Convolution powers tend to zero Let $\mu$ be a probability measure  on $[0,\infty)$ that is not degenerated ($\mu(0) &lt; 1$) 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.
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
$$\lim_{n\rightarrow\infty} \mu^{*n}(t) = 0$$
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
$$\lim_{n\rightarrow\infty} \mu^{*n}(t) = 0$$