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.${}{}{}{}{}$ $\endgroup$ – Davide Giraudo 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.