Real analysis · The Lumenary

Lens

MotivationH

When calculus goes wrong

Here is a famous series:

1 - \tfrac12 + \tfrac13 - \tfrac14 + \tfrac15 - \tfrac16 + \cdots = \ln 2 \approx 0.693.

Now add up exactly the same numbers in a different order — two positive terms, then one negative:

1 + \tfrac13 - \tfrac12 + \tfrac15 + \tfrac17 - \tfrac14 + \tfrac19 + \tfrac1{11} - \tfrac16 + \cdots = \tfrac32 \ln 2 \approx 1.040.

Same terms, different sum. Riemann proved in 1854 that a series like this one, which converges but not absolutely, can be rearranged to add up to any real number you like. Addition of infinitely many numbers is not addition.

The history of calculus is full of such surprises. In his Cours d'analyse of 1821, Augustin-Louis Cauchy "proved" that a convergent series of continuous functions has a continuous sum. Five years later Niels Henrik Abel pointed out that the Fourier series

\sin x - \tfrac12 \sin 2x + \tfrac13 \sin 3x - \cdots

converges for every $x$ to a function that jumps at $x = \pi$ . In 1872 Karl Weierstrass presented a continuous function that has a tangent line nowhere — at a time when many mathematicians believed a continuous function must be differentiable at most points.

Real analysis is the response: precise definitions of limit, continuity, convergence and integral, and proofs of everything calculus uses. All of it rests on one property of the real numbers, completeness: the real line has no holes. In this lesson you will prove, from that single axiom, the intermediate value theorem, the extreme value theorem, the Bolzano–Weierstrass theorem and the fundamental theorem of calculus, and you will learn exactly when two limiting processes can be swapped.

IntuitionH

No holes, and one N for everyone

The rational numbers are full of holes. The decimal approximations $1, 1.4, 1.41, 1.414, 1.4142, \dots$ to $\sqrt 2$ get closer and closer to each other, but in $\Q$ there is nothing for them to converge to. The set $\{x \in \Q : x^2 < 2\}$ is bounded above but has no least upper bound in $\Q$ . Completeness says that in $\R$ this never happens: every sequence that "ought to converge" does, and every bounded set has a sharpest bound. Each of the great theorems of this lesson fails over $\Q$ . The function $f(x) = x^2 - 2$ on $[0, 2] \cap \Q$ changes sign without ever being $0$ .

The second big idea is about the order of quantifiers. A sequence of functions $f_n$ converges to $f$ pointwise on $S$ if for each $x \in S$ the numbers $f_n(x)$ converge to $f(x)$ . It converges uniformly if one threshold $N$ works for all $x$ at once: once $n \ge N$ , the whole graph of $f_n$ lies inside a tube of width $\varepsilon$ around the graph of $f$ .

Uniform convergence: for $n \ge N$ the whole graph of $f_n$ (solid) lies in the tube of half-width $\varepsilon$ around the limit $f$ (dashed). Pointwise convergence only asks each vertical line to be eventually inside the tube — at its own pace.

The difference matters because limits are delicate. The continuous functions $x^n$ on $[0, 1]$ converge pointwise to a function that is $0$ on $[0, 1)$ and jumps to $1$ at $x = 1$ . The threshold $N$ that makes $x^n < \varepsilon$ gets larger and larger as $x$ approaches $1$ , and no single $N$ works. Uniform convergence is precisely the hypothesis that rescues continuity, integrals and (with care) derivatives.

DefinitionsH

Definitions

Throughout, $|x - y|$ is the distance between real numbers. We take $\R$ to be an ordered field (the familiar rules of arithmetic and inequalities) with one extra axiom.

Definition— supremum and the completeness axiom

An upper bound of $S \subseteq \R$ is a number $u$ with $x \le u$ for all $x \in S$ . A supremum (least upper bound) of $S$ is an upper bound $s$ such that $s \le u$ for every upper bound $u$ ; we write $s = \sup S$ . The completeness axiom: every non-empty subset of $\R$ that is bounded above has a supremum.

