Skip to content

KindiakMath

About
- Curriculum Vitae
- Review Quotes
Pre-University
Foundational Math
Theoretical Analysis
Unified Math
- Complex Analysis
- Abstract Algebra
Para-Mathematics

Category: Real Analysis

Dini’s Mutated Result
Problem 1. Let $\{f_n\}$ be a sequence of functions on $[0, 1]$ that satisfy the following properties:
- each $f_n$ is non-decreasing on $[0, 1]$ ,
- $f_n \to f$ pointwise,
- $f$ is continuous on $[0, 1]$ .
Prove that $f_n \to f$ uniformly. This is a modified convergence criterion from the more famous Dini’s theorem.

(Click for Solution)

Solution. Fix $\epsilon > 0$ . Using the (uniform) continuity of $f$ , for any $k_1 > 0$ , there exists $\delta > 0$ such that

$|u-v| < \delta \quad \Rightarrow \quad |f(u) - f(v)| < k_1 \cdot \epsilon.$

By the Archimedean property, find $N_1 \in \mathbb N$ so that $1/N_1 < \delta$ . Partition $[0, 1]$ into $\{x_0, x_1,\dots,x_{N_1}\}$ , $x_i = i/N_1$ . For each $i$ , point-wise convergence $f_n \to f$ yields that for any $k_2 > 0$ , there exists $N_2 \in \mathbb N$ such that whenever $n > N_2$ ,

$n > N_2 \quad \Rightarrow \quad |f_n(x_i) - f(x_i)| < k_2 \cdot \epsilon,\quad i=0,1,2,\dots,N_1.$

Finally, fix $x \in [0, 1]$ and suppose $x \in [x_0, x_1]$ without loss of generality. Then for $n > N_2 + 1$ , point-wise monotonicity yields

$\begin{aligned} |f_n(x) - f(x)| &\leq |f_n(x) - f_n(x_1)| + |f_n(x_1) - f(x_1)| + |f(x_1) - f(x)| \\ &< |f_n(x) - f_n(x_1)| + k_2 \cdot \epsilon + k_1 \cdot \epsilon \\&< |f_n(x) - f_n(x_1)| + (k_1 + k_2) \cdot \epsilon .\end{aligned}$

To bound the first term, we use monotonicity to obtain

$\begin{aligned} |f_n(x) - f_n(x_1)| &\leq |f_n(x_0) - f_n(x_1)|\\ &\leq |f_n(x_0) - f(x_0)| + |f(x_0) - f(x_1)| + |f(x_1) - f_n(x_1)| \\ &< k_2 \cdot \epsilon + k_1 \cdot \epsilon + k_2 \cdot \epsilon \\ &= (k_1 + 2 k_2 )\cdot \epsilon. \end{aligned}$

Combining the estimates yields

$\begin{aligned} |f_n(x) - f(x)| &< |f_n(x) - f_n(x_1)| + (k_1 + k_2) \cdot \epsilon \\ &< (2k_1 + 3k_2) \cdot \epsilon.\end{aligned}$

Finally, set $k_1 = 1/4$ , $k_2 = 1/6$ to obtain the desired result.

—Joel Kindiak, 31 Jan 25, 1422H
June 11, 2025
Strengthening Point-wise Convergence
Problem 1. Let $\{f_n\}$ be a sequence of differentiable functions on $[0, 1]$ with the following properties:
- $f_n \to f$ pointwise on $[0, 1]$ ,
- $f$ is continuous on $[0, 1]$ ,
- there exists $M > 0$ such that $\displaystyle \sup \{\|f_n'\|_\infty : n \in \mathbb N\} \leq M$ .
Prove that $f_n \to f$ uniformly.

(Click for Solution)

Solution. The key insight here is that $f$ being continuous on $[0, 1]$ strengthens it to being uniformly continuous, so that we can gain some global control over our estimates.

Fix $\epsilon > 0$ . Since $f$ is (uniformly) continuous on $[a, b]$ , for any $k_1 > 0$ , there exists $\delta > 0$ such that for any $u,v \in [0, 1]$ ,

$|u-v| < \delta \quad \Rightarrow \quad |f(u) - f(v)| < k_1 \cdot \epsilon.$

By the Archimedean property, find $N_1 \in \mathbb N$ so that $1/N_1 < \delta$ . Partition $[0, 1]$ into $\{x_0, x_1,\dots,x_{N_1}\}$ , $x_i = i/N_1$ . For each $i$ , point-wise convergence yields that for any $k_2 > 0$ , there exists $N_2 \in \mathbb N$ such that whenever $n > N_2$ ,

$n > N_2 \quad \Rightarrow \quad |f_n(x_i) - f(x_i)| < k_2 \cdot \epsilon,\quad i=0,1,2,\dots,N_1.$

Finally, fix $x \in [0, 1]$ and suppose $x\in [x_0, x_1]$ without loss of generality. Using the uniform bound on the derivatives, the mean value theorem yields

$\displaystyle |f_n(x) - f_n(x_0)| \leq M \cdot |x - x_0| \leq M \cdot \frac 1{N_1}.$

Then for any $n > N_2$ , combining the estimates yields

$\displaystyle \begin{aligned} |f_n(x) - f(x)| &\leq |f_n(x)-f_n(x_0)| + |f_n(x_0) - f(x_0)| + |f(x_0) - f(x)|\\ &< M \cdot \frac 1{N_1}+ k_2 \cdot \epsilon + k_1 \cdot \epsilon \\ &=M \cdot \frac 1{N_1} + (k_1 + k_2) \cdot \epsilon. \end{aligned}$

Taking care of the dependencies, set $k_1 = k_2 = 1/3$ and further stipulate $1/N_1 < \epsilon/(3M)$ to yield the desired result.

—Joel Kindiak, 31 Jan 25, 1221H
June 11, 2025
Functional Reverse-Engineering
Define $g : \mathbb R \to \mathbb R$ by $g(x) = x^2 - x + 1 = (x-1/2)^2 + 3/4$ .

Problem 1. Suppose there exists a function $f : \mathbb R \to \mathbb R$ such that $f \circ f = g$ . Evaluate $f(1)$ and $f(0)$ .

