On commutativity of convolution

[psie]

TL;DR Summary

My textbook claims the convolution is commutative. I think I have found a way to motivate this, but I don't quite understand the textbook's motivation.

I quote;

Throughout this section, we consider the measure space $(\mathbb R^d,\mathcal B(\mathbb R^d),\lambda)$ [$\lambda$ being Lebesgue measure]. Let $f$ and $g$ be two real measurable functions on $\mathbb R^d$. For $x\in\mathbb R^d$, the convolution $$f\ast g(x)=\int_{\mathbb R^d}f(x-y)g(y)\,\mathrm{d}y$$ is well defined provided $$\int_{\mathbb R^d}|f(x-y)g(y)|\,\mathrm{d}y<\infty.$$In that case, the fact that Lebesgue measure on $\mathbb R^d$ is invariant under translations and under the symmetry $y\to -y$ shows that $g\ast f(x)$ is also well defined and $g\ast f(x)=f\ast g(x)$.

I struggle with the last sentence and putting the pieces together. I know the abstract change of variables formula for $\nu$ on $(F,\mathcal D)$ being the pushforward measure of $\mu$ on $(E,\mathcal C)$ under $\varphi:E\to F$, i.e. $\nu(A)=\mu(\varphi^{-1}(A))$ for every $A\in\mathcal D$. Then for any $h:F\to\mathbb R$, $$\int_E h(\varphi(x))\,\mu(\mathrm{d}x)=\int_F h(y)\,\nu(\mathrm{d}y).$$I figured that $h(y)=f(x-y)g(y)$ and $\varphi(x)=x-y$, with $\mu=\lambda$ and $(E,\mathcal C)=(F,\mathcal D)=(\mathbb R^d,\mathcal B(\mathbb R^d))$. Since Lebesgue measure is invariant under translations, $\nu$ is just Lebesgue measure again. So I get indeed $f\ast g(x)=g\ast f(x)$. But what do they mean, or what's the relevance of, the invariance under the symmetry $y\to -y$? And how does one show Lebesgue measure is invariant under this symmetry?

 

[fresh_42]

I think your thoughts are way too complicated. Let $u=x-y.$
\begin{align*}
(f\ast g)(x)&=\int_{y\in \mathbb{R}^d} f(x-y)g(y)\,dy =- \int_{x-u\in \mathbb{R}^d} f(u)g(x-u) \,du \\[6pt]
&=\int_{u-x\in \mathbb{R}^d} f(u)g(x-u) \,du =\int_{u\in \mathbb{R}^d}g(x-u)f(u) \,du\\[6pt]
&=(g\ast f)(x)
\end{align*}