AN EXAMPLE IN POTENTIAL THEORY. 531 



intervals of S». We are evidently at liberty to suppose that n is the 

 smallest integer for which x\ has this property. Note that all the 

 parts of ^u(.r) above the intervals of S n are congruent. Hence for 



Ax 5= — , A/jl ^ (Ax) p . That the same inequality holds for all Ax 



with the above value of X\ follows from the congruence of the pieces of 

 n(x), the constancy of n(x) between the intervals of S n , and from the 

 fact that if we call S n ' the set obtained by shifting S n to the right a 

 distance X\, the total length of the intervals of S„ bet wren xi and x- is never 

 greater than the total length of the intervals of S„' between X] and x 2 . To 

 see this, we notice first that S„ and S„' begin at X\ with a common 



interval of length — . There follows a gap of length — between the 



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intervals of S n ', whereas a greater gap, of length g, say, follows for S n , 

 since n was the smallest integer for which .x\ began an interval of S n . 

 The corresponding gap g in S n ' must begin at a distance g from X\, 



i. e. a distance — before the gap g in S n terminates. Because of the 

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symmetry of S n to either side of a gap at least for a distance g, it 



follows that after the gap g in S n ', the two sets have again a common 



interval, and that before this point the accumulated length of intervals 



of Sn has exceeded, or at least equaled that of S n , because of -the 



earlier occurrence of the gap g in S n . The reasoning may now be 



repeated. The next gap in S„ f will be of length — , while the next gap 



in S n will be longer, for otherwise we should have intervals of S n -i 

 and S'n-i coinciding, which is contrary to the hypothesis that n is the 

 smallest integer for which X\ begins an interval of S n . 



Finally, if, instead of beginning an interval of S„, X\ lies in a gap 

 between two of them, the inequality A\x ^ {Ax) v holds a fortiori, since 

 the diminution of Xi will have increased Ax without changing A/i. 

 Because of the continuity of /x(.r), the inequality holds generally. 



The continuity of L(P) now follows without trouble if we separate 

 out from the integral a part corresponding to a small interval con- 

 taining the boundary point B, in whose neighborhood the continuity 

 of L(P) is to be investigated. The above inequality shows that this 

 interval can be made so small that the integral over it may be made 

 uniformly less than a preassigned positive e in a given neighborhood 

 of B. The rest of the integral is continuous at B, and hence so is 



UP). 



We now know that L(P) attains its negative minimum, —m, on S. 



