340 CHAPTER 13. LEBESGUE MEASURE

Let B∞ = ∪∞i=1Bi. In fact ∪B∞ = U . Clearly ∪B∞ ⊆U because every box of every Bi

is contained in U . If p ∈U , let k be the smallest integer such that p is contained in a boxfrom Ck which is also a subset of U . Thus

p ∈ ∪Bk ⊆ ∪B∞.

Hence B∞ is the desired countable disjoint collection of half open boxes whose union isU . The last assertion about the other type of half open rectangle is obvious. This provesthe lemma.

Now what does Lebesgue measure do to a rectangle, ∏ni=1(ai,bi]?

Lemma 13.1.3 Let R = ∏ni=1[ai,bi], R0 = ∏

ni=1(ai,bi). Then

mn (R0) = mn (R) =n

∏i=1

(bi−ai).

Proof: Let k be large enough that

ai +1/k < bi−1/k

for i = 1, · · · ,n and consider functions gki and f k

i having the following graphs.

ai +1k

ai

1bi− 1

k

bi

f ki ai− 1

k

ai

1

bi

bi +1k

gki

Let

gk(x) =n

∏i=1

gki (xi), f k(x) =

n

∏i=1

f ki (xi).

Then by elementary calculus along with the definition of Λ,

n

∏i=1

(bi−ai +2/k)≥ Λgk =∫

gkdmn ≥ mn(R)≥ mn(R0)

≥∫

f kdmn = Λ f k ≥n

∏i=1

(bi−ai−2/k).

Letting k→ ∞, it follows that

mn(R) = mn(R0) =n

∏i=1

(bi−ai).

This proves the lemma.