Colloquium publications.

62

THE CAMBRIDGE COLLOQUIUM.

It is necessary to show that this limit is independent of the choice
of the sequence of functions f, which approach (p.
Let the functions fn and gn form two such sequences, and
consider with them the sequences f, = f,, - 1/n and gn = g
- 1/n. We have, since fn - fn+i and gn gn+l, the inequalities
fn < fn+l, gn < gn+b, and also, as follows by an obvious calculation:
lim T[fn] = lim T[fn], lim T[gn] = lim T[gn].
n= oo n-=oo n= oo n=oo
Given fm we can take n great enough so that gn > fm. In fact,
since fm and gn are continuous functions of x, the values of x for
which g - fJm form a closed set En, and En, is included in E,
if n' > n. Hence if we cannot find n great enough so that there
are no points in En, there will be a point x0 such that*
lim gn (x): fm (xo),
n=oo
which is a contradiction, since fm(Xo) < <,(Xo).
We can then form a sequence of functions fm(x) < gml(x)
< fm2(x) < gm2(X)* *, approaching sp(x) as a limit. If we form
the functional T for this sequence, it will approach a limit which
cannot be different from lim T[fn] or lim T[gn] and will theren7=Coo n==
fore be the functional T[kp] already defined.
The functional T is linear in these functions, i. e.,
T[cl'pl + c2(P2] = ClT[cki] + C2T[(2]
if c1 and c2 are restricted to positive constants. In order to make
T completely distributive we need to define
(21) T[i- 2] = T[?]- T[P2],
a definition whose uniqueness is directly verifiable. We should
however prove also the inequality (7):
b
i T[p -- P2] I| M max I| p1 - 2 21
a
* If we have a sequence of point sets, each contained in the preceding, and
none of them a null set, then there will be a point common to all of them,
provided they are all closed sets.