i$. By uniform continuity: $M_i - m_i \lt \varepsilon/(b-a)$. Therefore:

By the Riemann integrability criterion, $f$ is integrable. $\blacksquare$

Theorem 6.3. Every monotone function on $[a,b]$ is Riemann integrable.

Proof. Assume $f$ is increasing (the decreasing case is analogous). Given $\varepsilon > 0$ Let $P_n$ be the uniform partition with $n$ subintervals of length $(b-a)/n$. On $[x_{i-1}, x_i]$: $M_i = f(x_i)$ and $m_i = f(x_{i-1})$. Then:

Choose $n$ large enough that $f(b) - f(a)/n \lt \varepsilon$. $\blacksquare$

Theorem 6.4. A bounded function with finitely many discontinuities on $[a,b]$ is Riemann integrable.

Proof (sketch). Let $f$ have discontinuities at $d_1, \ldots, d_m \in [a,b]$. Given $\varepsilon > 0$ Enclose each $d_j$ in a small interval $I_j$ of total length $\varepsilon/(2M)$ Where $M = \sup_{[a,b]} |f|$. On the remaining set (a finite union of closed intervals), $f$ is continuous, Hence uniformly continuous. Choose a partition fine enough that the oscillation of $f$ on each Subinterval outside the $I_j$ is less than $\varepsilon/(2(b-a))$. Then:

$\blacksquare$

Proposition 6.4a. The set of Riemann integrable functions on $[a,b]$ forms a vector space, and If $f$ and $g$ are integrable, then so are $|f|$, $f^2$ And $\max(f, g)$.

Theorem 6.4b (Lebesgue’s Criterion for Riemann Integrability). A bounded function $f : [a,b] \to \mathbb{R}$ Is Riemann integrable if and only if the set of its discontinuities has (Lebesgue) measure zero.

Remark. A set has measure zero if it can be covered by countably many intervals of arbitrarily Small total length. In particular, every countable set has measure zero. This means:

• Every continuous function is integrable (empty set of discontinuities).
• Every function with countably many discontinuities is integrable (Theorem 6.4 is a special case).
• The Dirichlet function $f(x) = 1$ for $x \in \mathbb{Q}$ and $f(x) = 0$ for $x \notin \mathbb{Q}$ is discontinuous everywhere (set of discontinuities = $[a,b]$ Measure $> 0$), hence not integrable.
• Thomae’s function $f(x) = 1/q$ if $x = p/q$ in lowest terms, and $f(x) = 0$ if $x$ is irrational, is continuous at every irrational and discontinuous at every rational. Since $\mathbb{Q}$ is countable (measure zero), Thomae’s function is Riemann integrable, with $\int_0^1 f = 0$.

Theorem 6.5 (Linearity). If $f$ and $g$ are integrable on $[a,b]$ and $\alpha, \beta \in \mathbb{R}$:

Theorem 6.6 (Monotonicity). If $f(x) \leq g(x)$ for all $x \in [a,b]$ Then $\int_a^b f \leq \int_a^b g$.

Theorem 6.7 (Triangle Inequality). $\left|\int_a^b f\right| \leq \int_a^b |f|$.

Theorem 6.8 (FTC Part 1). If $f$ is continuous on $[a,b]$ Then the function

Is differentiable on $(a,b)$ and $F'(x) = f(x)$.

Proof. Let $h > 0$ (the case $h \lt 0$ is similar). By the Mean Value Theorem for Integrals (which follows from the EVT), there exists $\xi \in [x, x+h]$ such that

As $h \to 0^+$ We have $\xi \to x^+$ (since $\xi \in [x, x+h]$). By continuity of $f$ $f(\xi) \to f(x)$. Hence $F'_+(x) = f(x)$. A similar argument gives $F'_-(x) = f(x)$. $\blacksquare$

Theorem 6.9 (FTC Part 2). If $F$ is differentiable on $[a,b]$ with $F' = f$ (and $f$ is integrable), Then

Proof. Let $P = \{x_0, \ldots, x_n\}$ be any partition of $[a,b]$. By the Mean Value Theorem, For each $i$ there exists $\xi_i \in [x_{i-1}, x_i]$ with $F(x_i) - F(x_{i-1}) = f(\xi_i)\Delta x_i$. Summing:

The right-hand side is a Riemann sum for $\int_a^b f$. As $\|P\| \to 0$ This converges to the Integral. Hence $F(b) - F(a) = \int_a^b f(x)\, dx$. $\blacksquare$

Problem. Compute $\int_0^1 x^2\, dx$ from the definition.

Solution. Let $P_n = \{0, 1/n, 2/n, \ldots, 1\}$. On $[x_{i-1}, x_i] = [(i-1)/n, i/n]$, $f(x) = x^2$ Has $M_i = (i/n)^2$ and $m_i = ((i-1)/n)^2$.

As $n \to \infty$: $\lim_{n \to \infty} U(f, P_n) = \lim_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2} = \frac{2}{6} = \frac{1}{3}$.

Similarly, $L(f, P_n) \to 1/3$. So $\int_0^1 x^2\, dx = 1/3$. $\blacksquare$

Definition. An improper integral is a Riemann integral where either the interval of integration Is unbounded or the integrand is unbounded.

Type I (Infinite Intervals). If $f$ is Riemann integrable on $[a, b]$ for every $b > a$ Define:

The integral converges if this limit exists as a finite number; otherwise it diverges.