The supremum is characterised by an approximation property: an upper bound $s$ of $S$ is $\sup S$ if and only if for every $\varepsilon > 0$ there is $x \in S$ with $x > s - \varepsilon$ . (Otherwise $s - \varepsilon$ would be a smaller upper bound.) Completeness implies the Archimedean property: $\N$ is not bounded above in $\R$ . Indeed, if $s = \sup \N$ existed, some $n > s - 1$ would give $n + 1 > s$ . Hence $1/n \to 0$ .

Definition— convergence and Cauchy sequences

A sequence $(a_n)$ converges to $L$ , written $a_n \to L$ , if for every $\varepsilon > 0$ there is $N$ such that $|a_n - L| < \varepsilon$ for all $n \ge N$ . It is a Cauchy sequence if for every $\varepsilon > 0$ there is $N$ such that $|a_m - a_n| < \varepsilon$ for all $m, n \ge N$ . A subsequence of $(a_n)$ is a sequence $(a_{\varphi(k)})_k$ with $\varphi : \N \to \N$ strictly increasing.

Definition— continuity and uniform continuity

Let $f : S \to \R$ with $S \subseteq \R$ . $f$ is continuous at $c \in S$ if for every $\varepsilon > 0$ there is $\delta > 0$ such that $|f(x) - f(c)| < \varepsilon$ for all $x \in S$ with $|x - c| < \delta$ ; it is continuous if it is continuous at every point of $S$ . It is uniformly continuous if for every $\varepsilon > 0$ there is a $\delta > 0$ that works at every point: $|f(x) - f(y)| < \varepsilon$ whenever $x, y \in S$ and $|x - y| < \delta$ .

Definition— pointwise and uniform convergence

Let $f_n, f : S \to \R$ . The sequence $(f_n)$ converges to $f$ pointwise on $S$ if $f_n(x) \to f(x)$ for each $x \in S$ : $\forall x \in S\ \forall \varepsilon > 0\ \exists N\ \forall n \ge N : |f_n(x) - f(x)| < \varepsilon$ . It converges uniformly on $S$ if $\forall \varepsilon > 0\ \exists N\ \forall n \ge N\ \forall x \in S : |f_n(x) - f(x)| < \varepsilon$ , equivalently if $\sup_{x \in S} |f_n(x) - f(x)| \to 0$ .

The only difference is where " $\forall x$ " stands. In the uniform definition $N$ is chosen before $x$ , so it cannot depend on $x$ .

Definition— Riemann integral (Darboux's formulation)

Let $f : [a, b] \to \R$ be bounded. A partition $P$ of $[a, b]$ is a finite list $a = x_0 < x_1 < \dots < x_n = b$ . With $M_i = \sup_{[x_{i-1}, x_i]} f$ and $m_i = \inf_{[x_{i-1}, x_i]} f$ , the upper and lower sums are

U(f, P) = \sum_{i=1}^n M_i\,(x_i - x_{i-1}), \qquad L(f, P) = \sum_{i=1}^n m_i\,(x_i - x_{i-1}).

$f$ is Riemann integrable if $\inf_P U(f, P) = \sup_P L(f, P)$ ; this common value is $\int_a^b f$ .

Adding points to a partition (a refinement) can only lower upper sums and raise lower sums. Comparing any two partitions with their common refinement shows that every lower sum is at most every upper sum. Consequently $f$ is integrable if and only if for every $\varepsilon > 0$ there is a partition with $U(f, P) - L(f, P) < \varepsilon$ (Riemann's criterion).

Worked examplesH

Examples, non-examples and warnings

Example— Limits, and how to prove them

$a_n = \frac{n}{n+1} \to 1$ : $|a_n - 1| = \frac{1}{n+1} < \varepsilon$ as soon as $n > 1/\varepsilon$ , which happens for large $n$ by the Archimedean property.
$a_n = (-1)^n$ does not converge: any limit $L$ would satisfy $|1 - L| < \tfrac12$ and $|-1 - L| < \tfrac12$ , so $2 = |1 - (-1)| < 1$ . But it has convergent subsequences ( $a_{2k} \to 1$ , $a_{2k+1} \to -1$ ), as Bolzano–Weierstrass promises for a bounded sequence.
$a_n = (1 + 1/n)^n$ is increasing and bounded above by $3$ , so it converges by the monotone convergence theorem below. Its limit is called $e$ .

