How to show this function is in $L^1$?

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$G$ be a locally compact abelian group. $\hat G$ denotes the group of characters on $G$.

$M(G)$ be the space of regular complex Borel measure on $G$.

Now let $\mu \in M(G)$ and $\hat{\mu}\in L^1(\hat G)$.

Define a function $f:G\to \mathbb C$ such that $$f(x)=\int_{\hat G}{\hat \mu}(\Lambda)\Lambda(x)\mathrm d\Lambda$$ where ${\hat \mu}:\hat G \to \mathbb C $ defined by $$ {\hat \mu}(\Lambda)=\int_G \Lambda(x^{-1})\mathrm d\mu(x) $$ such that $\mathrm dx$ is a haar measure on $G$ and $\mathrm d\Lambda$ is a haar measure on $\hat G$

Clearly the function is well defined at every point as we have $\hat{\mu}\in L^1(\hat G)$.

How do I show that $f \in L^1(G)$?

I tried taking modulus but didn't help . If $G$ were given compact then it is easy . But for locally compact I have no clue.