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Thread: Weak convergence of measures

  1. #1
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    Weak convergence of measures

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

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

    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 $\displaystyle f\in\mathcal{C}_b(\mathcal{R})$.

    Let $\displaystyle \varepsilon>0$.

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

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

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