Example— A Cauchy sequence of rationals with no rational limit

Define $q_1 = 1$ and $q_{n+1} = \frac{q_n}{2} + \frac{1}{q_n}$ (Newton's method for $x^2 = 2$ ). Every $q_n$ is rational, $q_n \to \sqrt 2$ in $\R$ , and so $(q_n)$ is Cauchy. In $\Q$ it has no limit: if $q_n \to L$ in $\Q$ then $L = \frac L2 + \frac1L$ , so $L^2 = 2$ , which has no rational solution. Being Cauchy is about the sequence; converging is about the space it lives in.

Example— Pointwise but not uniform

$f_n(x) = x^n$ on $[0, 1]$ : the pointwise limit is $f(x) = 0$ for $x < 1$ , $f(1) = 1$ . Convergence is uniform on $[0, r]$ for each $r < 1$ (the supremum of $|f_n|$ there is $r^n \to 0$ ) but not on $[0, 1)$ , where the supremum is $1$ for every $n$ .
$g_n(x) = \dfrac{nx}{1 + n^2x^2}$ converges pointwise to $0$ on $\R$ , and every $g_n$ and the limit are continuous. Yet $g_n(1/n) = \tfrac12$ , so the convergence is not uniform. Continuity of the limit does not imply uniform convergence.
$h_n(x) = 2n^2 x e^{-n^2x^2}$ converges to $0$ at every $x \in [0, 1]$ , but $\int_0^1 h_n = 1 - e^{-n^2} \to 1$ . The mass concentrates into a spike near $0$ , and limit and integral cannot be swapped.

Example— Functions that are and are not Riemann integrable

Dirichlet's function $\mathbf 1_\Q$ ( $1$ on rationals, $0$ on irrationals) is not integrable on $[0, 1]$ : every interval of positive length contains rationals and irrationals, so on every partition $U = 1$ and $L = 0$ .
Thomae's function ( $f(p/q) = 1/q$ in lowest terms, $f = 0$ at irrationals) is discontinuous at every rational and continuous at every irrational, yet it is integrable with integral $0$ : for $\varepsilon > 0$ only finitely many points have $f \ge \varepsilon$ , and they can be enclosed in subintervals of tiny total length.
Every monotone function on $[a, b]$ is integrable. With $n$ equal pieces the sum $U - L$ telescopes to $\big(f(b) - f(a)\big)(b - a)/n$ .

ProofH

Completeness: three theorems about sequences

Theorem— Theorem 1 (monotone convergence)

An increasing sequence that is bounded above converges, to $\sup_n a_n$ . (Likewise a decreasing sequence bounded below converges to its infimum.)

Proof

Let $s = \sup\{a_n : n \in \N\}$ , which exists by completeness. Given $\varepsilon > 0$ , the approximation property gives $N$ with $a_N > s - \varepsilon$ . For $n \ge N$ , monotonicity gives $s - \varepsilon < a_N \le a_n \le s$ , so $|a_n - s| < \varepsilon$ .

∎

Theorem— Lemma 2 (every sequence has a monotone subsequence)

Every real sequence has a subsequence that is increasing or decreasing.

Proof

Call an index $m$ a peak if $a_m \ge a_n$ for every $n > m$ — from $a_m$ you can see to the horizon. If there are infinitely many peaks $m_1 < m_2 < \cdots$ , then $a_{m_1} \ge a_{m_2} \ge \cdots$ is a decreasing subsequence. Otherwise let $n_1$ be larger than every peak. As $n_1$ is not a peak, some $n_2 > n_1$ has $a_{n_2} > a_{n_1}$ ; as $n_2$ is not a peak, some $n_3 > n_2$ has $a_{n_3} > a_{n_2}$ ; and so on. This gives an increasing subsequence.

∎

Theorem— Theorem 3 (Bolzano–Weierstrass)

Every bounded sequence of real numbers has a convergent subsequence.

Proof

By Lemma 2 it has a monotone subsequence, which is bounded and hence converges by Theorem 1.

∎

Theorem— Theorem 4 (Cauchy criterion)

A sequence of real numbers converges if and only if it is a Cauchy sequence.

Proof

( $\Rightarrow$ ) If $a_n \to L$ , choose $N$ with $|a_n - L| < \varepsilon/2$ for $n \ge N$ . Then for $m, n \ge N$ : $|a_m - a_n| \le |a_m - L| + |L - a_n| < \varepsilon$ .

