196 CHAPTER 8. MEASURES AND MEASURABLE FUNCTIONS

Then G is closed with respect to countable disjoint unions and complements. From whatwas just shown about rectangles, G ⊇K and so by Dynkin’s lemma, it follows that G =B (Rp) . Since m is arbitrary, this proves the theorem in case E is Borel.

From regularity, if E is only Lebesgue measurable, there is F an Fσ set and G a Gδ setsuch that F ⊆ E ⊆ G and mp (G\F) = 0. Then from what was just shown, it follows that

mp (E) = mp (F) = mp (z+F)≤ mp (z+E)

≤ mp (z+G) = mp (G) = mp (F)≤ mp (E)

By completeness of mp it follows that z+E is measurable because it lies between the Fσ

set z+F and the Gδ set z+G and

mp (z+G\ (z+F)) = mp (z+G\F) = mp (G\F) = 0 ■

Example 8.8.5 On R you could take an increasing function F and for the functional con-sider L( f )≡

∫f dF where this is the Riemann Stieltjes integral. This would give a measure

µF with all the properties of the above Theorem 8.8.2.

8.9 Exercises1. Show carefully that if S is a set whose elements are σ algebras which are subsets of

P (Ω) , then ∩S is also a σ algebra. Now let G ⊆P (Ω) satisfy property P if Gis closed with respect to complements and countable disjoint unions as in Dynkin’slemma, and contains /0 and Ω. If H ⊆ G is any set whose elements are subsets ofP (Ω) which satisfies property P, then ∩H also satisfies property P. Thus there is asmallest subset of G satisfying P. In other words, verify the details of the proof ofDynkin’s lemma.

2. The Borel sets of a metric space (X ,d) are the sets in the smallest σ algebra whichcontains the open sets. These sets are denoted as B (X). Thus B (X) = σ (open sets)where σ (F ) simply means the smallest σ algebra which contains F . Show that inRn, B (Rn) = σ (P) where P consists of the half open rectangles which are of theform ∏

ni=1[ai,bi).

3. Recall that f : (Ω,F )→ X where X is a metric space is measurable means thatinverse images of open sets are in F . Show that if E is any set in B (X) , thenf−1 (E) ∈ F . Thus, inverse images of Borel sets are measurable. Next considerf : (Ω,F ) → X being measurable and g : X → Y is Borel measurable, meaningthat g−1 (open) ∈ B (X). Explain why g ◦ f is measurable. Hint: You know that(g◦ f )−1 (U) = f−1

(g−1 (U)

). For your information, it does not work the other

way around. That is, measurable composed with Borel measurable is not necessarilymeasurable. In fact examples exist which show that if g is measurable and f iscontinuous, then g◦ f may fail to be measurable. An example is given later.

4. If you have Xi is a metric space, let X = ∏ni=1 Xi with the metric

d (x,y)≡max{di (xi,yi) , i = 1,2, · · · ,n}

Show that any set of the form ∏ni=1 Ei, Ei ∈ B (Xi) is a Borel set. That is, the

product of Borel sets is Borel. Hint: You might consider the continuous functions