Prove $(u\lambda^n)*(v\lambda^n)=(u*v)\lambda^n$

Question

Let $\lambda^n$ denote the $n$ dimensional Lebesgue measure and let $\mu$ , $\nu$ be two measures.

If $\mu$,$\nu$ have densities $u$, $v$ w.r.t. Lebesgue measure show that for all $B\in\mathcal B(\mathbb{R}^n)$

$$(u\lambda^n)*(v\lambda^n)=(u*v)\lambda^n$$

where * denotes the linear convolution.

What I've done so far:

$$(u\lambda^n)*(v\lambda^n)(B)=\int \int 1_B(x+y)u(x)v(y)\lambda^n(dx)\lambda^n(dy)$$ $$=\int v(y) \bigg( \int 1_B(x+y)u(x)\lambda^n(dx)\bigg)\lambda^n(dy)$$

$$=\int v(y) \bigg( \int 1_B(\tau_{+y}(x+y))u(\tau_{+y}x)\lambda^n(dx)\bigg)\lambda^n(dy)$$

where $\tau_{+y}$: $x\mapsto (x-y)$

then the latter equals

$$=\int \int 1_B(x)u(x-y)v(y)\lambda^n(dx)\lambda^n(dy)$$

Noticing that the integrand is positive we can now use Tonelli's theorem to change the order of integration, which yields to:

$$=\int \int 1_B(x)u(x-y)v(y)\lambda^n(dy) \lambda^n(dx)$$ $$=\int_B \bigg( \int u(x-y)v(y)\lambda^n(dy) \bigg)\lambda^n(dx)$$

the part inside the parenthesis is by definition the convolution of $u$ and $v$, hence we can write

$$\int_B (u*v)(x) \lambda^n(dx)$$

which resembles the equality that we want to prove but there's an extra integral. Honestly I don't understand how to go on from here.

Thanks in advance.

Answer 1

You have completed the proof! Saying that $\tau =(u*v)\lambda^{n}%$ is same as $\tau (B)=\int_B (u*v)d\lambda^{n}$ for all $B$.