Bourbaki

Exercise 9, § 8

Integration, V, p. 116

Let uu and vv be two increasing and right-continuous numerical functions on R\mathbf{R}, such that u(x)=v(x)=0u(x) = v(x) = 0 for x<0x < 0. Let ww be the increasing right-continuous function on R\mathbf{R} defined by w(t)=u(t)v(t)w(t) = u(t)v(t) for t≥0t \geq 0, w(t)=0w(t) = 0 for t<0t < 0; let λ,μ,ν\lambda, \mu, \nu be the Stieltjes measures associated with u,v,wu, v, w respectively (§6, Exer. 5).

To every function ff defined on R\mathbf{R}, with values in R‾\overline{\mathbf{R}} or in a Banach space FF, one makes correspond the function fˉ\bar{f} defined on R2\mathbf{R}^2 by the conditions fˉ(x,y)=f(x)\bar{f}(x, y) = f(x) if y < x , fˉ(x,y)=f(y)\bar{f}(x, y) = f(y) if y≥xy \geq x. Show that, for ff to be ν\nu-integrable, it is necessary and sufficient that fˉ\bar{f} be integrable for the product measure λ⊗μ\lambda \otimes \mu, in which case ∫f dν=∬fˉ dλ dμ\int f\, d\nu = \iint \bar{f}\, d\lambda\, d\mu (prove it first for the characteristic functions of intervals). From this, deduce the formula

∫f(x) dw(x)=∫f(x)v(x−)du(x)+∫f(x)u(x+)dv(x). \int f(x)\, dw(x) = \int f(x)v(x-) du(x) + \int f(x)u(x+) dv(x).

In particular, if uu and vv are continuous on R\mathbf{R}, one obtains the formula for integration by parts

∫abu(x) dv(x)=u(b)v(b)−u(a)v(a)−∫abv(x) du(x). \int_a^b u(x)\, dv(x) = u(b)v(b) - u(a)v(a) - \int_a^b v(x)\, du(x).
Solution

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Transcribed from the printed volume. Not translated.