Why is it true that $\int\int f(x-t)g(t)dtdx = \int f(x)dx \int g(x)dx$?

This appears in a proof on page 122 of Measure and Integral by Wheeden and Zygmund. The theorem is that

If $f,g$ are Lebesgue integrable functions from $\mathbb R^n$ to $\mathbb R$, then $(f*g)(x)$ exists for almost every $\mathbf x\in \mathbb R^n$ and is measurable. Morevover, $f*g$ is Lebesgue integrable and$$\int_{\mathbb R^n} |f*g|d\mathbf x \le \left(\int_{\mathbb R^n} |f|d\mathbf x\right)\left(\int_{\mathbb R^n} |g|d\mathbf x\right),$$ $$\int_{\mathbb R^n} (f*g)d\mathbf x \le \left(\int_{\mathbb R^n} fd\mathbf x\right)\left(\int_{\mathbb R^n} gd\mathbf x\right).$$

On the last few line of the proof, they say by Fubini's theorem, $\int f(\mathbf {x} -\mathbf t)g(\mathbf t) d\mathbf t$ exists for a.e. $\mathbf x$ and is measurable and integrable; also, $$\int\int f(x-t)g(t)dtdx = \int \left[\int f(x-t)g(t)dt \right] dx = \int f(x)dx \int g(x)dx.$$

Why does the last equality hold? I guessed they use change of variable? How do they change the variable?

• Change the order of integration. $\int f(x-t)\,dx = \int f(x)\,dx$ by the translation-invariance of the Lebesgue measure. – Daniel Fischer Jul 25 '18 at 21:43

It should be $$\int \int f(x-t) g(t) \, dt \, dx = \int \left( \int f(x-t)g(t) \, dx \right) dt = \int \left( \int f(x-t) \, dx \right) g(t) \, dt,$$ because $g(t)$ is independent of $x$. Lebesgue measure is translation-invariant, so $\int f(x-t) \, dx = \int f(x) \, dx$, and then this independent of $t$, so $$\int \left( \int f(x) \, dx \right) g(t) \, dt = \left( \int f(x) \, dx \right) \int g(t) \, dt$$ as desired.
Use Fubini to express it as $$\int g(t)\left( \int f(x-t) dx \right) dt.$$ Then because Lebesgue measure is translation invariant, the bracket is independent of $t$ and coincides with $\int f(x) dx$. Pull it out and you arrive at the desired expression.