Is $f:\mathbb{R}^2\to\mathbb{R}$ defined by $f(x,y)=xy$ Borel-measurable?

Question

Is $f:\mathbb{R}^2\to\mathbb{R}$ defined by $f(x,y)=xy$ a Borel-measurable function?

Sketch: simple proof

By definition of Borel sigma-algebra, $\mathcal{B}_{\mathbb{R}^2}$ is generated by the set of open sets in $\mathbb{R}^2$. Hence, it is the smallest sigma-algebra containing all open sets in $\mathbb{R}^2$.

One can show that $f: (\mathbb{R}^2,\mathcal{B}_{\mathbb{R}^2}) \to (\mathbb{R},\mathcal{B}_{\mathbb{R}})$ is measurable iff $\forall c\in\mathbb{R}, \{(x,y):f(x,y)<c\}\in \mathcal{B}_{\mathbb{R}^2}$.

Since $f$ is continuous, $\{f^{-1}(-\infty,c)\}$ is open and thus must be in $\mathcal{B}_{\mathbb{R}^2}$.

Do you see anything wront with this argument?

Answer 1

Yes. This is because every continuous function is automatically Borel-measurable.

Proof: Let $(X, \tau)$ and $(Y, \pi)$ be topological spaces, and consider continuous $f : X \to Y$. Let $B_Y$ be the Borel $\sigma$-algebra on $(Y, \pi)$, similarly for $B_X$.

Define $W = \{S \in B_Y : f^{-1}(S) \in B_X\}$. Since the inverse image function $f^{-1} : P(Y) \to P(X)$ preserves unions and complements, we see that $W$ is a $\sigma$-algebra.

Now note that if $U \in \pi$, then $f^{-1}(U) \in \tau \subseteq B_X$. Thus, $\pi \subseteq W$. Therefore, since $B_Y$ is the least $\sigma$-algebra containing $\pi$, we have $B_Y \subseteq W$. That is, the inverse image of every measurable set is measurable.

Answer 2

Continuous functions are always Borel measurable. In particular, $f(x,y)=xy$ is continuous.

For any Borel set $U\subset \mathbb{R}$, $f^{-1}(U)$ is a Borel set in $\mathscr{B}(\mathbb{R}\times\mathbb{R})$. Since $\mathbb{R}$ is separable, $\mathscr{B}(\mathbb{R})\otimes\mathscr{B}(\mathbb{R})=\mathscr{B}(\mathbb{R}\times\mathbb{R})$; hence $f$ is measurable with respect the product $\sigma$-algebra $\mathscr{B}(\mathbb{R})\otimes\mathscr{B}(\mathbb{R})$.