(Click for Solution)

Solution. We first observe that $(f \circ f)(1) = g(1) = 1$ . Applying $f$ on both sides,

$\begin{aligned}f((f \circ f)(1)) &= f(1) \\ (f \circ f)(f(1)) &= f(1) \\ f(1)^2 - f(1) + 1 &= f(1) \\ (f(1)-1)^2 &= 0.\end{aligned}$

Hence, $f(1) = 1$ . Similarly, $(f \circ f)(0) = g(0) = 1$ . Applying $f$ on both sides yields

$f(0)^2 - f(0) + 1 = 1 \quad \iff \quad f(0)(f(0) - 1) = 0.$

Hence, $f(0) = 0$ or $f(0) = 1$ . The former yields

$0 = f(0) = f(f(0)) = g(0) = 1,$

a blatant contradiction.

Problem 2. Define the sequence $\{x_n\}$ by $x_0 := 1/2$ , $x_1 := 5/8$ , and $x_{n+2} := g(x_n)$ for integers $n \geq 0$ . Prove that $x_n \to 1$ .
(Click for Solution)

Solution. We first make a few observations:
- $g$ is continuous.
- $g$ is increasing on $[1/2, \infty)$ .
- $g([1/2, 1]) = [3/4, 1]$ .
Thus, $g|_K$ is bijective to its image for any $K \subseteq [1/2, \infty)$ . By construction, $x_0 \leq x_1 \leq 3/4 = g(x_0) = x_2$ . Furthermore, for $n \geq 0$ ,

$x_{n+2} = g(x_n) \leq g(x_{n+1}) = x_{n+3}.$

Thus, by induction, $\{x_n\}$ is an increasing sequence. Since $\{x_n\}$ is bounded above by $1$ , by the monotone convergence theorem, $x_n \to r$ for some $r \in \mathbb R$ . By the continuity of $g$ ,

$\displaystyle r = \lim_{n \to \infty} x_{n+2} = \lim_{n \to \infty}g(x_n) = g \left( \lim_{n \to \infty} x_n \right) = g(r).$

But we have $g(r) = r \iff r = 1$ , so that $x_n \to 1$ , as required.
Problem 3. Construct a continuous function $f: \mathbb R \to \mathbb R$ such that $f \circ f = g$ .
(Click for Solution)

Solution. We follow the construction in this post by user pco. We will construct $f$ on the following subsets of $\mathbb R$ :
- $[1/2, 1)$ ,
- $\{1\}$ ,
- $(1, \infty)$ ,
- $(-\infty, 1/2)$ .
For the interval $[1/2, 1)$ , define the sequence $\{x_n\}$ according to Problem 2. Let $p_{a, b} : [0, 1) \to [a, b)$ denote the canonical continuous bijection defined by

$\displaystyle p_{a,b}(t) = a + t \cdot (b-a).$

Then define the sequence $\{h_n\}$ of bijections $h_n : [x_n, x_{n+1}) \to [x_{n+1}, x_{n+2})$ by

$h_0 = p_{x_1, x_2} \circ p_{x_0,x_1}^{-1},\quad h_{n+1} = g|_{[x_n, x_{n+1})} \circ h_n^{-1}.$

Thus, for $x \in [1/2, 1)$ , we define

$f(x) = h_n(x) \quad \iff \quad x \in [x_n, x_{n+1}).$

By construction,

$\begin{aligned} (f \circ f)(x) &= f(f(x)) \\ &= f(h_n(x)) \\ &= h_{n+1}(h_n(x)) \\ &= (h_{n+1} \circ h_n)(x) = g(x).\end{aligned}$

Thus, we have constructed $f$ on $[1/2, 1)$ . Since we require $f$ to satisfy

$f(f(x)) = x^2 - x + 1,$

Problem 1 stipulates that $f(1) = 1$ . Furthermore, this definition allows $f$ to be continuous by Problem 2. To define $f$ on $(1, \infty)$ , we first define $f$ on $[2, \infty)$ . Define the sequence $\{y_n\}$ similarly to $\{x_n\}$ as in Problem 2, except that we now have the initial conditions $y_0 := 2$ and $y_1 := (2+g(2))/2 = 5/2$ .

Just like before, $y_0 \leq y_1 \leq y_2$ and since $g$ is increasing, $\{ y_n \}$ is an increasing sequence by the argument in Problem 2. For divergence to infinity, we recall that

$y_{n+2} = y_n \cdot (y_n - 1) + 1 \geq y_n + 1.$

If $y_n$ converges to a limit, then $0 \geq 1$ , a blatant contradiction. Thus, $\{y_n\}$ diverges. Since $\{y_n\}$ is strictly increasing, it is unbounded. Thus, $y_n \to \infty$ .

Defining $\{ h_n \}$ similarly as in the $[1/2, 1)$ case, we define

$f(x) = h_n(x)\quad \iff \quad x \in [y_n, y_{n+1})$

and deduce $f \circ f = g$ on that domain too.

We next define $f$ on $(1, 2)$ . Define the sequence $\{z_n\}$ by $z_0 := 5/2$ , $z_1 := 2$ , and $z_{n+2} := g|_{(1,5/2)}^{-1}(z_n)$ . By a similar proof as in Problem 2, since $z_2 \leq z_1 \leq z_0$ and $g_{(1,2)}^{-1}$ is decreasing, $z_n$ decreases to $1$ .

Define the bijections $l_n : [z_{n+2},z_{n+1}) \to [z_{n+1}, z_n)$ in a similar manner to Problem 2, but in the reverse direction:

$l_0 = p_{z_1, z_0} \circ p_{z_2,z_1}^{-1}, \quad l_{n+1} = l_n^{-1} \circ g|_{[z_{n+3}, z_{n+2})}.$

Finally, define

$f(x) = l_n(x)\quad \iff \quad x \in [z_n, z_{n+1}).$

By construction,

$\begin{aligned} (f \circ f)(x) &=(l_n \circ l_{n+1})(x) = g(x).\end{aligned}$

Finally, for the domain $(-\infty, 1/2)$ , define $f(x) := f(1-x)$ , since

$\begin{aligned} (f \circ f)(x) &= f(f(x)) \\ &= f(f(1-x)) \\ &= (f \circ f)(1-x) \\ &= g(1-x) = g(x). \end{aligned}$
—Joel Kindiak, 11 Jun 25, 0008H
June 11, 2025
A Less-Dominated Convergence Theorem
Problem 1. Let $\{f_n\}$ be a sequence of functions defined on $[0,\infty)$ satisfying the following properties:
- for any interval of the form $[0, b]$ , $f_n \to f$ uniformly on $[a, b]$ ,
- the improper integrals $\displaystyle \int_0^\infty f_n$ and $\displaystyle \int_0^\infty f$ converge,
- there exists a nonnegative integrable function $g$ such that $|f_n| \leq g$ ,
- $\displaystyle \int_0^\infty g$ converges.
Prove that $\displaystyle \lim_{n \to \infty}\int_0^\infty f_n = \int_0^\infty f$ .

