Prove Theorem 25.6. Hint: Put m
¯ n(x) = ‑ An(x) m(z)µ(dz) µ(An(x)) ,
then, by Lemma 24.10,
m¯ n(x) → m(x) mod µ.
Moreover,
mn(x) = n i=1 YiI{Xi∈An(x)} nµ(An(x)) n i=1 I{Xi∈An(x)} nµ(An(x))
therefore, it is enough to show that
n i=1 YiI{Xi∈An(x)} nµ(An(x)) − m¯ n(x) → 0 a.s. mod µ
By Bernstein’s inequality
P n i=1 YiI{Xi∈An(x)} nµ(An(x)) − m¯ n(x) >
≤ 2e −n2 µ(An(x))2 2 Var(Y1I{X1∈An(x)})+4Lµ(An(x))/3
≤ 2e −n2 µ(An(x))2 2L2µ(An(x))+4Lµ(An(x))/3
≤ 2e −n2 µ(An(x)) λ(An(x)) λ(An(x)) 2L2+4L/3 ,
which is summable because of the condition and because of
lim inf n→∞ µ(An(x)) λ(An(x)) > 0 mod µ
(cf. Lemma 24.10).