Weak and vague convergence of restricted Lebesgue measure Let $\lambda$ be the Lebesgue measure on $\mathbb{R}$. For $n \in \mathbb{N}$ let $\mu_n=\lambda_{|[-n,n]}$ be the restriction of the Lebesgue measure on the set $[-n,n]$.
How can I prove that

*

*$\lambda$ is the vague limit of $\mu_n$

*$(\mu_n)_{n \in \mathbb{N}}$ does not converge weakly

I am trying with the indicator function on a compact and on an open set by applying the definitions of vague and weak convergence but with no results.
(This exercise could be found on Probability Theory by Achim Klenke (3rd version) on Chapter 13)
 A: I believe this can be solved as follows, for the first case this follows directly from the definition:
Take any $f \in C_c(\mathbf{R})$, then observe that $f$ attains a maximum on it's support, which is compact,  hence one is allowed to use DCT and thus
$$
lim_{n \to \infty} \int_{\mathbf{R}} f(x) d\mu_n(x) = \lim_{n \to \infty} \int_{\mathbf{R}} f(x) \mathbf{1}_{\text{supp}(f) \cap [-n,n]} d\lambda(x) \stackrel{DCT}{=} \int_{\mathbf{R}} f(x) d\lambda(x).
$$
Using the definition of vague convergence we are done. Secondly we want to prove that $(\mu_n)_{n \in \mathbf{N}}$ is not converging in the weak sense. Now take a function $f \in C_b(\mathbf{R})$ such that
$$
\int^{n+1}_{n} f(x) d\lambda(x) = 1/n, \quad \text{ and } \quad f(x) = 0 \text{ if } x < 0
$$
if you don't have intuition about this draw a picture, either for rescaled concatenated gaussian densities (sucht that you get 1/n) or piecewise linear functions and note that I just take it zero on the negative real axis for simplicity. Then we directly can compute
$$
\int_{\mathbf{R}} f(x) d\mu_n(x) = \sum^{n}_{k=1} 1/k,
$$
but this integral is divergent hence there cannot be any weak limit. This completes the proof. Do you have questions?
