Outline of example given and explain the following:

= = The assumption of completeness is essential for obtaining point mappings as in the theorems of this section. This is shown by the following example. There is a set A C[0, 1] with m* A = 1 and m*B = 1, where B = [0, 1] ~ A (see Halmos [5], p. 70). Let WR + R be defined by y(x) = -x. For each set ECR, let h(E) be its measurable hull (see Section 12.6). Then h(E) is determined modulo a set of measure zero and so h(E) can be taken to be an element of B/Mo, where is the class of Borel subsets of [-1, 1], and M. those of measure zero. Let X = A UV[B], and take Q to be the o-algebra of subsets of X of the form E = X NF, where F is a Borel set, and take N to be the subset of a consisting of sets of measure zero. Then h(X n F) is in the equivalence class of F, and so h is a o-isomorphism of A/N onto B/M, . If we define O by O(E)= X ny[h(E)], then O is a o-isomorphism of A/N onto itself. Let be the map of the Borel subsets of [-1, 1] onto A/N given by (F) = [F N X]. Then any map 0: X[-1, 1), which induces 0. must be equal to y a.e. But X and V[X] are disjoint except for 0. From this it follows that there is no point mapping 4: X X which induces Q. In this example we see that the associated point mapping ought to bey, but y does not map X into X, since X has "too many gaps. Completion of X under any suitable metric should fill in the gaps so that y can map points of X to the points added by completion. 99 a. Show that for each Borel subset F of [-1, 1] we have h(F n X) = [F], where [F] is the class of Borel sets that differ from F by a set of measure zero. Sec. 7] The Isometries of LP 415 0 b. Show that is a 6-isomorphism of A/N onto itself. c. Show that if 6: X[-1, 1] induces 0 , then 0 is equal to y a.e. d. Show that there is no point mapping 0: X - X that induces