( $\Leftarrow$ ) Let $(a_n)$ be Cauchy. It is bounded: with $\varepsilon = 1$ there is $N$ such that $|a_n - a_N| < 1$ for $n \ge N$ , so $|a_n| \le \max(|a_0|, \dots, |a_{N-1}|, |a_N| + 1)$ for all $n$ . By Bolzano–Weierstrass some subsequence $a_{\varphi(k)} \to L$ . The whole sequence converges to $L$ : given $\varepsilon$ , choose $N$ with $|a_m - a_n| < \varepsilon/2$ for $m, n \ge N$ , and $K$ with $|a_{\varphi(k)} - L| < \varepsilon/2$ for $k \ge K$ . For $n \ge N$ take $k = \max(K, n)$ . Since $\varphi(k) \ge k \ge n \ge N$ ,

|a_n - L| \le |a_n - a_{\varphi(k)}| + |a_{\varphi(k)} - L| < \tfrac\varepsilon2 + \tfrac\varepsilon2 = \varepsilon. \qquad \blacksquare

∎

Note

For an ordered field, the following are equivalent: the supremum property; the monotone convergence theorem; the Bolzano–Weierstrass theorem; "Cauchy sequences converge" together with the Archimedean property; the nested interval property together with the Archimedean property. Each fails in $\Q$ . Up to a unique isomorphism, $\R$ is the only ordered field with these properties — see constructing ℤ, ℚ and ℝ for its construction by Dedekind cuts or Cauchy sequences.

ProofH

Continuous functions on a closed interval

We first record the link between ε–δ continuity and sequences. If $f$ is continuous at $c$ and $x_n \to c$ (with $x_n$ in the domain), then $f(x_n) \to f(c)$ . Given $\varepsilon$ , take the $\delta$ of continuity, then $N$ with $|x_n - c| < \delta$ for $n \ge N$ ; then $|f(x_n) - f(c)| < \varepsilon$ for $n \ge N$ . Also, limits of sequences in $[a, b]$ stay in $[a, b]$ : if $x_n \to x$ and $x > b$ , then eventually $|x_n - x| < x - b$ , forcing $x_n > b$ .

Theorem— Theorem 5 (intermediate value theorem)

If $f : [a, b] \to \R$ is continuous and $f(a) \le 0 \le f(b)$ , then $f(c) = 0$ for some $c \in [a, b]$ . Consequently a continuous function on $[a, b]$ takes every value between $f(a)$ and $f(b)$ .

Proof

Let $S = \{x \in [a, b] : f(x) \le 0\}$ . It contains $a$ and is bounded above by $b$ , so $c = \sup S$ exists, and $a \le c \le b$ . We show that $f(c)$ is neither positive nor negative.

Suppose $f(c) > 0$ . By continuity (with $\varepsilon = f(c)$ ) there is $\delta > 0$ with $f(x) > 0$ for all $x \in [a, b]$ with $|x - c| < \delta$ . By the approximation property some $s \in S$ satisfies $c - \delta < s \le c$ . Then $|s - c| < \delta$ , so $f(s) > 0$ , contradicting $s \in S$ .

Suppose $f(c) < 0$ . Then $c \ne b$ (as $f(b) \ge 0$ ), so $c < b$ . By continuity there is $\delta > 0$ with $f(x) < 0$ whenever $x \in [a, b]$ and $|x - c| < \delta$ . The point $x = \min(c + \delta/2, b)$ lies in $[a, b]$ , satisfies $c < x$ and $|x - c| < \delta$ , so $f(x) < 0$ and $x \in S$ . But $x > c = \sup S$ , a contradiction.

Hence $f(c) = 0$ . For the general statement apply this to $f - y$ or $y - f$ .

∎

Theorem— Theorem 6 (extreme value theorem)

A continuous function $f : [a, b] \to \R$ is bounded and attains its maximum and its minimum.

Proof

Bounded: otherwise for each $n$ there is $x_n \in [a, b]$ with $|f(x_n)| > n$ . By Bolzano–Weierstrass a subsequence $x_{\varphi(k)}$ converges to some $x \in [a, b]$ , and by continuity $f(x_{\varphi(k)}) \to f(x)$ . A convergent sequence is bounded, but $|f(x_{\varphi(k)})| > \varphi(k) \ge k$ . Contradiction.

