In (3), some of us proved that Brownian paths in n-space have double points with probability 1 if rz = 2 or 3; but, for n > 4, there are no double points with probability 1. The question naturally arises as to whether or not Brownian paths in m-space (n = 2 or 3) have triple points. The case of paths in the plane is settled by (a), where it is shown that, with probability 1, Brownian paths in the plane have points of multiplicity k (k = 2,3,4, . . .). The purpose of the present paper is to settle the remaining case, n = 3. We prove that, with probability 1, Brownian paths in 3-dimensional space have no triple points. The general idea behind our proof is to show that there are not too many double points. That is, we show that the set of double points has sigma-finite linear measure, and therefore zero capacity, with probability 1.1. DeJinitions and preliminary results. Let (a, &,,u) be a probability space, i.e. I2 = (u> is a set of elements w, 8 = (E} is a Bore1 field of subsets of Q called events, and ,u is a countably additive measure defined on d and satisfying p(Q) = 1. ,,@E) is called the probability of the event E.A one-dimensional Brownian motion (see (I), (2), (7)) is a real-valued function x(&w) of the two variables t and W, defined for all non-negative real numbers t, 0 < t < co, and for all w E 8, which has the following properties: (a) x(0, w) = 0; (b) for any real numbers s, t with 0 6 s < t < co, the increment (x(t, w)-~(8, o)> is b-measurable in o and has a normal distribution with mean 0 and variance t -8, that is, if E(x, s, t, a) = (w: x(t, w) -x(s, w) denotes th e set of o having the properties following the colon, then E(x, s, t, a) is measurable and--m for every real number a; (c) for any real numbers si, ti (; = 1,2, . . ., m) with 0 < $1 < t, < s2 < t, < . . . < 5, < t, < Go, the increments (x(t,, w) -z(si, VU)>, i = 1,2, . . ., m, are independent in the sense of probability theory, i.e.for any real ~ii, i = 1,2, . . .,m.