Remark 1. This is a weak form of the dominated convergence theorem. The strong form which requires measure-theoretic machinery only requires point-wise convergence.

(Click for Solution)

Solution. Fix $\epsilon > 0$ . Our line of attack is to find sufficiently large $M > 0$ so that integrals of the form $\displaystyle \int_M^\infty$ can be controlled due to the integrals converging, while integrals of the form $\displaystyle \int_0^M$ can be controlled using the uniform convergence of $f_n \to f$ .

To that end, let’s first ascertain that $|f| \leq g$ point-wise. Fix $x \geq 0$ . Applying the uniform convergence $f_n \to f$ on $[0, x]$ , for any $k_1 > 0$ , there exists $N_1 \in \mathbb N$ such that

$n > N_1 \quad \Rightarrow \quad \|f_n - f\|_\infty < k_1 \cdot \epsilon.$

This implies the upper-found

$|f(x)| \leq |f_n(x)| + |f_n(x) - f(x)| < g(x) +k_1 \cdot \epsilon.$

Since $\epsilon > 0$ is arbitrary, $|f| \leq g$ point-wise. Therefore,

$|f_n - f| \leq |f_n|+|f| \leq 2g$

point-wise. The motivation for this step is to do some upper bounds on tail integrals. Since $\displaystyle \int_0^\infty g$ converges, for any $k_2 > 0$ , there exists $M > 0$ such that

$\displaystyle \int_M^\infty g < k_2 \cdot \epsilon.$

Employing our estimate from previous discussions,

$\displaystyle \int_M^\infty |f_n - f| \leq 2 \int_M^\infty g = (2 k_2) \cdot \epsilon.$

Applying the uniform convergence $f_n \to f$ on $[0, M]$ , for any $k_3 > 0$ , there exists $N_2 \in \mathbb N$ such that

$n > N_2 \quad \Rightarrow \quad \|f_n - f\|_\infty < k_3 \cdot \epsilon.$

This means for the infinite integrals, whenever $n > N_2$ ,

$\displaystyle \begin{aligned} \left| \int_0^\infty f_n - \int_0^\infty f \right| &\leq \int_0^\infty |f_n - f|\\ &\leq \int_0^M |f_n - f| + \int_M^\infty |f_n - f| \\ &\leq \int_0^M (k_3 \cdot \epsilon) + (2 k_2) \cdot \epsilon \\ &= (M \cdot k_3 + 2 \cdot k_2) \cdot \epsilon.\end{aligned}$

Taking care of the dependencies, setting $k_2 = 1/4$ , $k_3 = 1/(4M)$ yields the desired final result:

$\displaystyle \lim_{n \to \infty}\int_0^\infty f_n = \int_0^\infty f.$

—Joel Kindiak, 31 Jan 25, 1213H
June 4, 2025
The Integrable Limit Theorem

One of the most fascinating explorations of integrable functions pertains its limiting conditions. When would limits of sequences of functions be themselves (Riemann-)integrable?

A sufficient condition is that if $f_n \to f$ uniformly, then $f$ must be integrable.

Theorem 1. Let $\{f_n\}$ be a sequence of integrable functions $f_n : [a, b] \to \mathbb R$ that converge point-wise to some function $f : [a, b] \to \mathbb R$ . If $f_n \to f$ uniformly, then $f$ is integrable, and

$\displaystyle \int_a^b f \equiv \int_a^b \lim_{n \to \infty} f_n = \lim_{n \to \infty} \int_a^b f_n.$

Proof. Fix $\epsilon > 0$ . Since $f_n \to f$ uniformly, for any $k_1 > 0$ , there exists $N \in \mathbb N$ such that

$j > N\quad \Rightarrow \quad \|f_j - f\|_\infty < k_1 \cdot \epsilon.$

In particular, the estimate works for $j = N+1$ . Since $f_j$ is integrable, for any $k_2 > 0$ , there exists a partition $P \subseteq [a, b]$ such that

$\displaystyle \sum_{i=1}^n (M_i(f_j, P) - m_i(f_j, P))\Delta x_i < k_2 \cdot \epsilon.$

For any $x \in [a, b]$ , unraveling the estimate yields

$\displaystyle f_j(x) - k_1 \cdot \epsilon< f(x) < f_j(x) + k_1 \cdot \epsilon,$

which yields the estimates

$\begin{aligned} M_i(f, P) &< M_i(f_j, P) + k_1 \cdot \epsilon,\\ m_i(f, P) &> m_i(f_j, P) - k_1 \cdot \epsilon. \end{aligned}$

By unraveling the definitions and combining the estimates,

$\begin{aligned} U(f, P) - L(f, P) &< U(f_j, P) - L(f_j, P) + 2k_1(b-a) \cdot \epsilon \\ &< k_2 \cdot \epsilon + 2k_1(b-a) \cdot \epsilon \\ &= (2k_1(b-a) + k_2) \cdot \epsilon.\end{aligned}$

Setting $k_1 = 1/(4(b-a)), k_2 = 1/2$ yields the desired integrability result. Furthermore, by the triangle inequality for integrals, for any $n > N$ ,

$\displaystyle \begin{aligned}\left| \int_a^b f_n - \int_a^b f \right| &= \left| \int_a^b (f_n - f )\right|\\ &\leq \int_a^b | f_n - f | \\ &\leq \int_a^b \| f_n - f \|_\infty \\ &\leq \| f_n - f \|_\infty \cdot (b-a) \\ &< k_1(b-a) \cdot \epsilon. \end{aligned}$

Setting $k_1 = 1/(b-a)$ yields the desired limit property

$\displaystyle \int_a^b f \equiv \int_a^b \lim_{n \to \infty} f_n = \lim_{n \to \infty} \int_a^b f_n.$

This is a classic example of valid instances when we can push the limit into the integral and vice versa. However, uniform convergence is typically too strong of a condition to be met, apart from some toy examples in standard tutorial problems.