Attains its maximum: let $M = \sup f([a, b])$ , finite by the first part. For each $n$ choose $x_n$ with $f(x_n) > M - 1/n$ . A subsequence converges to some $c \in [a, b]$ , and $f(c) = \lim f(x_{\varphi(k)})$ . Since $M - 1/\varphi(k) < f(x_{\varphi(k)}) \le M$ , this limit is $M$ . Apply the same to $-f$ for the minimum.

∎

Both hypotheses matter. On $(0, 1]$ the continuous function $1/x$ is unbounded, on $[0, \infty)$ the function $x$ is unbounded, and on $(-1, 1)$ the function $x$ is bounded but has no maximum. On the closed interval $[0, 1]$ , the discontinuous function equal to $x$ for $x < 1$ and to $0$ at $x = 1$ is bounded, with supremum $1$ , but never reaches it. What the proofs used about $[a, b]$ is that every sequence in it has a subsequence converging in it — sequential compactness. In the language of point-set topology, the intermediate value theorem is about connectedness and the extreme value theorem about compactness.

Theorem— Theorem 7 (Heine–Cantor: uniform continuity)

A continuous function $f : [a, b] \to \R$ is uniformly continuous.

Proof

If not, there is $\varepsilon > 0$ such that for each $n$ some $x_n, y_n \in [a, b]$ satisfy $|x_n - y_n| < 1/n$ but $|f(x_n) - f(y_n)| \ge \varepsilon$ . By Bolzano–Weierstrass a subsequence $x_{\varphi(k)}$ converges to some $c \in [a, b]$ ; then $y_{\varphi(k)} \to c$ too, since $|y_{\varphi(k)} - x_{\varphi(k)}| < 1/\varphi(k)$ . By continuity at $c$ , $f(x_{\varphi(k)}) - f(y_{\varphi(k)}) \to f(c) - f(c) = 0$ , contradicting $|f(x_{\varphi(k)}) - f(y_{\varphi(k)})| \ge \varepsilon$ .

∎

On $(0, 1]$ the function $1/x$ is continuous but not uniformly continuous: $x = 1/n$ and $y = 1/(n+1)$ are very close, yet $|f(x) - f(y)| = 1$ .

ProofH

Uniform convergence and swapping limits

Theorem— Theorem 8 (uniform limit theorem)

If $f_n \to f$ uniformly on $S$ and every $f_n$ is continuous at $c \in S$ , then $f$ is continuous at $c$ . In particular, a uniform limit of continuous functions is continuous.

Proof

|f(x) - f(c)| \le |f(x) - f_N(x)| + |f_N(x) - f_N(c)| + |f_N(c) - f(c)| < \tfrac\varepsilon3 + \tfrac\varepsilon3 + \tfrac\varepsilon3 = \varepsilon. \qquad \blacksquare

∎

Where does the argument break for pointwise convergence? The first step would choose $N$ depending on $x$ , and then $\delta$ (chosen for $f_N$ ) depends on $x$ , which is circular. The example $x^n$ shows the theorem genuinely fails, and Theorem 8 gives a quick proof that $x^n \to 0$ is not uniform on $[0, 1]$ : the limit is discontinuous.

Theorem— Theorem 9 (integrals of uniform limits)

If $f_n \to f$ uniformly on $[a, b]$ and each $f_n$ is Riemann integrable, then $f$ is Riemann integrable and $\int_a^b f_n \to \int_a^b f$ .

Proof

Let $\varepsilon_n = \sup_{[a, b]} |f_n - f| \to 0$ . On every subinterval, $f_n - \varepsilon_n \le f \le f_n + \varepsilon_n$ , so for every partition $P$

U(f, P) - L(f, P) \le U(f_n, P) - L(f_n, P) + 2\varepsilon_n (b - a).

Given $\varepsilon$ , fix $n$ with $2\varepsilon_n (b - a) < \varepsilon/2$ , then $P$ with $U(f_n, P) - L(f_n, P) < \varepsilon/2$ : Riemann's criterion shows $f$ is integrable. Finally $\left|\int_a^b f_n - \int_a^b f\right| \le \int_a^b |f_n - f| \le \varepsilon_n (b - a) \to 0$ .

