6.3 The Selection Principle 91If the setIk−^1 (ω) is defined, find the smallest 1≤jk≤Nksuch that(fn(ω))n∈Ik− (^1) (ω)lies infinitely often inAkjk, and defineIk(ω)={n∈Ik−^1 (ω)|
fn(ω)∈Akjk}. Definingτk(ω)tobethek’th element ofIk(ω) it is straightfor-
ward to check thatτkis a well-definedN-valued measurable function. Clearly
(fτk(ω)(ω))∞k=1converges for eachω∈Ω.
Taking measurable subsequences of (fn)∞n=1inL^0 (Ω,F,P;K)worksjust
in the same way as taking subsequences of sequences (xn)∞n=1in the compact
spaceK. For example, consider the usual procedure of taking a subsequence
of a subsequence: in the present framework this means that we are given two
measurably parameterised subsequences (τk)∞k=1and (σj)∞j=1. Given (fn)∞n=1
we may extract the subsequencegk=fτkand the subsequencehj=gσj=
fτσj. Similarly, we may take diagonal subsequences etc., just as we are used
to do in analysis.
We often shall use the reasoning“by passing to a measurably parameterised
subsequence we may assume that(fn)∞n=1satisfies ...”which has, as usual, the
interpretation:there exists a measurably parameterised subsequence(gk)∞k=1=
(fτk)∞k=1which has the properties...
For later use we give some more properties which one may impose on a
suitably chosen measurably parameterised subsequence. We place ourselves
into the setting of Proposition 6.3.3.
Proposition 6.3.4.Under the assumptions of Proposition 6.3.3 we have in
addition:
(i) Letx 0 ∈Kand define
B={ω|x 0 is an accumulation point of(fn(ω))∞n=1}.
Then the sequence(τk)∞k=1in Proposition 6.3.3 may be chosen such that
lim
k→∞
fτk(ω)(ω)=x 0 ,foreachω∈B.
(ii)Letf 0 ∈L^0 (Ω,F,P;K)and define
C={ω|f 0 (ω)is not the limit of(fn(ω))∞n=1},
where the above means that either the limit does not exist or, if it exists,
it is different fromf 0 (ω). Then the sequence(τk)∞k=1in Proposition 6.3.3
may be chosen such that
lim
k→∞
fτk(ω)(ω)=f 0 (ω),foreachω∈C.
Proof.For (i) it suffices to choose the coverings (Anj)Nj=1n in the proof of Propo-
sition 6.3.3 such thatx 0 ∈An 1 ,foreachn≥1.