With tools in measure theory, which we will explore a bit of next time, we can obtain a much more lenient condition. We will state this result, but not prove it, since its real utility shines forth in the context of measure theory.

Dominated Convergence Theorem. Let $\{f_n\}$ be a sequence of Lebesgue-integrable functions $f_n : [a, b] \to \mathbb R$ that converge point-wise to some function $f : [a, b] \to \mathbb R$ .

If there exists a nonnegative Lebesgue-integrable function $g : [a, b] \to \mathbb R$ such that $|f_n| \leq g$ and $|f| \leq g$ , then $f$ is Lebesgue-integrable, and

$\displaystyle \int_a^b f \equiv \int_a^b \lim_{n \to \infty} f_n = \lim_{n \to \infty} \int_a^b f_n.$

Proof. Omitted. For further discussion in measure theory.

Clearly, we cannot prove what we don’t even know—what even is a Lebesgue-integrable function? The reason this condition is more relaxed than Theorem 1 is because bounded functions on $[a, b]$ that are Riemann-integrable are Lebesgue-integrable too. Then the limit function $f$ will be Lebesgue integrable.

From an applied perspective, probability theorists model random phenomena using random variables, which are defined to be Lebesgue-integrable (more technically, Lebesgue-measurable) functions. From probability theory arises all sorts of applications across various STEM fields.

For now, that’s it for undergraduate real analysis!

—Joel Kindiak, 24 Jan 25, 1452H

May 28, 2025
Relaxing the Fundamental Theorem

Problem 1. Let $f : [a, b] \to\mathbb R$ be differentiable such that $f'$ is Riemann-integrable. Prove that

$\displaystyle \int_a^b f' = f(b) - f(a),\quad \frac{\mathrm d}{\mathrm d x} \int_a^x f' = f'(x).$

(Click for Solution)

Solution. Fix $\epsilon > 0$ . For any $k > 0$ , let $P \subseteq [a, b]$ be any partition satisfying

$\displaystyle \sum_{i=1}^n (M_i(f', P) - m_i(f', P)) \Delta x_i < k \cdot \epsilon.$

For each $i = 1, \dots, n$ , use the mean value theorem to find $c_i \in (x_{i-1}, x_i)$ such that

$f(x_i) - f(x_{i-1}) = f'(c_i)\Delta x_i,$

and furthermore $f'(c_i) \in [m_i(f', P), M_i(f', P)]$ . This implies

$\displaystyle \begin{aligned} f(b) - f(a) - L(f', P)&= \sum_{i=1}^n (f(x_i) - f(x_{i-1})) \Delta x_i - L(f', P) \\ &= \sum_{i=1}^n (f'(c_i) - m_i(f',P))\Delta x_i \\ &\leq \sum_{i=1}^n (M_i(f', P) - m_i(f',P))\Delta x_i \in [0, k \cdot \epsilon).\end{aligned}$

Hence,

$L(f', P) \leq f(b) - f(a) < L(f', P) + k \cdot \epsilon \leq \mathcal L_a^b(f') + k \cdot \epsilon.$

Taking the supremum over $P$ ,

$\mathcal L_a^b(f') \leq f(b) - f(a) \leq \mathcal L_a^b(f') + k \cdot \epsilon.$

Setting $k = 1$ then taking $\epsilon \to 0^+$ ,

$\displaystyle f(b) - f(a) = \mathcal L_a^b(f') = \int_a^b f'.$

Adapting the proof to $[a, x] \subseteq [a, b]$ ,

$\displaystyle \int_a^x f' = f(x) - f(a)$

is differentiable with derivative

$\displaystyle \frac{\mathrm d}{\mathrm dx} \int_a^x f' = \frac{\mathrm d}{\mathrm dx}(f(x) - f(a)) = f'(x) - 0 = f'(x).$

—Joel Kindiak, 22 Jan 25, 0114H

May 21, 2025
The Popcorn Function

Problem 1. Prove that the popcorn function by $f : [0, 1] \to \mathbb R$ by

$\displaystyle f(x) = \begin{cases} 1/q, & x=p/q,\ p,q\ \text{coprime,} \\ 0, & \text{otherwise}. \end{cases}$

is Riemann-integrable.

(Click for Solution)

Solution. Fix $\epsilon > 0$ . The key insight is to observe that for any $k > 0$ , there are only finitely many rationals $x$ such that $1/f(x) < 1/(k \cdot \epsilon) \iff f(x) > k \cdot \epsilon$ . Denote these rationals in ascending order by $\{r_1,\dots,r_M\}$ . To “separate” these rationals, define

$\displaystyle \delta := \frac 13 \cdot \min_{2 \leq i \leq M} (r_i - r_{i-1})$

Define the partition

$\displaystyle P := \{0,1\} \cup \bigcup_{i=1} \{r_i - \delta, r_i + \delta\} \subseteq [0, 1].$

Then

$\begin{aligned} U(f, P) - L(f, P) &= \sum_{i = 1}^{|P|} (M_i(f, P) - m_i(f, P)) \Delta x_i \\ &\leq (1 - 0) \cdot (k \cdot \epsilon) + M \cdot 1 \cdot 2\delta\\ &= k \cdot \epsilon + 2M \cdot \delta. \end{aligned}$

To obtain the desired result, set $k = 1/4$ and refine the definition of $\delta$ to

$\displaystyle \delta := \min \left\{ \frac{\epsilon}{4M}, \frac 13 \cdot \min_{2 \leq i \leq M} (r_i - r_{i-1}) \right\}.$

—Joel Kindiak, 20 Jan 25, 2143H

May 14, 2025
Flavours of Integrability

In this post, we explore the various settings in which functions can be integrated. Here, we shorten “Riemann-integrable” by “integrable” for brevity.

Problem 1. Let $f : [a, b] \to \mathbb R$ be a bounded function. Suppose for any $c \in (a, b)$ , $f|_{[a,c]} : [a, c] \to \mathbb R$ is integrable. Prove that $f$ is integrable.

