Show $x \mapsto \int_G f(yx) \mu(dy)$ is continuous

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This is exercise 11.2 in Folland's book:

Let $\mu$ be a Radon measure on the locally compact group $G$ and $f \in C_c(G)$. Prove that $$x \mapsto \int_G f(yx) \mu(dy)$$ is continous.

Before showing my attempt, let me quote a result that might be relevant:

If $f \in C_c(G)$, then $f$ is left and right uniformly continuous, i.e. for every $\epsilon > 0$, there is a neighborhood $V$ of $e$ (= the identity of $G$) such that $\sup_{x \in G} |f(y^{-1}x)-f(x)| \leq \epsilon$ and $\sup_{x \in G} |f(xy)-f(x)| \leq \epsilon$ for $y \in V$.

Attempt:

Note that $\operatorname{supp}(y \mapsto f(yx)) = \operatorname{supp}(f)x^{-1}$, so the integral makes sense since $\mu$ is finite on compact sets.

Let $x \in G$, $\epsilon > 0$. Choose a compact set $K$ outside which $f$ vanishes. Choose a compact neighborhood $K'$ of $x$. Applying the result above, there is a symmetric neighborhood $V$ of $e$ such that $$v \in V \implies \sup_{z \in G} |f(z)-f(zv)| \leq \frac{\epsilon}{\mu(K(K')^{-1})+1} \quad (*)$$ Note for this that $K(K')^{-1}$ is compact, hence its measure is finite.

Let $W:= K' \cap xV$, which is a neighborhood of $x$. Then if $x \in W$ $$\left|\int_G f(yx)\mu(dy) -\int_G f(yx') \mu(dy)\right| $$ $$\leq \int_G |f(yx)-f(yx')| \mu(dy)$$ $$= \int_{K(K')^{-1}}|f(yx)-f(yx')|\mu(dy)$$ $$= \int_{K(K')^{-1}}|f(yx)-f((yx)(x^{-1}x'))|\mu(dy)\leq \epsilon$$

The equality in the third line follows because $$yx \in K \implies y \in Kx^{-1}\subseteq K(K')^{-1}$$ so $$y \notin K(K')^{-1} \implies yx \notin K \implies f(yx) = 0$$

Similarly, $yx' \in K \implies y \in K(x')^{-1} \subseteq K(K')^{-1}$ gives us that $f(yx') = 0$ if $y \notin K(K')^{-1}$

The last inequality follows because $x^{-1}x' \in x^{-1}xV = V$ so we can apply $(*)$. This shows that we have continuity at $x$ and thus ends the proof. $\quad \square$

Questions:

(1) Is this proof correct?

(2) Would it be correct that the statement is more generally true for Borel measures that are finite on compact sets (as my proof seems to suggest?)

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(1): Almost, the only error is that you need to choose $x' \in W$ before you estimate the difference of the integrals (this is probably a typo).

(2): Yes, any Borel measure that is finite on compact sets suffices here.