230 CHAPTER 11. ABSTRACT MEASURE AND INTEGRATION

Proof: Let U be an open set and let y ∈U. Since U is open, B(y,r)⊆U for some r > 0and it can be assumed r/

√n ∈Q. Let

x ∈ B(

y,r

10√

n

)∩Qn

and consider the cube, Qx ∈B defined by

Qx ≡n

∏i=1

(xi−δ ,xi +δ )

where δ = r/4√

n. The following picture is roughly illustrative of what is taking place.

yx

Qx

B(y,r)

Then the diameter of Qx equals(n(

r2√

n

)2)1/2

=r2

and so, if z ∈ Qx, then

|z−y| ≤ |z−x|+ |x−y|

<r2+

r2= r.

Consequently, Qx ⊆U. Now also,(n

∑i=1

(xi− yi)2

)1/2

<r

10√

n

and so it follows that for each i,|xi− yi|<

r4√

n

since otherwise the above inequality would not hold. Therefore, y ∈ Qx ⊆U . Now let BUdenote those sets of B which are contained in U. Then ∪BU =U.