Bourbaki

Exercise 5, § 2

Functions of a Real Variable, II, 85

a) Let f\mathbf{f} be a regulated function on a bounded open interval ]a,ba, b*; suppose that there exists a real function gg, decreasing on ]a,ba, b*, such that ∥f(x)∥⩽g(x)\| \mathbf{f}(x) \| \leqslant g(x) on ]a,ba, b* and such that the integral ∫abg(t) dt\int_a^b g(t)\,dt converges. Show that if (εn)(\varepsilon_n) is a sequence of numbers >0> 0, tending to 0, and such that inf⁡n⩾1nεn>0\inf_{n \geqslant 1} n \varepsilon_n > 0, one has

lim⁡n→+∞1n∑k=1n−1f(a+εn+kb−an)=∫abf(t) dt.(*) \lim_{n \to +\infty} \frac{1}{n} \sum_{k=1}^{n-1} \mathbf{f}\left(a + \varepsilon_n + k \frac{b-a}{n}\right) = \int_a^b \mathbf{f}(t)\,dt.\tag{*}
b) Give an example of a real regulated function f>0f > 0 on ]0, 1] such that the integral ∫01f(t) dt\int_0^1 f(t)\,dt converges, and yet the relation (*) does not hold for εn=1/n\varepsilon_n = 1/n (take ff so that its value for x=2−nx = 2^{-n} is 22n2^{2^n}).

c) With the same hypotheses as in a) show that

lim⁡n→+∞1n∑k=1n(−1)kf(a+kb−an)=0. \lim_{n \to +\infty} \frac{1}{n} \sum_{k=1}^n (-1)^k \mathbf{f}\left(a + k \frac{b-a}{n}\right) = 0.

Solution

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Transcribed from the printed volume. Not translated.