Is $C^1[0,1]$ a measurable subset of $C[0,1]$?

Question

I am wondering whether $C^1([0,1],\mathbb{R})$, the set of continuously differentiable functions, is an element of the Borel $\sigma$-algebra of $(C[0,1],\|\cdot \|_\infty)$. As far as I know, it is neither open nor closed, but I feel like it would be natural for this to be true.

If this is not the case, what about the set of (not necessarily continuously) differentiable functions?

Answer 1

Define $T: \mathbb R \times C[0,1] \to C[0,1]$ by $T(c,f)(x)=c+\int_0^{x} f(t)dt$. Then $T$ is continuous, hence Borel mesurable. Also, $T$ is injective. By Theorem 8.3.7 of Cohn's Measure Theory (see p. 276) it follows that the image of $T$ is Borel set. But the image is exactly $C^{1}[0,1]$.