It is classical that $L^1(G, m)$ is a Banach algebra when $G$ is a locally compact group with Haar measure $m$ by using the operation of convolution via the integral

$$(f*g)(y)=\int_Xf(x)g(yx^{-1})\,dm(x)$$

I'm interested in knowing if there is a natural Banach algebra structure on $L^1(X, \mathcal{B},\mu)$ for a locally compact, Hausdorff measure space $X$, with a Borel sigma algebra, $\mathcal{B}$ and radon measure, $\mu$ with the added assumption that the measure is induced by a (say normalized) Haar measure on a compact abelian group, $G$.

In particular, the situation I'm most interested in is

$$X=\coprod_{n\in\mathbb{N}} X_n$$

with each $X_n$ a compact space topologically a homogeneous $G$-space, i.e. a quotient space $G/H_n$ with $H_n$ closed subgroups of $G$. With these assumptions I can define just such a measure on $X$ by first taking $f\in C(X_n)$ and selecting any point $x_n\in X_n$ and using the normalized Haar measure $m$ on $G$, setting

$$I_n(F)=\int_G f(\tau x_n)\,dm(\tau)$$

Since $G$ acts transitively on its coset spaces, this integral is independent of choice of $x_n$. Then, on $X$, I define for $F\in C_c(X)$

$$I(F)=\sum_n I_n(F\cdot 1_{X_n})$$

Since $F$ has compact support, this sum is well-defined (being finite). I know that, since $I(F)$ is a linear functional, it is represented by integration against a Radon measure, $\lambda$ on $X$ and by construction and the transitivity on the pieces, this measure is invariant under the action of $G$, that is to say, there is a radon measure $\mu$ on $X$ and a $\sigma$-algebra, $\Sigma$, containing the Borel $\sigma$-algebra so that for every $F\in L^1(X,\Sigma,\mu)$

$$I(F)=\int_X F\,d\mu.$$

Moreover it is clear from the definition that this has the property that

$$\int_{X_n}F(x)\,d\mu(x)=\int_{X_n}F(\tau x)\,d\mu(x)$$

for every $\tau\in G$ and $F\in L^1(X)$. By writing

$$F=\sum_n F\cdot 1_{X_n}$$

we can see now that

$$\int_X F\,d\mu=\sum_n F(x)1_{X_n}\,d\mu(x)$$ $$=\sum_n\int_{X_n}F(x)\,d\mu=\sum_n \int_{X_n}F(\tau x)\,d\mu(x)$$ $$=\int_X F(\tau x)\,d\mu(x).$$

This establishes that the measure $\mu$ is invariant under $G$.

Now, $X$ itself *lacks* a group structure, but can we still somehow define a Banach algebra structure on $L^1(X)$ by using the fact that the measure $\mu$ is essentially pushing forward the measure $m$ on $G$ and $G$ *does* have the structure of a compact group?

Edit (Adam): I've added the fact about the $X_n$ being homogeneous spaces of $G$ by subgroups and clarified the invariance of $\mu$ under the $G$-action.