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We say that a family of measures $\mu_{t}\to \mu$ weakly if for any $g\in C_{0}$, $\int g d\mu_{t} \to \int g d\mu$. Show that if $\mu_{t}\to \mu$ weakly, then $\nu*\mu_{t}\to \nu*\mu$ weakly, where $\nu*\mu$ denotes the convolution of the measures $\mu$ and $\nu$.

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How is the convolution of measures defined? –  Dirk Apr 4 '12 at 11:00
    
convolution of two measures $\mu*\nu (E) = \int\mu(E-y)d\nu(y)$, where $E$ is a measurable set. –  user24367 Apr 4 '12 at 11:28

2 Answers 2

If the measures are finite, then for $g\in C_0$ $$\int g(x)\,(\nu*\mu_t)(dx)=\int\int g(x+y)\,\mu_t(dy)\,\nu(dx).$$ For each $x$, the weak convergence $\mu_t\to \mu$ gives
$$\int g(x+y)\,\mu_t(dy)\to \int g(x+y)\,\mu(dy).$$ Then by the dominated convergence theorem,
$$\int\int g(x+y)\,\mu_t(dy)\,\nu(dx)\to\int\int g(x+y)\,\mu(dy)\,\nu(dx).$$ By definition, this is the weak convergence $\nu*\mu_t\to \nu*\mu$.

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exactly, but what if the measures involved are not finite. –  user24367 Apr 4 '12 at 17:27
    
I'm not sure. ${}$ –  Byron Schmuland Apr 4 '12 at 17:34
    
I think the proof goes the same when the measure involved is not finite. Maybe one need to exclude a few cases. –  Bombyx mori Jan 13 '13 at 8:07

For a counterexample, consider the shifted $\delta$-masses on $\mathbb{R}$: $$ \int f(x) \ \delta_t(x) = f(t).$$ It's clear that $\delta_{t} \to 0$ vaguely as $t \to \infty$, while $\delta_t * dx = dx$ as measures by the translation invariance of Lebesgue measure.

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The question asks for $\mu_t \to \mu$ weakly, not vaguely. Weakly typically includes the assumption that the limiting measure is a probability measure. –  Nate Eldredge Jan 13 '13 at 7:25

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