Formally show that the set of continuous functions is not measurable Let $C(\mathbb{R})=\{ f:\mathbb{R}\to \mathbb{R} \colon \ f \text{ continuous}\}\subseteq \mathbb{R}^{\mathbb{R}} $.
How to prove formally that $C(\mathbb{R}) \notin \mathcal{B}(\mathbb{R}^{\mathbb{R}})$ (the product sigma algebra)?
The classic proof said that   $A \in \mathcal{B}(\mathbb{R}^{\mathbb{R}})$ iff exists $J\subseteq \mathbb{R}$ with $|J|\leq \aleph_0$ and $B\in \mathcal{B}(\mathbb{R^J})$ such that $A=B \times \mathbb{R}^{\mathbb{R}\setminus J}=\{f \in \mathbb{R}^\mathbb{R} \colon  \ (f(j) \colon \ t \in J) \in B\}$. In simpler terms,  $A$ only can have countable restrictions but the set of continuous functions have uncountable restrictions since every function on $C(\mathbb{R})$ needs to be continuous in every point $x \in \mathbb{R}$.
The last part of this proof always didn't satisfied me at all, and I don't really know how to do this proof formally. 
Any help or references will be appreciated.
 A: Let consider a mapping $T:C(R)\to R^{Q}$ defined by $T(f)=(f(q))_{q \in Q}$, where $Q$ denotes a set of all rational numbers of $R$. It is obvious that $T$ is injective which implies that $card(C(R))\le card(R^Q)=(2^{\aleph_0})^{\aleph_0}=
2^{\aleph_0\times \aleph_0}=2^{\aleph_0}$, where $\aleph_0$ denotes the cardinality of all natural numbers. 
If we assume that $C(R)$ is measurable with respect to product topology then there  will be a countable parameter set $J \subset R$ and a non-empty Borel subset $A \subseteq R^J$ such that
$C(R)=A \times R^{R \setminus J}$. Notice that $$card(A \times R^{R \setminus J})\ge 1 \times card(R^{R \setminus J})=card(R^R)={2^{\aleph_0}}^{2^{\aleph_0}}=2^{2^{\aleph_0}}>2^{\aleph_0}$$ and we get the contradiction. 
A: Here's a direct argument following tomasz's suggestion.  Let's take as given your characterization of the measurable sets $\mathcal{B}(\mathbb{R}^\mathbb{R})$.
Let $A \in \mathcal{B}(\mathbb{R}^\mathbb{R})$ be arbitrary.  We will show $A \ne C(\mathbb{R})$.  If $A = \emptyset$ we are done.  Otherwise, suppose $f \in A$.  If $f$ is not continuous, we are done.  Otherwise, suppose $f$ is continuous.  We know $A$ is of the form $A = B \times \mathbb{R}^{\mathbb{R} \setminus J}$, where $B \subset \mathbb{R}^J$ and $J$ is at most countable.  Since $\mathbb{R}$ is uncountable, $\mathbb{R} \setminus J$ is nonempty, so let $y \in \mathbb{R} \setminus J$.  Define $g$ by $g(y) = f(y)+1$ and $g(x) = f(x)$ for $x \ne y$.  Then $g \in A$, but $g$ is certainly not continuous.
