684 
Trigonometrical series 
[CH. YII 
Let U and L denote the upper, and the lower, limit of f(x + 2z), for all 
values of x in the interval (— tt, tt), and for all values of z in the interval 
(a, /3). Let the interval (L, U) be divided into p portions (c 0 , c/), (c 1? c 2 ) ... 
(c 9 _!, c q ) ... (Cp_i, c p ), where c 0 = L, c p = U, and where c q — c q -1< e, for every 
value of q. Let the function / 2 (x + 2z) be defined as follows:—For those 
values of x + 2z for which c 0 £ f(x + 2z) < c 1} let f x (x + 2z) = c 0 ; for those 
values of x + 2z for which c x ^ fix + 2 z) < c 2 , let /j (x + 2 z) = c x ; and generally, 
let /i (x + 2z) — c q _ 1} when c q _x ^ f(x + 2z) < c q ; when f(x + 2z) = c p , let 
fx(x+ 2z) = c p . For any particular value of x, it may, for example, happen 
that there are no values of 2 such that c 0 = f(x + 2z) < c x ; in that case there 
are no values of x + 2z, with the given value of x, for which (x + 2z) = c 0 . 
We have 
'ß rß 
/( x + 2z) x (z) sin mz dz — {x + 2z) x (z) sin mz dz 
< e {/3-a) x, 
where % is the upper limit of \x( z )\ i n the interval (a,/3); and this holds for 
all values of m, and of x in (— 7r, tt). 
We have also 
I fx{x + 2 z) x (z) sin mz dz= S cJ % (z) sin mz dz, 
■J « q=0 J e q 
where e q is that set of points z, at which c q ^ f(x + 2z) < c q -x; this set e q 
depending upon the value of x. 
In the interval (- 2tt, 2tt) of the variable x, let E q denote that set of 
points at each of which c q =f{x) < c q+1 . Let the set E q be enclosed in a 
finite, or enumerably infinite, set H q of non-overlapping intervals, such that 
m (Hq) — m (E q ) =7]. For any fixed value of x, the set e q consists of that part 
of E q which lies in the interval (x + 2a,x+ 2/3), contained in (— 2tt, 27r). Let 
that part of the set of intervals E q which lies in (x + 2a, x -f 2/3) be denoted 
by F q ; then it can be shewn that m (F q ) — m (e q ) ^ 97. For, if possible, let 
m (F q ) — m (e q ) = 97 + 7, where 7 is a positive number. Let the set e q be 
enclosed in the interiors of non-overlapping intervals of a set L q , all in the 
interval (x + 2a, x + 2/3), such that m (L q ) < m (e q ) + 7; and let H q denote 
that set of intervals which consists of L q together with that part of H q which 
is not in (x + 2a, x + 2/3). We have then 
m (Hq) = m (H q ) + m (L q ) - m (F q ) 
< m (H q ) - 97 
< m (E q ). 
As E q cannot be enclosed in a set of intervals H q , of measure less than m (E q ), 
it is impossible that the positive number 7 can exist; and therefore 
m (F q ) - m (e q ) ^ 97.