Math Infinitum

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Intuition

Two large families pass the criterion, by two different routes. A continuous function on a closed bounded interval is uniformly continuous, so a single fineness makes the oscillation small on every piece at once — and that is exactly the hypothesis Heine–Cantor was proved for. A monotone function may jump, and its oscillations need not be small; instead they telescope, so their total is the whole rise of the function, and the widths can be made small.

Two ways of keeping a bill small. Either every item is cheap, which is the continuous case, or the expensive items are few and together account for a fixed total, which is the monotone case. Both give a small bill and the arguments have nothing in common.

Two theorems, two routes

If $f$ is continuous on $[a,b]$ then it is integrable there. If $f$ is monotone on $[a,b]$ then it is integrable there. Neither family contains the other: a monotone function may have infinitely many jumps, and a continuous function may oscillate without ever being monotone on any interval.

\begin{aligned}&f\ \text{continuous or monotone on}\ [a,b]\\ &\implies\ f\in\mathcal{R}[a,b]\end{aligned}

What each proof spends

The continuous case spends Heine–Cantor: one fineness makes $M_i-m_i<\tfrac{\varepsilon}{b-a}$ on every piece at once.

A continuous function is integrable

f\ \text{continuous on}\ [a,b]\implies f\in\mathcal{R}[a,b]

Aim at the criterion. Uniform continuity supplies a single fineness at which any two nearby points have values closer than a chosen amount; choose that amount so that, multiplied by the total width of the interval, it comes under the tolerance. Take any partition finer than that. On each piece the supremum and the infimum are attained, by the Extreme Value Theorem, at two points of that piece, which are close together, so the oscillation is under the chosen amount. Summing the oscillations against the widths gives the tolerance.

Proof steps

\begin{aligned}&\text{Heine–Cantor: }\exists\delta\\ &\lvert x-y\rvert<\delta\implies\lvert f(x)-f(y)\rvert<\tfrac{\varepsilon}{b-a}\end{aligned}

Uniform continuity, with the amount chosen to suit the total width.

\text{Take}\ P\ \text{with every}\ \Delta x_i<\delta

Any partition finer than that fineness will do.

M_i=f(u_i),\ m_i=f(v_i)\ \text{for some}\ u_i,v_i\in[x_{i-1},x_i]

The Extreme Value Theorem makes the supremum and infimum values on each piece.

\lvert u_i-v_i\rvert<\delta\implies M_i-m_i<\tfrac{\varepsilon}{b-a}

The two points lie in one piece, so they are close, and uniform continuity applies.

U-L=\sum(M_i-m_i)\Delta x_i<\tfrac{\varepsilon}{b-a}\sum\Delta x_i=\varepsilon

The widths add to the length of the interval, and the criterion is met.

Applications

Practice

One Fineness for the Whole Interval

Uniform continuity is what makes a single fineness work everywhere at once.

\lvert x-y\rvert<\delta\implies\lvert f(x)-f(y)\rvert<\tfrac{\varepsilon}{b-a}

Try it

Which theorem does the proof for continuous functions spend?