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Let $\partial G$ be the boundary of the $n$-ball in $\mathbb{R}^n$ and $[a,b]$ an interval in $\mathbb{R}$.

Let $\mu$ be the product measure of the surface measure on $\partial G$ and the Lebesgue measure on $[a,b]$. I would like to show the vague convergence of a measure $\mu_N$ against $\mu$.

Let's say I can show for all $A\subset \partial G$ and $B\subset [a,b]$ the limit $$ \lim_{N\rightarrow \infty}\mu_N(A\times B) = \mu(A\times B). $$

My question is which subsets of $\partial G$ and $[a,b]$ do I actually need to show vague convergence. I know the following for the vague convergence on the interval $[a,b]$:

A Sequence $\{\mu_n,n\geq 1 \}$ of subprobability measures is said to converge vaguely to an subprobability measure $\mu$ if and only there exists a dense subset D of $[a,b]$ such that $$ \forall c \in D, d\in D, c<d \quad \mu_n((c,d))\rightarrow \mu((c,d)). $$

Is there a similar condition for the vague convergence on the boundary $\partial G$ of the $n$-ball in $\mathbb{R}^n$?

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Let $\mu_1$ and $\mu_2$ denote the measures on $\partial G$ and $[a,b]$, respectively, s.t. $\mu=\mu_1\times\mu_2$. The sequence of measures $\mu_N$ converges weekly to $\mu$ iff $\mu_N(A\times B)\to \mu(A\times B)$ for each $\mu_1$-continuity set $A$ and each $\mu_2$-continuity set $B$. (See, e.g., Theorem 2.7 here.)

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  • $\begingroup$ Your answer is correct. Unfortunately I asked the wrong question...Mea culpa! $\endgroup$
    – HyyFly
    Nov 23, 2021 at 13:15
  • $\begingroup$ @HyyFly It is better to ask a new question instead of modifying the old one. Now my answer seems irrelevant! $\endgroup$
    – d.k.o.
    Nov 23, 2021 at 13:48

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