How can $L^1(\Bbb R)$ be isomorphic to an ideal in $M(\Bbb R)$?

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Here is this week's POTW:

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Consider the Lebesgue space $L^1(\Bbb R)$ as an algebra with product given by convolution. Prove that $L^1(\Bbb R)$ is isomorphic as an algebra to an ideal in the algebra $M(\Bbb R)$ of complex Borel measures on $\Bbb R$, and identify the ideal. Note the product in $M(\Bbb R)$ is given by convolution of measures.-----

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I'm going to give members one extra week to solve this POTW. For a hint, consider the set of measures absolutely continuous with respect to the Lebesgue measure on $\Bbb R$. Use the Radon-Nikodym theorem to construct an algebra map.

 

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Honorable mention goes to vidyarth for an incomplete solution. You can read my solution below.

Spoiler

Let $A_m$ denote the set of measures absolutely continuous with respect the Lebesgue measure $m$. Then $A_m$ is an ideal of $M(\Bbb R)$. For given $\mu\in M(\Bbb R)$ and $\nu \in A$, $d\mu = f\, dm$ for some $f\in L^1(\Bbb R)$, and for every measurable set $E\subset \Bbb R$,

$$(\mu * \nu)(E) = \int \mu(E - t)\, d\nu(t) = \iint 1_{E-t}(s)f(s)\, dm(s)\, d\nu(t) = \iint 1_{E}(t + s)f(s)\, d\nu(t)\, dm(s)$$
$$ = \iint 1_E(s) f(s-t)\, d\nu(t)\, dm(s) = \int_E (f * \nu)(s)\, dm(s)$$

where $(f*\nu)(s) = \int f(s-t)\, d\nu(t)$. Thus $d(\mu * \nu) = (f * \nu)\, dm$, proving $\mu * \nu \in A_m$.

By the Radon-Nikodym theorem, the mapping $\Phi : A_m \to L^1(\Bbb R)$, $\mu \mapsto \frac{d\mu}{dm}$ is a one-to-one correspondence. For all $\mu, \nu \in A_m$,

$$\frac{d(\mu * \nu)}{dm} = \frac{d\mu}{dm} * \frac{d\nu}{dm}$$

whence $\Phi$ is an isomorphism of algebras.