Cylinder sets on $C[0,1]$

Question

Suppose $\mathscr{B}$ is the cylindrical $\sigma$-algebra on $\mathbb{R}^{[0,1]}$. Let $C:=C[0,1]$ and $\mathscr{A}:=\{B\cap C: B\in\mathscr{B}$. I am trying to show that $$ A:=\{f\in C[0,1]: \int_{[0,1]} f<1\}$$ is inside $\mathscr{A}$. We know that $\mathscr{A}$ is a $\sigma$-algebra but apart from this we have nothing. We need to construct the cylinder set $B$ such that $A=B\cap C$.

Answer 1

$$B := \bigcup_{k \ge 1} \bigcap_{N \ge 1}\bigcup_{n \ge N} \bigcup_{\substack{\alpha_0,\dots,\alpha_{n-1} \in \mathbb{Q} \\ \frac{1}{n}(\alpha_0+\dots+\alpha_{n-1}) \le 1-\frac{1}{k}}} \left\{f : [0,1] \to \mathbb{R} \hspace{1mm} | \hspace{1mm} \left|f\left(\frac{j}{n}\right)-\alpha_j\right| \le \frac{1}{2k} \hspace{2mm} \forall \hspace{1mm} 0 \le j \le n-1\right\}$$

Answer 2

Hints:

1. Recall (or show) that $\mathcal{A}$ is the Borel $\sigma$-algebra generated by the uniform topology on $C[0,1]$.
2. Show that the set $A$ is open in $(C[0,1],\|\cdot\|_{\infty})$. Since the Borel $\sigma$-algebra contains all open sets, this implies, by Step 1, that $A \in \mathcal{A}$.