How do I show this set is not in this $\sigma$-algebra? How do I show that set $$F= \{ f\in C[0,1]  \text{ such that  } \sup_{t \in [0,1]} f(t) < 1\} $$ is not in the $ \sigma$-algebra $ \mathcal{B} $ generated by the sets from   $$\{ f \in C[0,1], f(t_1) \in A_1, \ldots, f(t_n) \in A_n\} $$ where $ n \geq 0,t_i \in [0,1], A_i $ are intervals on $\mathbb{R}$ ?
Comment: Suppose that $ F \in \mathcal{B} $ so $ F^c \in \mathcal{B} $ also, i.e. set $$\{ f\in C[0,1], \exists t \in [0,1]: f(t) \geq 1 \} \in \mathcal{B}$$
But I don't image a set in $ \mathcal{B} $ have which form so that I can show a contradiction.
Thanks you for your help.
 A: Proof by contradiction. Each set in $\mathcal{B}$ is of the following form:
$$\{f \in C[0,1]: f(t_i) \in M_i \, \text{for $i \in \mathbb{N}$}\}$$
where $\{t_i\}_{i=1}^\infty$ is a sequence in $[0,1]$ and $\{M_i\}_{i=1}^\infty$ is a sequence of Borel measurable sets. Suppose that there exists $B \in \mathcal{B}$ such that $B = F$. Then $B \cap F = B$ so $M_i \subset (-\infty,1)$ for all $i$. In fact, $M_i = (-\infty,1)$ since $F$ contains the family of constant functions $f(t) = c$ with $c \in (-\infty,1)$. Thus $B$ must have the form
$$\{f \in C[0,1]: f(t_i) < 1 \, \text{for $i \in \mathbb{N}$}\}$$
and the statement $B = F$ amounts to the following: there exists a sequence $\{t_i\}_{i=1}^\infty$ such that $f(t_i) < 1$ for all $t_i$ implies $f(t) < 1$ for all $t \in [0,1]$. This is false and gives the desired contradiction, as we now show.
Claim: Given any sequence $\{t_i\}_{i=1}^\infty$, there is a continuous function $f$ such that $f(t_i) < 1$ for all $i$, but $f(t) = 1$ on a set of positive measure.
Proof: We use a construction inspired by the proof that the set of rationals has measure zero. We begin with the constant function $f = 1$ and subtract tent-shaped bumps centered on each $t_i$. The first bump has height $1$ and support $\epsilon/2$ for any $\epsilon > 0$ we choose, say $\epsilon = 0.1$. The bumps after it have support $\epsilon/2^i$ and height $1/4^i$, which ensures that the tent slope decays as $1/2^i$. The sequence of functions obtained by partial sums of this series is uniformly Lipschitz and has a pointwise limit everywhere, so the limiting function exists everywhere and is Lipschitz as well. This function is less than 1 on a set of measure $\leq \epsilon$ encompassing all the $t_i$, and is equal to 1 everywhere else.
