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Math Help - Weak convergence of measures

  1. #1
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    Weak convergence of measures

    Let (X,d) a metric space, \mathcal{B}(X) the algebra of its Borel sets, (\mathbb{P}_n)_n a sequence of probability measures on (X,\mathcal{B}(X)) and C_b(X) the space of bounded continuous functions on X.
    We say the sequence \mathbb{P}_n converges weakly (e.g. in distribution) to the probability measure \mathbb{P} if, for each f\in C_b(X):
    \lim_{n\rightarrow\infty}\int_Xf(x)d\mathbb{P}_n(x  )=\int_Xf(x)d\mathbb{P}(x)

    Now suppose that, given a sequence of measures (\mathbb{P}_n)_n on (\mathbb{R},\mathcal{B}(\mathbb{R})), we have:
    \lim_{n\rightarrow\infty}\int_{\mathbb{R}}f(x)d\ma  thbb{P}_n(x)=\int_{\mathbb{R}}f(x)d\mathbb{P}(x) for every infinitely differentiable function f with compact support.
    How to you prove this implies that (\mathbb{P}_n)_n converges weakly to \mathbb{P}?

    The Weierstrass theorem only applies to continuous functions on a compact, so it can't be used here. I wouldn't be surprised if this is a classic, but I would greatly appreciate some help.
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  2. #2
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    Quote Originally Posted by akbar View Post
    The Weierstrass theorem only applies to continuous functions on a compact, so it can't be used here.
    Yes it can. Let f\in\mathcal{C}_b(\mathcal{R}).

    Let \varepsilon>0.

    There exists a segment K=[a,b] such that \mu(K^c)\leq\varepsilon. Choose \theta to be a [0,1]-valued \mathcal{C}^\infty function that equals 1 on K and 0 on (K')^c where K'=K+[-1,1]=[a-1,b+1] (for instance). Then, since \theta\in\mathcal{C}^\infty_c, using the assumption there exists n_0 such that, for n\geq n_0, \int (1-\theta) d\mu_n\leq \varepsilon+\int (1-\theta) d\mu\leq \varepsilon+\mu(K^c)\leq 2\varepsilon.

    Now write f=f\theta+f(1-\theta). We have \left|\int f(1-\theta) d\mu_n\right|\leq 2 \|f\|_\infty \varepsilon. Same with respect to \mu, hence \left|\int f(1-\theta)d\mu_n-\int f(1-\theta)d\mu\right|\leq 4\|f\|_\infty \varepsilon.
    And for the first term, Stone-Weierstrass provides a polynomial P such that |f(x)-P(x)|\leq \varepsilon for x\in K', hence |f(x)\theta(x)-P(x)\theta(x)|\leq \varepsilon, and \psi=P\theta is in \mathcal{C}^\infty_c. Hence \left|\int f\theta d\mu_n - \int \psi d\mu_n\right|\leq \varepsilon, and we also have \left|\int \psi d\mu_n - \int \psi d\mu\right|<\varepsilon for n\geq n_1(\geq n_0) using the assumption. Finally, \left|\int \psi d\mu -\int f\theta d\mu\right|\leq \|\psi-f\theta\|_\infty\leq \varepsilon like with \mu_n.

    Glue together the above pieces, and you get what you want for n\geq n_1: \left|\int f d\mu_n-\int f d\mu\right|\leq C\varepsilon
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  3. #3
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    Finally made some sense out of it. The trick with f=f\theta+f(1-\theta) was hard to improvise...
    Thanks again.
    Last edited by akbar; December 13th 2009 at 02:26 PM.
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