226 DR. W. H. YOUNG ON THE GENERAL THEORY OF INTEGRATION. 



longer hold, that is to say, it is no longer true that definite limits, independent of the 

 mode of division, &c., exist, when sets of points are substituted for intervals : 



Example 2. Let y he a function of x, which is zero everywhere, in the segment 

 (0, 1 ), except at every point of a perfect set G, nowhere dense, whose content is I, 

 and at every point of G let y = 1. 



If we divide the segment (0, 1) into n equal parts, each of length less than d, the 

 upper summation, that is, the summation which corresponds to the upper integral, is 

 always greater than I, but approaches the value I as d is indefinitely decreased. If 

 I is zero (in which case the function is integrable in RIEMANN'S sense), the limit is 

 always zero, and in any case it is less than 1. We can, however, arrange so that we 

 get 1 as limit. 



Let the semi-external points of the intervals be arranged in countable order 

 PI, P 2 , . . . ; take the same division as before, abstracting, however, the points 

 PI, P 2 , . . . P., and then adding these points singly to the single parts ; thus we have 

 a division into n measurable sets, each of content less than d, in each of which the 

 function has the value unity, so that the summation always has the value unity 

 however small d may be ; therefore we get 1 as limit. 



5. If, on the other hand, we retain the intervals, but drop the restriction that 

 they should be finite in number, there remains over a complementary set of points G, 

 of content, say, I. The following example shows that the upper summations over a 

 set of intervals have not then in general a definite limit, so that this extension of the 

 definition of upper integral cannot be made without some restriction : 



Example 3. Take a perfect set, nowhere dense, of content I in the segment (0, 1), 

 and let the function be zero everywhere except at the points of this set where the 

 value of the function is unity. 



The value of the upper integral as defined by DAKBOUX is I. 



Now let d be any assigned norm, and let an odd integer m be determined so that 



Divide the segment (0, 1) into m equal parts, and blacken the middle part. Then 

 divide each of the (m 1) remaining equal parts of the segment into m 2 equal parts, 

 and blacken each of the middle parts, and so on, in the usual manner. The set of 

 intervals, each less than d, which we thus obtain, have content 



... 



m m m m m a 



which is less than d. The complementary set of points is a perfect set nowhere 

 dense of content greater than (1 d). The upper summation of the given function 

 over these intervals is then at most equal to d, and is, therefore, less than the upper 

 integral as soon as the norm is less than I, and has the limit zero. 



The principle of this example shows that, if we omit the condition that the number 