∎

For series this becomes the most useful test in analysis.

Theorem— Theorem 10 (Weierstrass M-test)

If $|u_k(x)| \le M_k$ for all $x \in S$ and $\sum_k M_k < \infty$ , then $\sum_k u_k$ converges uniformly on $S$ . If moreover every $u_k$ is continuous, so is the sum.

Proof

Let $s_n(x) = \sum_{k \le n} u_k(x)$ . For $m > n$ , $|s_m(x) - s_n(x)| \le \sum_{k = n+1}^m M_k$ , which is small for large $n$ independently of $x$ . So each $(s_n(x))$ is Cauchy and converges (Theorem 4) to some $s(x)$ . Letting $m \to \infty$ gives $|s(x) - s_n(x)| \le \sum_{k > n} M_k$ for all $x$ , which tends to $0$ : the convergence is uniform. Continuity follows from Theorem 8.

∎

Derivatives are the most delicate. The functions $\sin(nx)/n$ converge uniformly to $0$ (the supremum is $1/n$ ), but their derivatives $\cos(nx)$ do not converge at all — at $x = \pi$ they alternate between $\pm 1$ . Uniform convergence of the derivatives is what is needed.

Theorem— Theorem 11 (derivatives of limits)

Let $f_n : [a, b] \to \R$ be continuously differentiable, suppose $f_n' \to g$ uniformly on $[a, b]$ and $f_n(x_0)$ converges for some $x_0 \in [a, b]$ . Then $f_n$ converges uniformly to a differentiable function $f$ with $f' = g$ .

Proof

By Theorem 8, $g$ is continuous. By the fundamental theorem of calculus (Theorem 15 below), $f_n(x) = f_n(x_0) + \int_{x_0}^x f_n'$ . By Theorem 9 the right-hand side converges, uniformly in $x$ (the error is at most $|f_n(x_0) - L| + (b - a)\sup|f_n' - g|$ ), to $f(x) := L + \int_{x_0}^x g$ , where $L = \lim f_n(x_0)$ . By Theorem 14 below, $f' = g$ .

∎

Without some such hypothesis, anything can happen.

Theorem— Theorem 12 (Weierstrass's function is not differentiable at 0)

Let $W(x) = \sum_{k \ge 0} a^k \cos(b^k \pi x)$ with $a = \tfrac12$ and $b = 13$ . Then $W$ is continuous on $\R$ but not differentiable at $0$ .

Proof

Continuity: the M-test with $M_k = a^k$ . Now put $h_m = b^{-m}$ . For $k \ge m$ , $b^k h_m = 13^{k - m}$ is an odd integer, so $\cos(b^k \pi h_m) = -1$ . Since $W(0) = \sum a^k = \frac{1}{1-a}$ ,

\frac{W(h_m) - W(0)}{h_m} = b^m \sum_{k < m} a^k \big(\cos(\pi b^{k - m}) - 1\big) \;-\; \frac{2\,(ab)^m}{1 - a}.

Using $0 \le 1 - \cos t \le t^2/2$ , the first term has absolute value at most $\frac{\pi^2}{2} \sum_{k < m} a^k b^{2k - m} = \frac{\pi^2}{2} (ab)^m \sum_{j = 1}^m (ab^2)^{-j} \le \frac{\pi^2 (ab)^m}{2(ab^2 - 1)}$ . Hence

\left| \frac{W(h_m) - W(0)}{h_m} \right| \ge (ab)^m \left( \frac{2}{1 - a} - \frac{\pi^2}{2(ab^2 - 1)} \right) = 6.5^m \,(4 - 0.059\ldots) \to \infty.

If $W'(0)$ existed, these difference quotients would converge to it.

∎

The same function has no derivative at any point. Theorem (Weierstrass, 1872; stated, not proved here). If $0 < a < 1$ , $b$ is an odd integer and $ab > 1 + \frac{3\pi}{2}$ , then $W$ is continuous and nowhere differentiable. G. H. Hardy (1916) showed that $ab \ge 1$ suffices. A proof of a closely related example is in Stein and Shakarchi, Fourier Analysis (2003), chapter 4. Every partial sum of $W$ is infinitely differentiable; the limit is differentiable nowhere. The derivatives of the partial sums, of size about $(ab)^k$ , are as far from converging uniformly as possible.

