Bourbaki

Exercise 13, § 1

Functions of a Real Variable, II, p. 31

Bourbaki marks this one as harder than the rest.

Let ff be a regulated numerical function on a compact interval [a,b][a, b]. We put

r(n)=b−an∑k=1nf(a+kb−an)−∫abf(t) dt. r(n) = \frac{b - a}{n} \sum_{k=1}^{n} f\left(a + k \frac{b - a}{n}\right) - \int_a^b f(t) \, dt.
a) Now prove that, if ff is increasing in [a,b][a, b], one has
0≤r(n)≤b−an(f(b)−f(a)). 0 \leq r(n) \leq \frac{b - a}{n} (f(b) - f(a)).
b) If ff is continuous and admits in [a,b][a, b] a regulated and bounded right derivative, now prove that one has
lim⁡n→∞nr(n)=b−a2(f(b)−f(a))(by setting xk=a+kb−an, notice that one has \lim_{n \to \infty} n r(n) = \frac{b - a}{2} (f(b) - f(a)) \quad \text{(by setting } x_k = a + k \frac{b - a}{n}, \text{ notice that one has}
r(n)=∑k=0n−1∫xkxk+1(f(xk+1)−f(t)) dt, r(n) = \sum_{k=0}^{n-1} \int_{x_k}^{x_{k+1}} (f(x_{k+1}) - f(t)) \, dt,
and apply Proposition 5 of II, p. 7). c) Give an example of an increasing and continuous function ff in [a,b][a, b] such that nr(n)n r(n) does not tend to b−a2(f(b)−f(a))\frac{b - a}{2} (f(b) - f(a)) when nn increases indefinitely [take for ff the limit of a decreasing sequence (fn)(f_n) of increasing functions whose representative curves are broken lines, such that
fn(a+kb−a2n)=f(a+kb−a2n)for 0≤k≤2n f_n\left(a + k \frac{b - a}{2^n}\right) = f\left(a + k \frac{b - a}{2^n}\right) \quad \text{for } 0 \leq k \leq 2^n
and
(b−a)∑k=12nfn(a+kb−a2n)−2n∫abfn(t) dt≥14(b−a)(fn(b)−fn(a))]. (b - a) \sum_{k=1}^{2^n} f_n\left(a + k \frac{b - a}{2^n}\right) - 2^n \int_a^b f_n(t) \, dt \geq \frac{1}{4}(b - a)(f_n(b) - f_n(a))].

Solution

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Machine translation of the English, by gpt-5-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.