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Proceedings of the Royal Society of Edinburgh. [Sess. 
first factor cos x, and it applies in each subsequent interval of the same 
length, omitting in succession 2,3,.... factors of the infinite product. 
This shows that the product is a continuous function of x for all positive 
finite values and lies between —1 and +1. But the expression is un- 
altered by changing x into — x, so that the same is true for all negative 
finite values, which proves the required result. 
§ 8. The use of the test in theoretical investigations is exemplified by 
the following proof of the continuity of an improper integral defined in 
the mode of De la Vallee-Poussin. 
For brevity, it will be assumed that the function f(x) to be integrated 
has a finite lower bound. The argument would require a repeated applica- 
tion if the lower as well as the upper bound were infinite. 
Let , M 2 , . . . . be any set of constantly increasing positive constants 
having infinity as limit, and let the function f n (x) differ only from f(x) at 
the points where f(x) > M n , and at these points f n (x ) = M n . 
Assuming the functions f n (x ) to be integrable, and denoting J* f n (x)dx by 
F n (cc), De la Vallee-Poussin defines the improper integral F(af) of f(x) to 
be the limit of F n (&), provided this latter limit exist and is finite and 
independent of the particular sequence M 1 , M 2 , . . . . 
Now we know that the proper integral ¥ n {x) is a continuous function of 
x, and hence F 1 (a?), F 2 (x), .... form a monotone increasing sequence of 
continuous functions, having F(^) as limit, since the functions f ± (x), f 2 (x), 
. . . . form a monotone increasing sequence having /( x) as limit. 
Denote by F (b) the integral from a to b of f(x), and let 
G n (x) = F (b)- \lf n (x)dx . 
Then it is plain that the functions G 1 (x), G 2 (x), .... form a monotone 
decreasing sequence, having F($) for limit. Hence F(cc) is continuous. 
It will be noticed that the argument depends on F (b) being finite. In 
fact, if we admit integrals with infinite values, the theorem of continuity 
no longer holds. We still have the first part of the argument holding 
good, viz. that F^sc), F 2 (cc), .... form a monotone increasing sequence, 
whence it follows that F(x) is, at a point of discontinuity, lower semi- 
continuous. 
Suppose, for definiteness, that f(x ) is a positive function, so that F(x) 
is a monotone increasing function of x. Then if F(x) is first of all finite, 
and then becomes infinite, the only way in which this can happen is 
that up to and including some value x = c, F(ce) is finite, and then 
suddenly becomes infinite. Moreover, this is the only kind of discontinuity 
possible, since F(a?) cannot jump up to a finite value. 