Solution. Since $f$ is bounded, there exists $M > 0$ such that for any $x \in [a, b]$ , $|f(x)| \leq M$ .

Fix $\epsilon > 0$ . Fix $\delta \in (0, b-a)$ to be tuned later. Since $f_{[a,b-\delta]}$ is integrable, for any $k > 0$ there exists a partition $P := \{x_0 = a, x_1,\dots,x_n = b-\delta\}$ such that

$U(f_{[a,b-\delta]}, P) - L(f_{[a,b-\delta]}, P) < k \cdot \epsilon.$

Define the partition $Q := \{x_0, x_1,\dots, x_n, x_{n+1} = b\}$ of $[a, b]$ . Expanding the definition,

$\begin{aligned} &U(f, Q) - L(f, Q)\\ &= U(f, P) - L(f, P) + (M_{n+1}(f, Q) - m_{n+1}(f, Q))\Delta_{n+1} \\ &< k \cdot \epsilon + (M+M) \cdot \delta \\ &< k \cdot \epsilon + 2M \cdot \delta. \end{aligned}$

Hence, set $k := 1/2$ and $\delta := \epsilon/(4M)$ to yield the result.

Henceforth, we will say $f : [a, b] \to \mathbb R$ is integrable on $K \subseteq [a, b]$ to mean that $f|_K : K \to \mathbb R$ is integrable.

Problem 2. Let $f : [a, b] \to \mathbb R$ be bounded. Suppose there exists $c \in (a, b)$ such that $f$ is integrable on $[a, c]$ and $[c, b]$ . Then $f$ is integrable on $[a, b]$ . Furthermore,

$\displaystyle \int_a^b f = \int_a^c f + \int_c^b f.$

Solution. Fix $\epsilon > 0$ . For any $k_1 > 0$ , find a partition $P$ of $[a, c]$ such that

$U(f, P) - L(f, P) < k_1 \cdot \epsilon.$

For any $k_2 > 0$ , find a partition $Q$ of $[c, b]$ such that

$U(f, Q) - L(f, Q) < k_2 \cdot \epsilon.$

Then $P \cup Q$ is a partition of $[a, b]$ , and

$\begin{aligned}&U(f, P \cup Q) - L(f, P \cup Q)\\ &= (U(f, P) - L(f, P)) + (U(f, Q) - L(f, Q)) \\ &< k_1 \cdot \epsilon + k_2 \cdot \epsilon \\ &= (k_1 + k_2) \cdot \epsilon.\end{aligned}$

Setting $k_1 = k_2 = 1/2$ yields the desired result.

Finally, we prove the integral identity in two directions. Fix $\epsilon > 0$ . Find a partition $P$ such that

$\displaystyle \int_a^b f < L(f, P) + \epsilon.$

Define the partitions

$\begin{aligned} P_- &:= \{x \in P : x < c\} \cup \{c\} \subseteq [a, c],\\ P_+ &:= \{x \in P : x > c\} \cup \{c\} \subseteq [c, b]. \end{aligned}$

Then

$\displaystyle \int_a^b f < L(f, P) + \epsilon \leq L(f, P_-) + L(f, P_+) + \epsilon \leq \int_a^c f + \int_c^b f+ \epsilon .$

Taking $\epsilon \to 0^+$ yields the inequality

$\displaystyle \int_a^b f \leq \int_a^c f + \int_c^b f.$

For the other direction, fix $\epsilon > 0$ . For $k_1,k_2 > 0$ , find partitions $P_- \subseteq [a, c]$ , $P_+ \subseteq [c, b]$ , such that

$\displaystyle\int_a^c f < L(f, P_-) + k_1 \cdot \epsilon \quad \int_c^b f < L(f, P_+) + k_2 \cdot \epsilon.$

Defining $P := P_- \cup P_+$ ,

$\begin{aligned} \int_a^c f + \int_c^b f & < (L(f, P_-) + L(f, P_+)) + (k_1 + k_2) \cdot \epsilon \\ & = L(f, P) + (k_1 + k_2) \cdot \epsilon \\ &\leq \int_a^b f + (k_1 + k_2) \cdot \epsilon. \end{aligned}$

Setting $k_1 = k_2 = 1/2$ ,

$\displaystyle \int_a^c f + \int_c^b f < \int_a^b f + \epsilon.$

Taking $\epsilon \to 0^+$ ,

$\displaystyle \int_a^c f + \int_c^b f \leq \int_a^b f.$

Combining the results yields the desired result.

Problem 3. Let $f : [a, b] \to \mathbb R$ be continuous except at some $c \in (a, b)$ . Prove that $f$ is integrable.

Solution. For any $u \in (a, c)$ , $f$ is continuous on $[a, u]$ and thus integrable on $[a, u]$ . By Problem 1, $f$ is integrable on $[a, c]$ . Similarly, $f$ is integrable on $[c, b]$ . By Problem 2, $f$ is integrable on $[a, b]$ .

Problem 4. Let $f : [a, b] \to \mathbb R$ be increasing. Prove that $f$ is integrable.

Solution. We prove $[a, b] = [0,1]$ for simplicity. We observe that for any $x \in (a, b)$ , $f(a) < f(x) < f(b)$ so that $f$ is bounded. Fix $\epsilon > 0$ . For any $k > 0$ , use the Archimedean property to find $n \in \mathbb N$ such that $1/n < k \cdot \epsilon$ . Define $P := \{x_0, x_1, \dots, x_n\} \subseteq [0, 1]$ by $x_i = i/n$ . Then

$\begin{aligned} U(f, P) - L(f, P) &= \sum_{i=1}^n (M_i(f, P) - m_i(f, P)) \Delta x_i \\ &= \sum_{i=1}^n (f(x_i) - f(x_{i-1})) \cdot \frac 1n \\ &= \frac 1n \sum_{i=1}^n (f(x_i) - f(x_{i-1})) \\ &= \frac 1n (f(x_n) - f(x_0)) \\ &= \frac {f(1) - f(0)}n \\ & < (f(1) - f(0)) \cdot k \cdot \epsilon. \end{aligned}$

Since $f(1) - f(0) > 0$ , set $k := 1/(f(1) - f(0)) > 0$ to yield the desired result.

