# Measurability of product of Borel measurable functions with different domains?

Suppose we are in the measure space $(\mathbb{R}, \Sigma(m), m)$ ($m$ is Lebesgue measure). Also, suppose $f, g \in L^{1}(dm)$.

We define the convolution of $f$, $g$, by $(f * g)(y) = \int \limits_{\mathbb{R}} f(x)g(y - x) \,dm$.

I want to show that $L^{1}(dm)$ is closed under the operation of taking convolutions. I can do this by using Fubini's (or I think Tonelli's) Theorem, as long as $|f(x)g(y - x)|$ is measurable with respect to the $\sigma$-algebra $\overline{\Sigma(m) \times \Sigma(m)}$. But actually, I want to start this off by first showing the function is Borel measurable (i.e., measurable with respect to $\mathcal{B}(\mathbb{R}^{2})$, I think -- or should it be $\overline{\mathcal{B}(\mathbb{R}) \times \mathcal{B}(\mathbb{R})}$?).

The problem is, I don't understand how the reasoning my professor used to show this. Here is what he said:

The function $\phi(x,y): \mathbb{R}^{2} \to \mathbb{R}$ defined by $\phi(x,y) = y -x$ is continuous and hence Borel measurable. If $g$ is also Borel measurable, then since the composition of Borel measurable functions is Borel measurable, we have $\psi(x,y) = g(y - x)$ is Borel measurable, where $\psi: \mathbb{R}^{2} \to \mathbb{R}$.

Then, if $f: \mathbb{R} \to \mathbb{R}$ is Borel measurable, the product $|f(x)g(y - x)|$ is Borel measurable because the product of Borel measurable functions is Borel measurable.

I'm confused by this because the function $\phi(x,y) = g(y - x)$ has a different domain than $f(x)$, so when we multiply the two together, how can we say the product is Borel measurable?

## 1 Answer

First, we can show that $\overline{\mathcal B(\Bbb R)\times\mathcal B(\Bbb R)} = \mathcal B(\Bbb R^2)$, as every open disk -hence every open set in $\Bbb R^2$- can be covered by open rectangles (with sides parallel to the axes).

Second, $f(x)$ can also be considered as a function in two variables (but constant in $y$). More precisely we consider $$\varphi(x,y):=f(x)$$ which is also Borel measurable if $f$ is, then so is the product $\phi\cdot\psi$, and finally, composing it with $x\mapsto |x|\,:\Bbb R\to\Bbb R$ still gives a Borel measurable function.

• Why is $f$ Borel measurable with respect to $\mathcal{B}(\mathbb{R}^{2})$? I have a suspicion it has something to do with the fact that $f^{-1}(B) \times \emptyset$ is a Borel set for any Borel set $B$... – layman Aug 25 '14 at 22:38
• Actually, it's because if $B$ is a Borel set in $\mathbb{R}$, then defining $F(x,y) = f(x)$ gives that $F^{-1}(B) = f^{-1}(B) \times \mathbb{R}$, which is in $\mathcal{B}(\mathbb{R}^{2})$. – layman Aug 26 '14 at 20:10