Question

Let f be a nonnegative measurable function on E

(1) Show there is an increasing sequence of nonnegative simple functions on E, each of finite support, which converges pointwise on E to
(2) Show that, f=sup φ simple, of finite supportand 0 os fonE}

Answer 1

If f is F-measurable and non-negative on E ∈ F and µ(E) = 0

thenREf dµ = 0.ProofLet 0 ≤ s ≤ f be a simple, F-measurable function. So s =PNn=1 anχAnfor some an ≥ 0,

An ∈ F. Then IE(s) = PNn=1 anµ(An ∩ E). But µ ismonotone which means that µ(An ∩ E) ≤ µ(E) = 0 for all n and so IE(s) =0 for all such simple functions.

Hence I(f, E) = {0} and so REf dµ =sup I(f, E) = 0. ¥Lemma 4.7 If g ≥ 0 and REgdµ = 0 thenµ{x ∈ E : g(x) > 0} = 0.Proof Let A = {x ∈ E : g(x) > 0} and An = {x ∈ E : g(x) >1n}.

Thenthe sets An = E ∩ {x : g(x) >1n} ∈ F satisfy A1 ⊆ A2 ⊆ A3 ⊆ ... withA =S∞n=1 An. By lemma 4.1 µ(A) = limn→∞ µ(An). Usingsn(x) = ½ 1nif x ∈ An0 otherwise,so sn ≤ g on An we have1nµ(An) = IAn(sn)≤RAngdµ by the definition of RAn≤REgdµ

Thereom 4.4(iii)= 0 by assumption.So µ(An) = 0 for all n and hence µ(A) = 0. ¥Definition If a property P holds on all points in E \ A for some set A withµ(A) = 0 we say that P holds almost everywhere (µ) on E, written as a.e.(µ)on E.

(*It might be that P holds on some of the points of A or that the set ofpoints on which P does not hold is non-measurable.

This is immaterial. Butif µ is a complete measure, such as the Lebesgue-Steiltje’s measure µF, thenthe situation is simpler.

Assume that a property P holds a.e.(µ) on E.

Thedefinition says that the set of points, D say, on which P does not hold can becovered by a set of measure zero, i.e. there exists A : D ⊆ A and µ(A) = 0.

Yet if µ is complete then D will be measurable of measure zero.In this section we are not assuming that µ is complete.)