ProofH

The Riemann integral and the fundamental theorem

Theorem— Theorem 13 (continuous functions are integrable)

Every continuous $f : [a, b] \to \R$ is Riemann integrable. The indicator function $\mathbf 1_\Q$ of the rationals is not.

Proof

Let $\varepsilon > 0$ . By Heine–Cantor there is $\delta > 0$ with $|f(s) - f(t)| < \varepsilon/(b - a)$ whenever $|s - t| < \delta$ . Take a partition with all $x_i - x_{i-1} < \delta$ . On each piece, the extreme value theorem gives $M_i = f(s)$ and $m_i = f(t)$ with $|s - t| < \delta$ , so $M_i - m_i < \varepsilon/(b - a)$ , and

U(f, P) - L(f, P) = \sum_i (M_i - m_i)(x_i - x_{i-1}) < \frac{\varepsilon}{b - a} \sum_i (x_i - x_{i-1}) = \varepsilon.

For $\mathbf 1_\Q$ : every piece $[x_{i-1}, x_i]$ of positive length contains a rational and an irrational, so $M_i = 1$ and $m_i = 0$ ; every upper sum is $b - a$ and every lower sum is $0$ .

∎

We use three elementary properties of the integral, all proved directly from the definition: it is linear; it is additive over adjacent intervals, $\int_a^c f = \int_a^b f + \int_b^c f$ (with the convention $\int_b^a = -\int_a^b$ ); and $\left|\int_a^b g\right| \le (b - a) \sup_{[a, b]} |g|$ .

Theorem— Theorem 14 (fundamental theorem of calculus, part 1)

Let $f$ be Riemann integrable on $[a, b]$ and $F(x) = \int_a^x f$ . If $f$ is continuous at $x_0 \in (a, b)$ , then $F$ is differentiable at $x_0$ and $F'(x_0) = f(x_0)$ .

Proof

F(x_0 + h) - F(x_0) - h f(x_0) = \int_{x_0}^{x_0 + h} \big(f(t) - f(x_0)\big)\,dt,

whose absolute value is at most $|h| \cdot \varepsilon$ . Dividing by $|h|$ : $\left|\frac{F(x_0 + h) - F(x_0)}{h} - f(x_0)\right| \le \varepsilon$ .

∎

Theorem— Theorem 15 (fundamental theorem of calculus, part 2)

If $F$ is differentiable on $[a, b]$ (one-sided at the ends) and $F' = f$ is Riemann integrable, then $\int_a^b f = F(b) - F(a)$ .

Proof

Let $P$ be any partition. By the mean value theorem, on each piece $F(x_i) - F(x_{i-1}) = f(t_i)(x_i - x_{i-1})$ for some $t_i \in (x_{i-1}, x_i)$ , and $m_i \le f(t_i) \le M_i$ . Summing, the sum telescopes:

L(f, P) \le \sum_i f(t_i)(x_i - x_{i-1}) = F(b) - F(a) \le U(f, P).

So $F(b) - F(a)$ lies between every lower sum and every upper sum. Since $f$ is integrable, the only such number is $\int_a^b f$ .

∎

The hypothesis that $F'$ is integrable cannot be dropped: Vito Volterra (1881) constructed a differentiable function whose derivative is bounded but not Riemann integrable. The Lebesgue integral repairs this partly (a bounded derivative is always Lebesgue integrable, and part 2 then holds), which is one of many reasons to continue to measure theory.

Formal mathematicsF

Real analysis in Lean and Mathlib

Mathlib states limits with filters: Tendsto a atTop (𝓝 L) says that $a$ eventually lands in every neighbourhood of $L$ . That one notion covers limits of sequences, of functions at a point, from the left, at infinity, and much more. To connect it with the lesson, we define the ε–N notions directly and prove that they agree with Mathlib's.

ConvergesToLean 4Formally verified

/-- `ConvergesTo a L`: the sequence `a` converges to `L` — for every `ε > 0` there is an index `N`
such that `|aₙ - L| < ε` for every `n ≥ N`. -/
def ConvergesTo (a : ℕ → ℝ) (L : ℝ) : Prop :=
  ∀ ε > 0, ∃ N : ℕ, ∀ n ≥ N, |a n - L| < ε