Consequences: integration and numerical computation

Approximation by step functions

Let $f$ be continuous on $[a \,;\, b]$ and $\varepsilon > 0$. Heine's theorem gives a uniform $\delta > 0$. Choose a regular subdivision $a = t_0 < t_1 < \cdots < t_n = b$ with step $h = \dfrac{b - a}{n} \leq \delta$.

On each piece $[t_i \,;\, t_{i+1}]$, any two points are at most $\delta$ apart, so the values of $f$ vary by less than $\varepsilon$ there. The step function equal to $f(t_i)$ on $\left[t_i \,;\, t_{i+1}\right[$ therefore approximates $f$ to within $\varepsilon$ on the whole interval.

This is the deep meaning of the theorem: on a compact set, a single subdivision is enough to control the function everywhere.

Riemann integrability

With the same subdivision, let $M_i$ and $m_i$ denote the maximum and the minimum of $f$ on $[t_i \,;\, t_{i+1}]$. We have $M_i - m_i \leq \varepsilon$, so the gap between the upper and lower Darboux sums satisfies

$$S(f, \sigma) - s(f, \sigma) = \sum_{i=0}^{n-1} (M_i - m_i)\,(t_{i+1} - t_i) \leq \varepsilon\,(b - a)$$

This gap becomes as small as we like: every function continuous on a closed bounded interval is Riemann integrable. This result, used in every integral computation, therefore rests directly on Heine.

Controlling the error in numerical computation

The modulus of continuity gives an explicit bound for the left rectangle method:

$$\left\lvert \int_a^b f(x)\,\mathrm{d}x - \frac{b - a}{n}\sum_{i=0}^{n-1} f(t_i) \right\rvert \leq (b - a)\,\omega_f\!\left(\frac{b - a}{n}\right)$$

Heine ensures that the right-hand side tends to $0$: the method converges for any continuous function, even without a differentiability assumption. If moreover $f$ is $k$-Lipschitz, we obtain the quantitative bound

$$\left\lvert \int_a^b f(x)\,\mathrm{d}x - \frac{b - a}{n}\sum_{i=0}^{n-1} f(t_i) \right\rvert \leq \frac{k\,(b - a)^2}{2n}$$

For $f(x) = \sin x$ on $[0 \,;\, \pi]$, with $k = 1$, this gives an error less than $\dfrac{\pi^2}{2n}$, that is, about $0.005$ for $n = 1000$.

Extension by continuity

Another consequence, in the other direction: if $f$ is uniformly continuous on $\left]a \,;\, b\right[$, it extends by continuity to $a$ and to $b$. The proof uses the Cauchy criterion — the image of a Cauchy sequence under a uniformly continuous function is still Cauchy, which is false under the mere assumption of continuity.

This sheds the right light on $x \mapsto \dfrac{1}{x}$ on $\left]0 \,;\, 1\right]$: this function does not extend to $0$, so it could not be uniformly continuous there.

Classic pitfall

Heine provides no explicit $\delta$. When an exercise asks for a quantitative bound — the number of points in a subdivision, the error of a numerical method —, the theorem only proves that the constant exists. To compute it, you need additional information: a Lipschitz constant, a bound on $\lvert f' \rvert$, or an explicit modulus of continuity.