Reason: Proof of mean-square completeness via a geometrically spaced subsequence, Markov and Borel-Cantelli, an everywhere-defined G-measurable limit on the G-measurable Cauchy set, and Fatou. Approved by Aaron.
Step 1 (a subsequence with geometrically small increments). Using the Cauchy hypothesis with ε=2−j, choose recursively indices n1<n2<… such that ∥Xn−Xm∥2<2−j for all n,m≥nj. Set Dj=Xnj+1−Xnj; then Dj is G-measurable and square-integrable with ∥Dj∥2<2−j, and more generally ∥Xni−Xnj∥2<2−j for all i≥j.
Step 2 (almost sure absolute convergence along the subsequence). Let Aj={∣Dj∣≥(3/4)j}={Dj2≥(9/16)j} (the two events coincide because squaring preserves the order of nonnegative reals). By Markov's inequality applied to the nonnegative random variable Dj2,
P(Aj)≤(916)jE[Dj2]≤(916)j(41)j=(94)j.
The series ∑j(4/9)j converges: by the telescoping identity (1−r)(r+r2+⋯+rm)=r−rm+1 with r=4/9, its partial sums are nondecreasing and bounded by r/(1−r)=4/5, hence converge to their least upper bound, which exists by the least upper bound property (for any ε>0 some partial sum exceeds the supremum minus ε, and all later partial sums lie between it and the supremum, which is the definition of the limit). By comparison (partial sums of ∑jP(Aj) are nondecreasing and bounded by those of ∑j(4/9)j, hence by 4/5), ∑jP(Aj) converges. By the first Borel-Cantelli lemma, the event A=⋂J∈N⋃j≥JAj (occurrence of infinitely many Aj) has P(A)=0.
Fix ω∈/A. There is J0 with ∣Dj(ω)∣<(3/4)j for all j≥J0. For J0≤j<j′, the telescoping bound gives
Step 3 (a G-measurable limit defined everywhere). Let
Ω∗={ω∈Ω:(Xnj(ω))j∈Nis a Cauchy sequence}=k∈N⋂J∈N⋃j,j′≥J⋂{Xnj−Xnj′≤k1},
a set built from the G-measurable random variables Xnj−Xnj′ by countable intersections and unions, hence Ω∗∈G. By Step 2, Ω∖Ω∗⊆A, so P(Ω∗)=1 by monotonicity and complementation. Define
X(ω)=j→∞limXnj(ω)(ω∈Ω∗),X(ω)=0(ω∈/Ω∗),
the limit existing on Ω∗ by Every Cauchy Sequence of Real Numbers Converges. Put Zj=Xnj1Ω∗, where 1Ω∗ is 1 on Ω∗ and 0 off it; 1Ω∗ is G-measurable because Ω∗∈G, so each Zj is G-measurable, and Zj(ω)→X(ω) for everyω∈Ω (on Ω∗ this is the definition of X; off Ω∗ both sides are 0). For every real a,
{X>a}=k∈N⋃J∈N⋃j≥J⋂{Zj>a+k1}.
Indeed, if X(ω)>a, choose k with X(ω)>a+2/k; convergence gives J with Zj(ω)>a+1/k for all j≥J. Conversely, if Zj(ω)>a+1/k for all j≥J, then the limit satisfies X(ω)≥a+1/k>a by the order properties of limits. Hence {X>a}∈G for every a, and by the generator criterion of Measurable Function and Real-Valued Measurable Function, X is G-measurable; in particular X is a random variable, as G⊆F.
Step 4 (square-integrability and mean-square convergence). A nonnegative random variable W with W=0 on the complement of an event of probability 0 has E[W]=0: every nonnegative simple functions≤W has each of its nonzero values attained on a subset of that null event, so its integral is 0 by monotonicity of the measureP, and the integral of W is the supremum of these. We refer to this as the null-support principle.
Fix j. For i≥j consider the nonnegative random variables fi=(Xnj−Xni)21Ω∗. For every ω, fi(ω)→(Xnj(ω)−X(ω))21Ω∗(ω) as i→∞ (on Ω∗ by the algebra of limits applied to the square of a convergent sequence; off Ω∗ both sides are 0). By Fatou's lemma and monotonicity (fi≤(Xnj−Xni)2 pointwise), together with Step 1,
The function (Xnj−X)21Ω∖Ω∗ is nonnegative and vanishes on Ω∗, an event of probability 1, so its expectation is 0 by the null-support principle. Adding the two pieces (pointwise (Xnj−X)2=(Xnj−X)21Ω∗+(Xnj−X)21Ω∖Ω∗, and expectation is additive on nonnegative random variables by Linearity and Monotonicity of the Lebesgue Integral),
Finally, let ε>0. Choose N with ∥Xn−Xm∥2<ε/2 for n,m≥N, and choose j with nj≥N and 2−j<ε/2 (possible since 2−j→0 as in Step 2). For every n≥N, the triangle inequality of Cauchy-Schwarz and Triangle Inequalities for the Mean-Square Norm applied to Xn−X=(Xn−Xnj)+(Xnj−X) gives