462 
DISCONTINUOUS FUNCTIONS 
ifflPMi; 
We show now that F is discontinuous at each point of For 
let e m be an end point of one of the intervals of D m+l but not of 
ZL. Then 
/iCO = ^ 
' f m(prf) ,1 
m a 
Hence 
fm+p(.&m)— 0 ? P — 1, 2 ••• 
F( e m) — H m = ro + *” 4 9 
m 1 
As the points A are pantactic in 21, there exists a sequence in 
A which = e. For this sequence F = H. Hence 
Disc F= H— II m = II m . 
X=e m 
Similarly, if r\ m is not an end point of the intervals D m+1 , but a 
limiting point of such end points, 
Disc = II m . 
X=Vm 
The function F is jR-integrable in 21 since its points of discon 
tinuity © form a null set. 
467. Let Q£ = je 4 . } be an enumerable set of 'points lying in the 
limited or unlimited set 21, which lies in Si m . For any x in 21 and 
any e i in Gs, let x — e l lie in 23. Let g(x x • •• x m ) be limited in 23 and 
continuous, except at x = 0, where 
Disc g(x) = b. 
Let C= ... converge absolutely. Then 
F(xj — x m ) = 'Zc l g(x- c t ) 
is continuous in A — 21 — (5, and at x= e L , 
Disc = <? t b. 
For when x ranges over 21, x — e L remains in 23, and g is limited 
in 23. Hence F is uniformly and absolutely convergent in 21. 
Now each g(x — e t ) is continuous in A; hence F is continuous 
in A by 147, 2.