Consider a locally compact group $\mathrm{G}$ and a left-invariant Haar measure $\lambda$ on it. Let $\mu$ be a probability measure. Suppose $f$ is a function continuous and bounded. Denote by $\Delta$ the modulus of the group $\mathrm{G}.$ I want to show that the function
$$(f \ast \mu)(x) = \int\limits_\mathrm{G} f(xs^{-1}) \Delta(s^{-1})\ d\mu(s)$$
is (1) defined everywhere, (2) continuous and (3) bounded.
EDIT: I am looking at the integral on a locally compact space, the usual way they handle this is by means of the Daniell integral. So, I guess the measures here are called Radon measures. In particular, they are regular (both inner and outer) and finite on every compact set. Also, if $\mathscr{A}_\mu$ denotes the $\mu$-integrable sets, then $\mathscr{A}_\mu$ contains the topology of $\mathrm{X}$ (remark $\mu$ is a probability measure, so there are no issues of infinite measure here).
In the main text the proof of the same statement is given for the convolution $\mu \ast f$ which is given by $$(\mu \ast f)(x) = \int\limits_\mathrm{G} f(s^{-1} x)\ d\mu(s).$$
I tried to immitate the proof, but the are problems that arise. The easiest way to illustrate this is when they show boundedness. For the case $\mu \ast f$ the author uses a well-known inequality and boundedness of $f$
$$|(\mu \ast f)(x)| \leq \int\limits_\mathrm{G} |f(s^{-1} x)|\ d\mu(s) \leq \|f\|,$$
since $\mu$ is a probability measure. Obviously, the same proof can't proceed for the convolution $f \ast \mu$ since one would get $$|(\mu \ast f)(x)| \leq \|f\| \int\limits_\mathrm{G} \Delta(s^{-1})\ d\mu(s)$$
and I don't think the modulus function is integrable.
Any suggestions is greatly appreciated.
Best Answer
Too long for a comment:
You need some more assumptions on $f$ or on the measure $\mu$.
For example, for the function $f$ take the constant function $f \equiv 1$. Of course, this is continuous and bounded. For this choice of $f$ your question amounts to showing that the function $s \mapsto \Delta(s^{-1})$ is integrable with respect to any (suitable) probability measure.
The map $s \mapsto \Delta(s^{-1})$ is a group morphism to the multiplicative group of positive reals. Hence it is either constant (in this case the question is trivial) or unbounded. But for any unbounded function $h$ one can always find a probability measure on $G$ such that $h$ is not integrable with respect to this measure. The idea is to take a sequence $x_n \in G$ such that $|h(x_n)| > 2^n$, and define the measure $\mu = \sum_{n=1}^{\infty} 2^{-n} \delta_{x_n}$, where $\delta_x$ denotes the Dirac measure at $x$ (see e.g. here for some more details). Hence for such a probability measure the function $f * \mu$ as you defined it is identically infinite.
Maybe you wanted $f$ to be compactly supported? Or integrable with respect to $\lambda$?