Proof.
Firstly, it is not difficult to check
(2.2){k(ξ^kp-ξp)≤σx}={Fk(ξp+σxk)≥p}=:{1k∑i=1kYi,k≤ωk},
where
(2.3)Yi,k=E1{Xi≤ξp+σx/k}-1{Xi≤ξp+σx/k},ωk=k(E1{Xi≤ξp+σx/k}-p).
From the Taylor's formula, it follows
(2.4)E1{Xi≤ξp+σx/k}=F(ξp+σxk)=F(ξp)+F′(ξp)σxk+o(1k)=p+f(ξp)σxk+o(1k),
which implies
(2.5)ωk=f(ξp)σx+o(1).
By the Lindeberg's central limit theorem, we can get
(2.6)1f(ξp)σk∑i=1kYi,k⟶dN(0,1), as k⟶∞.
Hence, (2.1) is equivalent to
(2.7)limn→∞1logn∑k=1n1k1{(1/f(ξp)σk)∑i=1kYi,k≤x+o(1)}=Φ(x), a.s.

Throughout the following proof, C denotes a positive constant, which may take different values whenever it appears in different expressions.

Put that
(2.8)Zi,k:=1f(ξp)σYi,k.
Let g be a bounded Lipschitz function bounded by C, then from (2.6), we have
(2.9)Eg(1k∑i=1kZi,k)⟶Eg(N), as k⟶∞,
where N denotes the standard normal random variable. Next, we should notice that (2.7) is equivalent to
(2.10)limn→∞1logn∑k=1n1kg(1k∑i=1kZi,k)=Eg(N) a.s.
from Section 2 of Peligrad and Shao [8] and Theorem 7.1 of Billingsley [9]. Hence, to prove (2.7), it suffices to show that as n→∞,
(2.11)Rn=1logn∑k=1n1k[g(1k∑i=1kZi,k)-Eg(1k∑i=1kZi,k)]=:1logn∑k=1n1kTk⟶0, a.s.
It is obvious that
(2.12)ERn2=1log2n[∑k=1n1k2ETk2+2∑k=1n-1 ∑j=k+1n1kjETkTj].
Since g is bounded, we have
(2.13)1log2n∑k=1n1k2ETk2≤Clogn.
Furthermore, for 1≤k<j≤n, we have
(2.14)|ETkTj|=|Cov(g(1k∑i=1kZi,k),g(1j∑i=1jZi,j))|=|Cov(g(∑i=1kZi,kk),g(∑i=1jZi,jj)-g(∑i=k+1jZi,jj))|≤CjE|∑i=1kZi,j|≤Ckj(EZ1,j2)1/2,
where
(2.15)EZ1,j2=1+O(1j).
Therefore, we have
(2.16)1log2n∑k=1n-1∑j=k+1n1kj|ETkTj|≤Clog2n∑k=1n-1 ∑j=k+1n1k1/2j3/2(EZ1,j2)1/2=Clog2n∑j=2n ∑k=1j-11k1/2j3/2(EZ1,j2)1/2≤Clogn.
From the above discussions, it follows that
(2.17)ERn2≤Clogn.
Take nk=ekτ, where τ>1. Then by Borel-Cantelli lemma, we have
(2.18)Rnk⟶0, a.s. as k⟶∞.
Since g is bounded function, then for nk<n≤nk+1, we obtain
(2.19)|Rn|≤1lognk|∑l=1nk1l[g(1l∑i=1lZi,l)-Eg(1l∑i=1lZi,l)]|+1lognk∑l=nk+1nk+11l|g(1l∑i=1lZi,l)-Eg(1l∑i=1lZi,l)|≤|Rnk|+Clognk∑l=nk+1nk+11l⟶0, a.s., as n⟶∞,
where we used the fact
(2.20)lognk+1lognk=(k+1)τkτ⟶1, as k⟶∞.
So, the proof of the theorem is completed.