Let 
INTEGltABILITY 565 
© = Qdv {(£,}, 
then © = 36 
by 410, 6. 
Let £) = Di>((E, ©), 
then 3) = 36, (1 
as we proceed to show. For if 6r = 36 — ©, 
© = ©) + Dt>(®, (7) = © + D<©, (7). 
But (7 is a null set. Hence Meas Z)y(©, (7) = 0, and thus 
(| = 36 = ¿>, which is 1). 
Let now £ be a point of 3), let it lie in © <a ••• where £j, £ 2 ••• 
form a monotone sequence = t. Then since 
/(i, 
there is an m such that 
I e(|, 0|<| , for any w>m. (2 
But £ lying in $D, it lies in © and ©, n . 
Thus |*0)-*(f)|<§. 
I/O, <„)-/«, f„)l<§> 
for any a; in Vs(£). Hence 
| e(>, t n ) - e(£, t n ) | » * in ( 3 
Now 
e(®, O = £ n ) - e(£, £„) + e(£, £„). 
Hence from 2), 3), 
| e(>, £ n ) | <e , for any x in Vs(g). 
Thus associated with the point f, there is a cubeT lying in 
having f as center. As D = £ — 3) is a null set, each of its points 
can be enclosed within cubes (7, such that the resulting enclosure