Fall 2020 - Problem 3

construction, weak convergence

Let dμn\diff\mu_n be a sequence of probability measures on [0,1]\br{0, 1} so that

f(x)dμn(x)\int f\p{x} \,\diff\mu_n\p{x}

converges for every continuous function f ⁣:[0,1]R\func{f}{\br{0,1}}{\R}.

  1. Show that

    g(x,y)dμn(x)dμn(y)\iint g\p{x, y} \,\diff\mu_n\p{x} \,\diff\mu_n\p{y}

    converges for every continuous g ⁣:[0,1]2R\func{g}{\br{0,1}^2}{\R}.

  2. Show by example that under the above hypotheses, it is possible that

    {0xy1}dμn(x)dμn(y)\iint_{\set{0 \leq x \leq y \leq 1}} \,\diff\mu_n\p{x} \,\diff\mu_n\p{y}

    does not converge.

Solution.
  1. By Stone-Weierstrass, the set

    A={k=1nfk(x)gk(y)|nN, fk,gkC([0,1])}A = \set{\sum_{k=1}^n f_k\p{x}g_k\p{y} \st n \in \N,\ f_k, g_k \in C\p{\br{0,1}}}

    is dense in C([0,1]2)C\p{\br{0,1}^2}. Hence, by applying weak convergence and Fubini's theorem, we see that μnμn\mu_n \otimes \mu_n converges weakly-* to μμ\mu \otimes \mu.

  2. Let μn=δ0\mu_n = \delta_0 if nn is odd and μn=1nk=1nδk/n2\mu_n = \frac{1}{n} \sum_{k=1}^n \delta_{k/n^2} for nn even. Then for nn even and gg is continuous,

    g(x)dμn(x)=1nk=1ng(kn2)ng(0),\int g\p{x} \,\diff\mu_n\p{x} = \frac{1}{n} \sum_{k=1}^n g\p{\frac{k}{n^2}} \xrightarrow{n\to\infty} g\p{0},

    by continuity, and it's clear that the same holds for the odd terms. However, for nn od, we have

    {0xy1}dμn(x)dμn(y)=0\int_{\set{0 \leq x \leq y \leq 1}} \diff\mu_n\p{x} \,\diff\mu_n\p{y} = 0

    and for nn even,

    {0xy1}dμn(x)dμn(y)=010ydμn(x)dμn(y)=j=1n2[j1n2,jn2)0ydμn(x)dμn(y)=j=1n[j1n2,jn2)jndμn(y)+j=n+1n2[j1n2,jn2)nndμn(y)=j=1njn2=n+12nn12,\begin{aligned} \int_{\set{0 \leq x \leq y \leq 1}} \diff\mu_n\p{x} \,\diff\mu_n\p{y} &= \int_0^1 \int_0^y \diff\mu_n\p{x} \,\diff\mu_n\p{y} \\ &= \sum_{j=1}^{n^2} \int_{\pco{\frac{j-1}{n^2}, \frac{j}{n^2}}} \int_0^y \diff\mu_n\p{x} \,\diff\mu_n\p{y} \\ &= \sum_{j=1}^n \int_{\pco{\frac{j-1}{n^2}, \frac{j}{n^2}}} \frac{j}{n} \,\diff\mu_n\p{y} + \sum_{j=n+1}^{n^2} \int_{\pco{\frac{j-1}{n^2}, \frac{j}{n^2}}} \frac{n}{n} \,\diff\mu_n\p{y} \\ &= \sum_{j=1}^n \frac{j}{n^2} \\ &= \frac{n + 1}{2n} \xrightarrow{n\to\infty} \frac{1}{2}, \end{aligned}

    so the sequence diverges.