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Let $\mu$ be a probability measure on $[0,\infty)$ that is not degenerated ($\mu(0) < 1$) and $f$ be a bounded function on $[0,\infty)$. Show that pointwise $f * \mu^{*n} \rightarrow 0$ where $*$ denotes convolution $(f * \mu)(t) = \int_0^t f(t-u) d \mu(u)$ and $\mu^{*n} = \mu * \dots * \mu\quad n\text{ times}$

Edit: Added that $\mu$ is not degenerated. Edit2: Clarified that $f$ is defined on the positive reals.

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    oh thanks, I edited the question for clarification.2012-09-04

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Okay, here is my attempt, please correct me if I'm wrong. Let's assume that $\nu(t) := \lim_{k\rightarrow\infty} \sum_{n=0}^k \mu^{*n}(t)$ defines a measure that is finite for bounded sets.

Then for every $t \ge 0$ $\lim_{k\rightarrow\infty} \sum_{n=0}^k (f*\mu^{*n})(t)= (f*\nu)(t)$ is finite, and hence $\lim_{n\rightarrow\infty} (f*\mu^{*n})(t) = 0$ For the special case $f \equiv 1$ this means $\lim_{n\rightarrow\infty} \mu^{*n}(t) = 0$

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    This is wro$n$g: the sentence starting "Let's assume..." simply invents a fact without any justification and then uses it to prove the result.2013-05-23