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 ?



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.

  • $\begingroup$ thanks for the quick reply, if $\endgroup$
    – jed
    May 3 '13 at 16:00
  • $\begingroup$ thanks for the quick reply, if take $\delta_{\frac{1}{n}}$ then I get the weak convergence since $\int f(x)\, d\delta_{\frac{1}{n}} \stackrel{n\rightarrow\infty}{\to} \int f(x)\, \delta_{0}(dx)$ right ? $\endgroup$
    – jed
    May 3 '13 at 16:11
  • $\begingroup$ Yes.${}{}{}{}{}$ $\endgroup$ May 3 '13 at 16:14
  • $\begingroup$ Let $\mu_n = N(0, \frac{1}{n})$ then I have $\frac{\sqrt{n}}{\sqrt{2\pi}} \int f(x)\,e^{-\frac{x²n}{2}} dx$ as $n \rightarrow\infty$ the integral goes to $0$. Does this proof that it does not converge weakly ? $\endgroup$
    – jed
    May 3 '13 at 17:44
  • 1
    $\begingroup$ thanks, let $ \mu_n=N(0,\frac{1}{n})$ then and substitute $x=\frac{t}{\sqrt{n}}$ then we have $\lim\limits_{n\rightarrow \infty} \int\limits_{\mathbb{R}} f_c(\frac{t}{\sqrt{n}}) e^{-\frac{t²}{2}}\, dt = \int\limits_{\mathbb{R}} f_c(0)\, e^{-\frac{t²}{2}}\, dt = f_c(0) = \int\limits_{\mathbb{R}} f_c d\delta_0$ hence it converges weakly $\endgroup$
    – jed
    May 5 '13 at 14:19

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