—Joel Kindiak, 20 Jan 25, 1941H

May 7, 2025
Technical Integration

In integral calculus, we usually assume that continuous functions $f : [a, b] \to \mathbb R$ can be integrated. Here, we will formally prove this really works, and in doing so, formulate Riemann integration more rigorously.

Lemma 1. Let $f : [a, b] \to \mathbb R$ be continuous and nonnegative (i.e. $f \geq 0$ ). For any partition $P = \{x_0, x_1, ..., x_n\}$ of $[a, b]$ (i.e. $x_0 = a$ , $x_n = b$ , and $P$ is non-decreasing), and for any $x \in [a, b]$ ,

$\displaystyle \sum_{i=1}^n m_i(f, P) \cdot \mathbb{I}_{[x_{i-1}, x_i)} \leq \sum_{i=1}^n f \cdot \mathbb{I}_{[x_{i-1}, x_i)}\leq \sum_{i=1}^n M_i(f, P) \cdot \mathbb{I}_{[x_{i-1}, x_i)},$

where $x_i := i/n$ , $P := \{x_0, x_1, ..., x_n\}$ , and we make the notations

$\displaystyle \begin{aligned} M_i(f, P) := \sup_{x \in [x_{i-1}, x_i] } f(x),\quad m_i (f, P) := \inf_{x \in [x_{i-1}, x_i] } f(x). \end{aligned}$

Recall that for sets $L \subseteq K$ , we define the indicator function $\mathbb{I}_L : \mathbb R \to \{0, 1\}$ by

$\mathbb{I}_L(x) = \begin{cases} 1, & x \in L, \\ 0, & x \notin L. \end{cases}$

Lemma 2. Let $f : [a, b] \to \mathbb R$ be continuous and nonnegative. Then for any $\epsilon > 0$ , there exists a partition $P = \{x_0,x_1,\dots,x_n\}$ such that for any $i = 1,\dots,n$ ,

$M_i (f, P) - m_i(f, P) < \epsilon,$

where we follow the notations in Lemma 1.

Proof. We prove the special case $[a, b] = [0, 1]$ for simplicity and leave the general case as an exercise in bookkeeping.

Fix $\epsilon > 0$ . Since $f$ is continuous on $[0, 1]$ , it is uniformly continuous on $[0, 1]$ . Hence, there exists $\delta > 0$ such that for any $u,v \in [0, 1]$ ,

$|u - v| < \delta \quad \Rightarrow \quad |f(u) - f(v)| < k \cdot \epsilon.$

By the Archimedean property, find $n \in \mathbb N$ such that $1/n < \delta$ . Define $P = \{x_0,x_1,\dots,x_n\}$ where $x_i = i/n$ .

Fix $i = 1,\dots,n$ . By the extreme value theorem, find $c_i^{\pm} \in [x_{i-1}, x_i]$ such that $f(c_i^-) = m_i(f, P)$ and $f(c_i^+) = M_i(f, P)$ . Since $|c_i^+ - c_i^-| \leq x_i - x_{i-1} = 1/n < \delta$ ,

$M_i (f, P) - m_i(f, P) = | f(c_i^+) - f(c_i^-) | < k \cdot \epsilon.$

Setting $k = 1$ yields the desired result.

Theorem 1. Let $f : [a, b] \to \mathbb R$ be continuous and nonnegative. Then for any $\epsilon > 0$ , there exists a partition $P = \{x_0, x_1, \dots, x_n\}$ such that

$\displaystyle \sum_{i=1}^{n} (M_i (f, P) - m_i (f, P)) \Delta x_i < \epsilon,$

where $\Delta x_i := x_i - x_{i-1}$ and we follow the notations in Lemma 1. Consequently,

$\displaystyle \sum_{i=0}^n m_{i}(f, P) \Delta x_i \leq \sum_{i=0}^n f(x_i)\Delta x_i\leq \sum_{i=0}^n m_{i}(f,P) \Delta x_i + \epsilon.$

Proof. Fix $\epsilon > 0$ . By Lemma 2, for any $k > 0$ , find a partition $P = \{x_0,x_1,\dots,x_n\}$ such that for any $i = 1,\dots,n$ ,

$M_{i}(f, P) - m_{i}(f, P) < k \cdot \epsilon,$

Summing over $i = 1, \dots, n$ ,

$\displaystyle \begin{aligned} \sum_{i=1}^{n} (M_i (f, P) - m_i (f, P)) \Delta x_i &< \sum_{i=1}^{n} (k \cdot \epsilon) \cdot \Delta x_i \\ &= k \cdot \epsilon \cdot \sum_{i=1}^{n} \Delta x_i \\ &= k \cdot \epsilon \cdot 1 \\ &= k \cdot \epsilon. \end{aligned}$

Setting $k = 1$ yields the desired result.

What has Theorem 1 got to do with integration? Intuitively, $\epsilon \to 0^+$ should induce $n \to \infty$ , so that in the limit, $\displaystyle \sum_{i=0}^n f(x_i)\Delta x_i$ yields the area under the graph of $y = f(x)$ on $[0, 1]$ , yielding the familiar (approximate) definition

$\displaystyle \int_0^1 f(x)\, \mathrm dx = \lim_{n \to \infty} \sum_{i=1}^n f(x_i) \Delta x.$

We will now formulate this limiting process precisely using Riemann integration. For the rest of this post, let $f : [a, b] \to \mathbb R$ be bounded. For simplicity, we will occasionally assume $[a, b] = [0, 1]$ to elucidate the underlying analysis.

Definition 1. A partition $P$ of $[a, b]$ is an increasing sequence

$P = \{x_0 = a, x_1,\dots, x_n = b\}$

of $n+1$ terms. For any partition $P$ with $n+1$ terms, define

$\displaystyle U(f, P) := \sum_{i = 1}^n M_i(f, P)\Delta x_i,\quad L(f, P) := \sum_{i = 1}^n m_i(f, P)\Delta x_i,$

where $\Delta x_i := x_i - x_{i-1}$ and

$\displaystyle \begin{aligned} M_i(f, P) := \sup_{x \in [x_{i-1}, x_i] } f(x),\quad m_i (f, P) := \inf_{x \in [x_{i-1}, x_i] } f(x). \end{aligned}$

Finally, define the upper and lower integrals by

$\displaystyle \mathcal U_a^b (f) := \inf_P U(f, P),\quad \mathcal L_a^b (f) := \sup_P L(f, P).$

We say that $f$ is Riemann-integrable on $[a, b]$ if $\mathcal U_a^b (f) = \mathcal L_a^b (f)$ . In this case, we write

$\displaystyle \mathcal L_a^b (f) =: \int_a^b f = \int_a^b f(x)\, \mathrm dx,$

and the right-hand side is used when emphasising the underlying variables.

