Bourbaki

Exercise 2, § 1

Integration, IX

Let T be a set and p an encumbrance on T. A function f∈F+(T)f \in \mathscr{F}_+(T) (resp. a subset A of T) is said to be p-negligible if p(f)=0p(f) = 0 (resp. p(A)=0p(A) = 0). a) Let f and g be in F+(T)\mathscr{F}_+(T) and let t be a positive real number such that f≤t⋅gf \leq t \cdot g. If g is p-negligible, then so is f. The sum and upper envelope of a sequence of p-negligible functions are p-negligible. b) Every subset of a p-negligible set is p-negligible, as is the union of a sequence (finite or infinite) of p-negligible sets. c) Let f∈F+(T)f \in \mathscr{F}_+(T). For every finite real number a>0a > 0, denote by TaT_a the set of t∈Tt \in T such that f(t)≥af(t) \geq a. Show that p(Ta)≤a−1⋅p(f)p(T_a) \leq a^{-1} \cdot p(f). From this, deduce that f is p-negligible if and only if the set of t∈Tt \in T such that f(t)≠0f(t) \neq 0 is p-negligible. d) Let f∈F+(T)f \in \mathscr{F}_+(T). If p(f)p(f) is finite, the set of t∈Tt \in T such that f(t)f(t) is infinite is p-negligible. e) Let f and g be in F+(T)\mathscr{F}_+(T). If the set of t∈Tt \in T such that f(t)≠g(t)f(t) \neq g(t) is p-negligible, then p(f)=p(g)p(f) = p(g).

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