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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 $\nu$.

  • 2
    How is the convolution of measures defined?2012-04-04
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    convolution of two measures $\mu*\nu (E) = \int\mu(E-y)d\nu(y)$, where $E$ is a measurable set.2012-04-04

2 Answers 2