Lemma 3. For partitions $P \subseteq Q$ of $[a, b]$ ,

$U(f, Q) \leq U(f, P),\quad L(f, Q) \geq L(f, P).$

Proof. We will prove the case $Q = P \cup \{c\}$ where $c \notin P$ and leave the general result as an exercise in induction. Find $i = 1,\dots,n$ such that $x_{i-1} < c < x_i$ . Then

$\displaystyle \begin{aligned} &\sup_{x \in [x_{i-1}, c]} f(x) \cdot (c - x_{i-1}) + \sup_{x \in [c, x_i]} f(x) \cdot (x_i - c) \\ &\leq \sup_{x \in [x_{i-1}, x_i]} f(x) \cdot (c - x_{i-1}) + \sup_{x \in [x_{i-1}, x_i]} f(x) \cdot (x_i - c) \\ &\leq \sup_{x \in [x_{i-1}, x_i]} f(x) \cdot (c - x_{i-1} + x_i - c) \\ &\leq \sup_{x \in [x_{i-1}, x_i]} f(x) \cdot (x_i - x_{i-1} ). \end{aligned}$

The result $U(f, Q) \leq U(f, P)$ follows by summing on both sides. The result for $L(f, Q) \geq L(f, P)$ follows similarly.

Lemma 4. $\mathcal L_a^b(f) \leq \mathcal U_a^b(f)$ .

Proof. For any partition $P$ ,

$\displaystyle L(f, P) = \sum_{i = 1}^n m_i(f, P) \Delta x_i\leq \sum_{i = 1}^n M_i(f, P)\Delta x_i = U(f, P).$

Now fix partitions $P, Q$ . By Lemma 3,

$L(f, P) \leq L(f, P \cup Q) \leq U(f, P \cup Q) \leq U(f, Q).$

Thus, $U(f, Q)$ is an upper bound for $L(f, P)$ regardless of $P$ . By definition of the supremum,

$\displaystyle \mathcal L_a^b(f) = \sup_P L(f, P) \leq U(f, Q)$

Since $\mathcal L_a^b(f)$ is a lower bound for $L(U, Q)$ regardless of $Q$ , by definition of the infimum,

$\displaystyle \mathcal L_a^b(f) \leq \inf_Q U(f, Q) = \mathcal U_a^b(f),$

as required.

Theorem 2 (Riemann Integrability Criterion). $f$ is Riemann-integrable if and only if for any $\epsilon > 0$ , there exists a partition $P = \{x_0,x_1,\dots,x_n\}$ of $[a, b]$ such that

$\displaystyle U(f, P) - L(f, P) < \epsilon.$

Proof. We will prove in two directions. For $(\Rightarrow)$ , suppose $f$ is Riemann-integrable. Then $\mathcal U_a^b (f) = \mathcal L_a^b (f)$ . Expanding the meaning of $\mathcal U_a^b (f)$ , for any $k_1 > 0$ , there exists a partition $Q$ of $[a, b]$ such that

$U(f, Q) < \mathcal L_a^b (f) + k_1 \cdot \epsilon.$

Expanding the definition of $\mathcal L_a^b (f)$ , for any $k_2 > 0$ , there exists a partition $R$ of $[a, b]$ such that

$L(f, R) > \mathcal L_a^b (f) - k_2 \cdot \epsilon.$

Combining the estimates,

$U(f, Q) - L(f, R) < (k_1 + k_2) \cdot \epsilon.$

By Lemma 3, the partition $Q \cup R$ satisfies

$\begin{aligned} U(f, Q \cup R) - L(f, Q \cup R) &\leq U(f, Q) - L(f, R) \\ &< (k_1 + k_2) \cdot \epsilon. \end{aligned}$

Set $k_1 = k_2 = 1/2$ and define the partition $P:= Q \cup R$ .

For $(\Leftarrow)$ . We need to prove that $\mathcal L_a^b(f) = \mathcal U_a^b(f)$ . By Lemma 4, $\mathcal L_a^b(f) \leq \mathcal U_a^b(f)$ unconditionally. To prove $\mathcal U_a^b(f) \leq \mathcal L_a^b(f)$ , the hypothesis yields that, for any $\epsilon > 0$ , there exists a partition $P$ of $[a, b]$ such that

$U(f, P) - L(f, P) < \epsilon.$

Taking the infimum over $P$ on the left-hand side,

$\displaystyle \inf_P (U(f, P) - L(f, P)) \leq 0.$

On the other hand,

$\begin{aligned} \mathcal U_a^b(f) - \mathcal L_a^b(f) &= \inf_P U(f, P) - \sup_P L(f, P) \\ &= \inf_P U(f, P) + \inf_P (- L(f, P)) \\ &= \inf_P (U(f, P) + (- L(f, P))) \\ &= \inf_P (U(f, P) - L(f, P)) \leq 0. \end{aligned}$

Thus, $\mathcal U_a^b(f) \leq \mathcal L_a^b(f)$ . Therefore, $\mathcal L_a^b(f) = \mathcal U_a^b(f)$ , and $f$ is Riemann-integrable, as required.

This tells us, finally, that continuous functions are Riemann-integrable.

Corollary 1. If $f : [a, b] \to \mathbb R$ is continuous, then $f$ is Riemann-integrable.

Proof. Combine the results in Theorem 1 and Theorem 2 to obtain the desired result.

Technically, we have formulated Darboux-integrability that focuses on the “maximal” and “minimal” areas, since Riemann-integrability is more closely connected with general areas. Yet, Darboux integrability basically yields upper and lower estimates on the Riemann integral, and so in the limit, both terms match up and the existence of one implies the other.

Is the limit of Riemann-integrable functions itself Riemann-integrable? Yes with an asterisk, and we explore this result the next time.

