10 Proceedings of tlie Royal Iris]t Acadeiiiij. 



TJieorem. — The content of a plane open domain = \ I (x) . dx. 



The projection of the domain on the sj-axis consists of a set of non- 

 overlapping open intervals. T^t ,these intervals, aiTanged in countable order, 

 be Ci, Cj, . . . &c. 



Let A be the content of the domain. By drawing parallels to the axes 

 we can find a rectilinear figure R inscribed in D and of area > A - (, 

 where e is any assigned positive number. 



The projection of B consists of a finite number of intervals contained in 

 the c intervals. Choose n so that Ci, c., . . . c„ contain the projection of R. 

 The part of the given domain associated with the remaining intervals 

 c,i+i, c„+2, . . . &c., has content < f. 



Let 6',. be the content of the part associated witli c,- Divide c,. into 

 elementary intervals 9,, di, . . . 9„„ and let /,. be the minimum of / (.-'.) in o^. 

 Then C, > 2 (/, - e,) S,. 



by Lemma 2 for all positive values of ei 



Proceeding to the limit we have 



C, > f I{x) . dx. 



On the other hand, /,. is greater than the ordinal section of 2?, and therefore 



S Irdr > Cr - i, :. I I{cc) dx $ Cr- 



J Cr 



Therefore, finally 6',. = I{x) dx. 



]Cr 



.'. A equals the lower integral of I(:c) taken over the set of open 

 intervals Ci, Cj, . . . &c. 



Cm: — There is no difficulty in modifying the above proof so as to show 

 that the lower integral may be taken over the whole interval A£. 



13, I do not find that the usual inequalities can be asserted for the double 

 integral in an open domain of the second kind. However, by an adaptation 

 of Jordan's method I have obtained the following results : — 



^^,t.dc$^dy^f.dx, 

 and ~ 



fde < \ dy \f.dr. 



