Subdividing the Interior of a Rectifiable Curve 381 
for the corresponding /] and Then in Ri we have 7]^ > 0 as 
in the original rectangle R we had ^>0. 
Thus for every integer v there is a rectangle R^ whose longest 
side is y/2^ when 7 is the largest side of R. The integral 
satisfies the inequality. 
[ 1 ] \JA-V. 
This dissection process determines a definite point z ^ ^ which 
lies in every rectangle and is either interior to or oni?. For 
every point 2: of the rectangle Rv one has the inequality 
We set for points 2: interior to or on R 
f{z) =/(r) + (^-r)/' (r) + A(^). 
By introducing a positive number e subject to later determina- 
tion, one has interior to or on R^ 
[2] |A(2 )|ge| 2 -f| Se72~’'V2 
provided that p is taken sufficiently large, say greater than p^. 
We now integrate along the rectangle Rj,. This gives 
Jp = \f \ dz+f'iO r zdz+ r A(z)dz 
L J ty Rj/ tJ Rp «_/ Rp 
which in turn gives 
[3] Jv= i A (2) dz 
C Rp 
in view of the fact that J dz and ^ zdz taken along Rp are seen 
without difficulty to vanish. 
From [2] and [3] we obtain 
\Jp\=^n.^^2ey2-^Xp, 
