An Invitation to Random Schr¨odinger operators - FernUniversität in ...

probability one. We do not know whether this intersection has full measure, in fact
we even don’t know whether this set is measurable.
THEOREM 5.5. The measures ν L converge vaguely to the measure ν P-almost
surely, i.e. there is a set Ω 0 of probability one, such that
∫
∫
ϕ(λ) dν L (λ) → ϕ(λ) dν(λ) (5.10)
for all ϕ ∈ C 0 (R) and all ω ∈ Ω 0 .
REMARK 5.6. The measure ν is non random by definition.
PROOF: Take a countable dense set D 0 in C 0 (R) in the uniform topology. With
Ω ϕ being the set of full measure for which (5.10) holds, we set
Ω 0 = ⋂
Ω ϕ .
ϕ∈D 0
Since Ω 0 is a countable intersection of sets of full measure, Ω 0 has probability one.
For ω ∈ Ω 0 the convergence (5.10) holds for all ϕ ∈ D 0 .
By assumption on D 0 , if ϕ ∈ C 0 (R) there is a sequence ϕ n ∈ D 0 with ϕ n → ϕ
uniformly. It follows
∫
|
∫
≤ |
∫
+ |
∫
+ |
∫
ϕ(λ) dν(λ) −
∫
ϕ(λ) dν L (λ)|
ϕ(λ) dν(λ) − ϕ n (λ) dν(λ)|
∫
ϕ n (λ) dν(λ) − ϕ n (λ) dν L (λ)|
∫
ϕ n (λ) dν L (λ) − ϕ(λ) dν L (λ)|
31
≤
|| ϕ − ϕ n || ∞ · ν(R) + || ϕ − ϕ n || ∞ · ν L (R)
∫
∫
+ | ϕ n (λ) dν(λ) − ϕ n (λ) dν L (λ)| . (5.11)
Since both ν(R) and ν L (R) are bounded by 1 (in fact are equal to one) the first two
terms can be made small by taking n large enough. We make the third term small
by taking L large.
□
REMARKS 5.7.
(1) As we remarked already in the above proof both ν L and ν are probability
measures. Consequently, the measures ν L converge even weakly to ν, i. e.
when integrated against a bounded continuous function (see e. g. [12]).
Observe that the space of bounded continuous functions C b (R) does not
contain a countable dense set, so the above proof does not work for C b
directly.