322 CHAPTER 11. REGULAR MEASURES

contains K . Then it is obvious G is closed with respect to countable disjoint unions. Thecase of complements maybe is not as obvious.

(x+Rn)\ (x+Rn∩E) = x+Rn∩EC.

Then

mp(x+Rn∩EC) = mp (x+Rn)−mp (x+Rn∩E)

= mp (Rn)−mp (Rn∩E) = mp(Rn∩EC)

Thus by Dynkin’s lemma, G = B (Rp) . Thus for all E Borel,

mp (E ∩Rn) = mp (x+E ∩Rn) .

Now let n→ ∞. It follows that mp is translation invariant for all Borel sets.In general, if E is Lebesgue measurable, it follows from Proposition 11.1.2 that there

are sets F ⊆ E ⊆ G where F,G are Borel and mp (F) = mp (E) = mp (G). Then

mp (E) = mp (F) = mp (F +x)≤ mp (E +x)≤ mp (G+x) = mp (G) = mp (E)

and so all the inequalities are equal signs. Hence mp (E +x) = mp (E). ■The following is a useful lemma. In this lemma, Xi will be some metric space or more

generally a topological space. It is useful in recognizing a Borel measurable set when yousee it.

Lemma 11.3.2 If Ei is a Borel set in Xi, then ∏pk=1 Eik is a Borel set in ∏

pk=1 Xik .

Proof: Let π ir : ∏pk=1 Xik → Xir be the projection map. That is π ir (x) = xir when

x =(xi1 ,xi2 , ...,xip

). Obviously this is continuous. Therefore, if U is an open set in

Xir ,π−1ir (U) = Xi1 ×Xi2 × ·· · ×U × ·· · ×Xip . Is an open set. Let Bir be the Borel sets

of Xir E such that π−1ir (E) = Xi1 ×Xi2 ×·· ·×E×·· ·×Xip is a Borel set in ∏

pk=1 Xik . Then

Bir is a σ algebra and it contains the open sets. Therefore, it contains the Borel sets of Xir .It follows that ∏

pk=1 Eik = ∩

pk=1π

−1ik

(Eik

)is a finite intersection of Borel sets in ∏

pk=1 Xik

and so it is also a Borel set. ■

Example 11.3.3 Let A≡ {(x,y) : y < g(x)} for g a Borel measurable real valued function.Then A is a Borel set.

To see this, partition R into equally spaced points{

rnk

}∞

k=−∞,rn

k < rnk+1,r

nk+1− rn

k =

2−n and let gn (x) ≡ ∑∞k=−∞

rnk−1Xg−1((rn

k−1,rnk ])

(x) so that gn (x)→ g(x) for each x. Let

An ≡ {(x,y) : y < gn (x)} . Now each Ak is Borel by the above Lemma. Then thanks toconvergence, A = ∩∞

m=1∩k≥m Ak so A is Borel.

Example 11.3.4 Let A ≡ {(x,y) : y≤ g(x)} for g a Borel measurable function. Then A isa Borel set.

This follows from observing that if An ≡ {(x,y) : y < g(x)+2−n} then A = ∩∞n=1An.

Thus sets of the form [a,b]×{(x,y) : y≤ g(x)} for g Borel measurable are Borel measur-able. These examples justify the usual calculus manipulations involving iterated integrals,the next example being an illustration of this.