Is the set of continuous functions a Borel measurable subset of $L^2$?

Question

Let $C([0,1])$ be the set of continuous real-valued functions on $[0,1]$ and $L^2([0,1])$ the Hilbert space of (equivalence classes) of square-integrable realvalued functions on $[0,1]$.

Then $C([0,1])$ can be identified with a subset of $L^2([0,1])$. I am wondering if $C([0,1])$ is in the Borel sigma-algebra of $L^2([0,1])$.

(It is well-known that $C([0,1])$ is not an element of the product sigma-algebra of the product space $\mathbb R^{[0,1]}$, because the latter only sees a countable set of values of a function).

Answer 1

Check L. Schwartz "Radon measures on arbitrary topological spaces and cylindrical measures," p.101.

Let $X$ be a Hausdorff space. Then every Lusin subspace $A$ (with the induced topology) of $X$ is Borel in $X.$

Image of $C([0,1])$ in $L^2([0,1])$ is a Lusin space as $C([0,1])$ is complete, metric and separable. See definition of Lusin space.

Answer 2

Both $C[0,1]$ and $L^{p}[0,1]$ are polish. The embedding map $$f: C[0,1] \hookrightarrow L^{p}[0,1]$$ is continuous (therefore borel measurable) and injective. So, $f(C[0,1])$ is Borel subset of $L^{p}[0,1]$.