1 2 1 1
convention F = F = F (which may or may not equal F = VF or
F2 = VF2)

 t 1 2
 We then set Ft = F F and verify that this family satisfies
 st s
the usual conditions.
 1 2
 1) Each Fst contains the P-negligible sets, since Fs, F contain

 them for all s,t.

 2) If (s,t) S (s',t'), then Ft C F t..
 1 1 2 2
 In fact, F1 C F since sSs', F2 C F, since t5t', hence
 s s t t
 1 2 1 F 2
 Fst F (- F Fl C F OF Fs
 st s t S t st
 3) We have F = ( F ,. (Note: In two parameters,
 st st
 (s,t)<(s',t')
 our usual definition of right limit is for (s,t) S (s',t'),

 (s,t) (s',t'), as we shall see later. The statement we

 shall prove is somewhat stronger.)

 Proof. One containment is evident: since FstC Fs t, for

 each (s',t') > (s,t) by (2), Fs C f Fst
 t (s,t)<(s',t')
 For the other containment, let

 A c n F = (F nFi).
 (s,t)<(s',t') s'>s
 t'>t

 Then A E F for all s'>s, hence A E Sn F = F
 SS '>s

 Similarly, A c F2 for all t'>t, hence A E } F2 = F2
 t <t
 Then A E (F1 (F) =2 Ft, hence F ( ( F.).
 s t stt st s, s t
 s'>s
 t'>t

 Putting the two containments together gives the equality.