Showing a set is in the product $\sigma$-algebra Show that $C=\{(x,x^2)\}_{x \in \mathbb{R}}$ is in the product $\sigma$-algebra $\mathcal{B}_\mathbb{R} \times \mathcal{B}_{\mathbb{R}}$.
Attempt: I know it is not enough to show that the sections of $C$ are measurable in their respective components. I am reading this post Is the Graph of a Measurable Function Measurable?, and trying to model an explanation for this particular problem.
Define $\gamma_f:\mathbb{R} \rightarrow \mathbb{R} \times \mathbb{R}$ by $\gamma_f(x)=(x,x^2)$. Then $\psi:\mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R} \times \mathbb{R}$ given by $\psi(x,y)=y-x^2$ is a measurable function and $\psi^{-1}(\{0\})=\{(x,y) \mid \psi(x,y)=0\}=\gamma_f(\mathbb{R})=C \in \mathcal{B}_{\mathbb{R}} \times \mathcal{B}_{\mathbb{R}}$ since $\psi$ is measurable.
Is there a better way to do this? Is it correct? I would have never thought to try this and I wonder if there is a more methodical approach to this type of problem.
 A: The problem with your attempt is that you need to still prove that $\psi$ is measurable.
Let $C=\{(x,x^2):x\geq 0\}$ (getting to your $C$ is then a trivial step).
Note that for any $x$ we have $I_\epsilon(x)=[x,x+\epsilon)\times[x^2,x^2+\epsilon^2 +2x\epsilon)$ does contain enough of the set $C$ (which is in fact a graph) so that we can cover $C$ by a countable family of such rectangles. If $U_\epsilon$ is a union of such rectangles that contains $C$ (i.e. set $U_\epsilon = \bigcup_{k\in\mathbb Z} I_\epsilon(k\epsilon)$) then if $(x,y_1),(x,y_2)\in U_\epsilon$ then there is some $k$ so that $(x,y_1),(x,y_2)\in I_\epsilon(k\epsilon)$.
Then $k^2\epsilon^2 \leq y_1,y_2 < (k^2 + 1 + 2k)\epsilon^2$.
Then
$$ |y_1-y_2| \leq \epsilon^2 (1+2k)$$
Note that as $x\in[k\epsilon,(k+1)\epsilon)$ that $k$ must be the floored value of $x/\epsilon$, so $|k|\leq |x/\epsilon|$. Thus
$$ |y_1-y_2| \leq \epsilon^2 + |x|\epsilon $$
Now take $D=\bigcap U_{1/n}$. We have $C\subset D$ (as $C\subset U_\epsilon$). On the other hand, if $(x,y)\in C$ then we get for all $n$ that
$$ |y-x^2| \leq 1/n^2 + |x|/n $$
so in fact $y=x^2$.
Now, for your $C$ (call the $C$ above $C_+$) we have $C=C_+\cup C_-$ and $C_-$ is the complement of $C_+$.
Thus $C$ is in the product $\sigma$-Algebra.
A: You don't need to introduce $\gamma_f$, but you do have to be careful of exactly why $\psi$ is measurable.
$\psi \colon \mathbb{R}^2 \to \mathbb{R}$ defined by $\psi(x, y) = y - x^2$ is continuous, hence $(B_{\mathbb{R}^2}, B_{\mathbb{R}})$ measurable. We have $C = \psi^{-1}\{0\}$. Since $\{0\} \in B_{\mathbb{R}}$, it follows that $C \in B_{\mathbb{R}^2}$. Now since $\mathbb{R}$ is second countable, $B_{\mathbb{R}^2} = B_{\mathbb{R}} \otimes B_\mathbb{R}$. Hence $C \in B_{\mathbb{R}} \otimes B_\mathbb{R}$.
A similar argument shows that if $(X, M)$ is a measurable space and $f \colon X \to \mathbb{C}$ is measurable, then $\{(x, f(x)) : x \in X\} \in M \otimes B_\mathbb{C}$, i.e. the graph of $f$ is product measurable.
