Continuous functions are integrable proof

How to use

• Visualize the proof idea: for every $\varepsilon>0$ we construct staircase functions $\psi$ and $\varphi$ such that $\psi \le f \le \varphi \quad\text{and}\quad I(\varphi)-I(\psi)=\varepsilon.$
• Drag the blue points $a$ and $b$ to choose the interval $[a,b]$.
• Use the slider $\varepsilon$ to choose the tolerance.
• Use the slider $n$ to choose the number of subintervals in the uniform partition $Z$.
• The blue curve is the continuous function $f$.
• The partition $Z$ is uniform with mesh size $\Delta x=\frac{b-a}{n}.$
• The applet computes $\tilde\varepsilon=\frac{\varepsilon}{2(b-a)}.$ The vertical bracket at $x=b$ has height $2\tilde\varepsilon$.
• The orange staircase is $\varphi$ (intended upper bound), defined on each open interval $(x_{i-1},x_i)$ by $\varphi(x)=f(x_i)+\tilde\varepsilon.$
• The purple staircase is $\psi$ (intended lower bound), defined on each open interval $(x_{i-1},x_i)$ by $\psi(x)=f(x_i)-\tilde\varepsilon.$
• The filled band between them is colored green/red depending on whether the proof’s size condition is satisfied.
• The proof requires choosing $n$ so that $\Delta x < \delta$, where $\delta$ comes from uniform continuity (condition (7.2)).
• In the applet, this is checked via a sufficient bound (using an estimate of $\sup|f'|$ for the fixed function $f$). Concretely, it computes a number $\delta$ and checks whether $\frac{b-a}{n} < \delta.$
• When you change $\varepsilon$, $a$, or $b$, the $n$ slider automatically jumps to the minimum value required for the proof condition to hold (green state).
• You can manually drag the $n$ slider below the required minimum to see what happens when the proof condition fails:
  • Everything turns red (band, staircases, labels).
  • A warning message appears: “⚠️ Bounds fail: $\psi \not\le f$ or $f \not\le \varphi$”.
  • You can visually observe the blue curve escaping the red band — the purple staircase may appear above $f$, or the orange staircase below $f$.
  • This demonstrates why the condition $(b-a)/n < \delta$ is necessary: without it, the construction does not produce valid bounds.
• To restore the green (valid) state, you can:
  • Increase $n$ (finer partition → smaller $\Delta x$),
  • Increase $\varepsilon$ (larger tolerance → easier condition),
  • Or shrink the interval $[a,b]$ (smaller range).
• Once the display turns green, the intended implication of (7.2) is in effect: $|x-x_i|<\delta \Rightarrow |f(x)-f(x_i)|<\tilde\varepsilon,$ which gives $\psi(x)\le f(x)\le\varphi(x)$ on each $(x_{i-1},x_i)$, matching the proof idea.

Description

Link to code.

Reference: Lecture Notes Calculus 1 (May 22, 2022) Theorem 7.5 page 129

This applet is quite experimental and visualizes the proof that a continuous function $f:[a,b]\to\mathbb{R}$ is Riemann integrable by constructing two staircase functions $\psi\le f\le\varphi$ whose integrals differ by at most $\varepsilon$. The function is fixed in the code as const f = (x: number) => 0.5 * Math.sin(x) * x + 2.5;.

A good way to use the applet is the same order as the proof. First fix an interval $[a,b]$ by dragging the gliders gliderA and gliderB. Then choose an error tolerance $\varepsilon>0$ with epsSlider. After selecting the $\varepsilon>0$, the $n$ slider will auto-adjust to the minimum required value. The student can then decrease $n$ manually to observe the failure case (red state), or increase $n$ further to see the construction with a finer partition.

The applet computes $\widetilde{\varepsilon}=\varepsilon/(2(b-a))$ via const eTilde = eps / (2 * range);. The dashed bracket at $x=b$ has vertical length $2 \widetilde{\varepsilon}$ (objects epsBracket and epsBracketLabel), and it is colored green/red together with the construction.

Next choose $n$ (the number of subintervals) with nSlider. The applet creates the uniform partition $Z: a=x_0<\dots<x_n=b$ with $x_k=a+k(b-a)/n$, implemented by uniformPartition(a,b,n), where dx=(b-a)/n is the mesh size, as in the proof.

const uniformPartition = (a: number, b: number, n: number) => {
    const dx = (b - a) / n;
    const cuts = Array.from({ length: n + 1 }, (_, k) => a + k * dx);
    cuts[cuts.length - 1] = b;
    return { cuts, dx };
};