Bourbaki

Exercise 6, § 4

Integration, IV, p. 121

Bourbaki marks this one as harder than the rest.

a) Let ff be a numerical function such that μ∗(f)\mu^*(f) (Exer. 5) is finite. Show that there exists an integrable function f1≥ff_1 \geq f such that μ(f1)=μ∗(f)\mu(f_1) = \mu^*(f); if f2f_2 is a second integrable function such that f2≥ff_2 \geq f and μ(f2)=μ∗(f)\mu(f_2) = \mu^*(f), then f1f_1 and f2f_2 are equivalent.

b) For a numerical function ff to be integrable, it is necessary and sufficient that μ∗(f)\mu^*(f) and μ∗(f)\mu_*(f) be finite and equal.

c) Let ff be a numerical function such that μ∗(f)\mu^*(f) and μ∗(f)\mu_*(f) are both finite; let gg and hh be two integrable functions such that g≤f≤hg \leq f \leq h and μ(g)=μ∗(f)\mu(g) = \mu_*(f), μ(h)=μ∗(f)\mu(h) = \mu^*(f). Show that

μ∗(f−g)=μ∗(h−f)=0andμ∗(f−g)=μ∗(h−f)=μ∗(f)−μ∗(f). \mu_*(f - g) = \mu_*(h - f) = 0 \quad \text{and} \quad \mu^*(f - g) = \mu^*(h - f) = \mu^*(f) - \mu_*(f).

d) Let f1,f2f_1, f_2 be two numerical functions such that the numbers μ∗(f1),μ∗(f2),μ∗(f1)\mu^*(f_1), \mu^*(f_2), \mu_*(f_1) and μ∗(f2)\mu_*(f_2) are all finite; show that

μ∗(f1+f2)≤μ∗(f1)+μ∗(f2)≤μ∗(f1+f2) \mu_*(f_1 + f_2) \leq \mu_*(f_1) + \mu_*(f_2) \leq \mu^*(f_1 + f_2)

(if g2g_2 is an integrable function such that f2≤g2f_2 \leq g_2 and μ∗(f2)=μ(g2)\mu^*(f_2) = \mu(g_2), note that, for every integrable function hh such that h≤f1+f2h \leq f_1 + f_2, one has h−g2≤f1h - g_2 \leq f_1). From this, deduce that if f1f_1 is integrable then

μ∗(f1+f2)=μ(f1)+μ∗(f2),μ∗(f1+f2)=μ(f1)+μ∗(f2), \mu^*(f_1 + f_2) = \mu(f_1) + \mu^*(f_2), \quad \mu_*(f_1 + f_2) = \mu(f_1) + \mu_*(f_2),

and μ∗(f1+f2)=μ∗(sup⁡(f1,f2))+μ∗(inf⁡(f1,f2))\mu^*(f_1 + f_2) = \mu^*(\sup(f_1, f_2)) + \mu^*(\inf(f_1, f_2)) (for the latter relation, make use of Exer. 1 of §1).

e) Let ff be an integrable function. In order that a function gg, such that μ∗(g)\mu^*(g) is finite, be integrable, it is necessary and sufficient that μ(f)=μ∗(g)+μ∗(f−g)\mu(f) = \mu^*(g) + \mu^*(f - g) (if g1g_1 is an integrable function such that g≤g1g \leq g_1 and μ∗(g)=μ(g1)\mu^*(g) = \mu(g_1), note that f−g1≤f−gf - g_1 \leq f - g).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.