I am trying to understand 'vague/weak convergence' and need to decide whether or not a measure converges vaguely or weakly. Weak convergence implies vague convergences. However, I don't really understand whole thing.
The measure $\delta_n$ converges vaguely but not weakly, but I cannot see how that works if I take the definition of weak/vague convergence
weak convergence : $\int f(x)\, \mu_n(dx) \stackrel{n\rightarrow\infty}{\to} \int f(x)\, \mu(dx)\, f \in C_b$
vague convergence : $\int f(x)\, \mu_n(dx) \stackrel{n\rightarrow\infty}{\to} \int f(x)\, \mu(dx)\, f \in C_0$
If I take the $\delta_n$ from above I have
$\int f(x)\, \delta_n\,(dx) \stackrel{n\rightarrow\infty}{\to} \int f(x)\, \delta_{\infty}(dx)$
How do I take the limit ?
Do I calculate the integral at every n while $n\rightarrow \infty$, say if $f(x) \in C_b$ (e.g. $f \equiv 1)$ I get always $1$ but at some $n$ I get $0$ since $f(x)$ is bounded and, hence, it does not converge weakly ?
How does the vague case look like ?
jed
Best Answer
Notice that $\int fd\delta_n=f(n)$ for all $n$. There are continuous bounded functions on the real line for which the sequence $(f(n),n\geqslant 1)$ does not converge (like $f(x):=\cos(\pi x)$).
However, when $f$ vanishes at infinity, $f(n)\to 0$ by definition, which proves vague convergence to the null measure.