—Joel Kindiak, 19 Jan 25, 1554H

April 30, 2025
Kronecker Approximation

After class one day, one of my students shared with me about this fascinating result. The source material comes from this video, and the writeup is my interpretation of its key ideas.

Let $\mathbb K$ be an ordered field.

Problem 1. Prove that for $\mathbb K$ -sequences $( x_n ), ( y_n )$ , if $x_n \to 0$ and $y_n$ is bounded, then $x_n \cdot y_n \to 0$ .

(Click for Solution)

Solution. Since $(y_n)$ is bounded, find $M > 0$ such that $|y_n| < M$ . Fix $\epsilon > 0$ . Since $x_n \to 0$ , for any $k > 0$ , there exists $N \in \mathbb N$ such that if $n > N$ , then $|x_n| < k \cdot \epsilon$ . By construction,

$|x_n \cdot y_n| = |x_n| \cdot | y_n | < k \cdot \epsilon \cdot M.$

Setting $k := 1/M > 0$ yields the result $x_n \cdot y_n \to 0$ .

For the rest of this exercise, let $\alpha$ be an irrational number.

Definition 1. For any $x \in \mathbb R$ , define the fractional part of $x$ by $\{ x \} := x - \lfloor x \rfloor$ .

Problem 2. Prove that for distinct positive integers $p, q$ , $\{p \alpha\} \neq \{q \alpha\}$ .

(Click for Solution)

Solution. Suppose for a contradiction that there exist distinct positive integers $p, q$ such that $\{p \alpha\} = \{q \alpha \}$ . By the definition of the fractional part,

$\begin{aligned}p \alpha - \lfloor p\alpha \rfloor &= q \alpha - \lfloor q \alpha \rfloor\quad \Rightarrow \quad \alpha = \frac{ \lfloor p\alpha \rfloor - \lfloor q \alpha \rfloor}{p - q} \in \mathbb Q, \end{aligned}$

a contradiction.

Problem 3. Prove that for any $n \in \mathbb N$ , there exist integers $p_n, q_n$ such that

$0 < \{ p_n \alpha \} - \{ q_n \alpha \} < 1/n.$

(Click for Solution)

Solution. Fix $n \in \mathbb N$ . By Problem 2, consider the $n+1$ distinct values

$\{\alpha \},\{2\alpha\},\dots, \{n \alpha\}, \{(n+1)\alpha\}$

and the $n$ equally-spaced sub-intervals of $[0, 1)$ :

$[0, 1/n), [1/n, 2/n), \dots, [(n-1)/n, 1).$

This means there exists integers $i_n, p_n, q_n$ such that

$\{p_n \alpha\}, \{q_n\alpha\} \in [i_n/n, (i_n+1)/n)\quad \Rightarrow \quad 0 < \{ p_n \alpha \} - \{ q_n \alpha \} < 1/n.$

Problem 4. Define

$A := \mathbb Z + \alpha \mathbb Z \equiv \{ m + n \alpha : m, n \in \mathbb Z\}.$

Prove that for any $x \in \mathbb R$ , there exists an $A$ -sequence $( x_n )$ such that $x_n \to x$ .

(Click for Solution)

Solution. Fix $x \in \mathbb R$ . Use Problem 3 to define the sequence $(y_n)$ by $y_n := \{ p_n \alpha \} - \{ q_n \alpha \}$ . It is clear that $y_n \to 0$ and each $y_n \neq 0$ . Define $x_n := y_n \cdot \lfloor x/y_n \rfloor$ . Then

$\displaystyle x_n = y_n \cdot \lfloor x/y_n \rfloor = y_n \cdot \left( \frac x{y_n} - \left\{ \frac x{y_n} \right\}\right)= x - y_n \left\{ \frac x{y_n} \right\}.$

By the definition of the fractional part, the sequence $( \{x / y_n \} )$ is bounded. Since $y_n \to 0$ , by Problem 1, we have $y_n \cdot \{x / y_n \} \to 0$ . Hence, $x_n \to x$ .

It remains to verify that $x_n \in A$ for each $n$ . Indeed,

$\begin{aligned} x_n &= y_n \cdot \lfloor x/y_n \rfloor \\ &= (\{ p_n \alpha \} - \{ q_n \alpha \}) \cdot \lfloor x/y_n \rfloor \\ &= \{ p_n \alpha \} \cdot \lfloor x/y_n \rfloor - \{ q_n \alpha \} \cdot \lfloor x/y_n \rfloor \\ &= (p_n \alpha - \lfloor p_n \alpha \rfloor) \cdot \lfloor x/y_n \rfloor - (q_n \alpha - \lfloor q_n \alpha \rfloor) \cdot \lfloor x/y_n \rfloor \\ &= \underbrace{ (\lfloor q_n \alpha \rfloor - \lfloor p_n \alpha \rfloor) \cdot \lfloor x/y_n \rfloor }_{\in \mathbb Z} +\, \alpha \cdot \underbrace{ (p_n - q_n) \cdot\lfloor x/y_n \rfloor }_{\in \mathbb Z} \\ &\in \mathbb Z + \alpha \mathbb Z = A, \end{aligned}$

as required.

Problem 5. Define

$B := \sin(\mathbb N) \equiv \{ \sin(n) : n \in \mathbb N\}.$

Prove that for any $y \in [-1, 1]$ , there exists a $B$ -sequence $( y_n )$ such that $y_n \to y$ .

(Click for Solution)

Solution. For any $y \in [-1, 1]$ , find $u \in (0, \infty) \subseteq \mathbb R$ such that $\sin(u) = y$ . Define

$C := \mathbb Z + 2\pi \mathbb Z.$

By Problem 4, there exists a $C$ -sequence $(u_n)$ such that $u_n \to u$ . Denote $u_n = s_n + 2\pi \cdot t_n$ with $s_n \in \mathbb N$ . Defining $y_n = \sin(s_n) \in B$ ,

$y_n = \sin(s_n) = \sin(s_n + 2\pi \cdot t_n) = \sin(u_n) \to \sin(u) = y.$

—Joel Kindiak, 24 Apr 25, 1851H

April 24, 2025

Previous Page Next Page

Blog at WordPress.com.

Subscribe Subscribed
- KindiakMath
- Already have a WordPress.com account? Log in now.