Type II (Unbounded Integrands). If $f$ is unbounded near $a$ but integrable on $[c, b]$ for every $c \in (a, b]$:

Theorem 6.10 (Comparison Test for Improper Integrals). If $0 \leq f(x) \leq g(x)$ for $x \geq a$:

• If $\int_a^{\infty} g$ converges, then $\int_a^{\infty} f$ converges.
• If $\int_a^{\infty} f$ diverges, then $\int_a^{\infty} g$ diverges.

Theorem 6.11 (Absolute Convergence). If $\int_a^{\infty} |f(x)|\, dx$ converges, then $\int_a^{\infty} f(x)\, dx$ converges.

Theorem 6.12 ($p$-Test for Improper Integrals).

• Type I: $\int_1^{\infty} \frac{1}{x^p}\, dx$ converges if and only if $p > 1$.
• Type II: $\int_0^1 \frac{1}{x^p}\, dx$ converges if and only if $p < 1$.

Proof. For Type I with $p \neq 1$:

This converges when $1 - p < 0$ I.e., $p > 1$. For $p = 1$: $\int_1^{\infty} 1/x\, dx = \lim_{b \to \infty} \ln b = \infty$.

For Type II: $\int_0^1 x^{-p}\, dx = \lim_{c \to 0^+} \frac{1 - c^{1-p}}{1-p}$. This converges when $1 - p > 0$ I.e., $p < 1$. $\blacksquare$

Remark. The $p$-test for Type I integrals mirrors the $p$-series test: $\sum 1/n^p$ converges Iff $p > 1$. This is not a coincidence --- the integral test establishes the connection.

Why Dirichlet fails and Thomae succeeds. Every upper sum for Dirichlet’s function is $1$ (every interval contains a rational) and every lower sum is $0$ (every interval contains an irrational), so the upper and lower integrals differ by $1$. For Thomae, the discontinuities are the rationals — countable, hence of measure zero — so the upper and lower sums converge to the same value.

That is the Lebesgue criterion in action, and it is the reason the Riemann integral had to be replaced. A function that is $1$ on a dense set and $0$ on another dense set ought to have integral $0$ if the first set is small, and Riemann’s integral cannot say so.

The improper integral is a different object. $\int_0^1 \frac{1}{\sqrt x} dx = 2$ even though $1/\sqrt x$ is unbounded. The “integral” there is defined as a limit of proper integrals, not as an upper/lower sum, and the two notions agree only when the proper integral exists.

Mix these with material from Sequences and Limits, Continuity and Measure Theory.

1. (Sequences.) Show that $f(x) = x$ is integrable by computing the upper and lower sums for the uniform partition and taking the limit.
2. (Continuity.) Show that a continuous function on $[a,b]$ is uniformly continuous, and use that to prove it is Riemann integrable.
3. (Measure theory.) Show that a countable set has measure zero, and deduce that a bounded function with finitely many discontinuities is integrable.
4. (Measure theory.) Show Thomae’s function and Dirichlet’s function have the same set of discontinuities as each other’s complement, and explain why one is integrable and the other is not.
5. (Probability.) Let $X$ be uniform on $[0,1]$. Show that $P(\text{Dirichlet}(X) = 1) = 0$, and explain why Dirichlet’s function is Riemann integrable in the Lebesgue sense with integral $0$.

Mistake 1: Confusing Riemann integrability with Lebesgue integrability Every Riemann integrable function is Lebesgue integrable, but not vice versa. The Dirichlet function is not Riemann integrable but is Lebesgue integrable (with integral zero). The Riemann integral requires the set of discontinuities to have measure zero, while the Lebesgue integral handles a much broader class of functions.

Mistake 2: Interchanging limits and integrals without justification Swapping $\lim$ and $\int$ requires a convergence theorem such as the dominated convergence theorem or the monotone convergence theorem. Pointwise convergence alone is insufficient: the sequence $f_n = n\chi_{(0, 1/n)}$ converges pointwise to $0$ on $[0, 1]$, yet $\int_0^1 f_n = 1$ for all $n$. Always verify the hypotheses of a convergence theorem before interchanging limits and integrals.

Mistake 3: Applying the Fundamental Theorem of Calculus to discontinuous integrands The FTC Part 1 requires $f$ to be continuous. If $f$ has a jump discontinuity, the integral function $F(x) = \int_a^x f(t)\,dt$ is continuous but not differentiable at the discontinuity. The FTC Part 2 requires $F' = f$ everywhere on $[a, b]$, which fails if $f$ is discontinuous. Check continuity before applying the FTC.

The Riemann integral answers the question: how do we add up infinitely many infinitely thin slices? Consider the area under a curve f(x) from a to b. The Riemann sum approximates this by dividing the interval into subintervals and summing f(x) times the width of each subinterval — a collection of rectangles whose total area approaches the true area as the partition becomes finer. This is the same logic as computing distance by summing velocity over time, or charge by summing current over time: the integral accumulates a quantity that varies continuously.

The Fundamental Theorem of Calculus reveals the deep connection between accumulation (integration) and rate of change (differentiation). If F(x) represents the accumulated area from a to x, then F’(x) = f(x): the rate at which area accumulates is exactly the height of the curve. Conversely, integrating a rate function recovers the total change. This duality is why the FTC is the most important theorem in calculus: it converts the hard problem of computing areas into the easy problem of finding antiderivatives. The improper integral extends this to infinite intervals or unbounded functions by taking limits, but the core intuition — accumulation of slices — remains the same.

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