Is the space $C([0, 1])$ with the $L^1$ metric completely metrizable? (or is $C([0,1])$ a $G_\delta$ in $L^1([0, 1])$?) Let $C([0, 1])$ be the continuous functions from $[0, 1]$ to $\mathbb{R}$.
Define the metric $d_1$ on $C([0, 1])$ as follows:
$$
d_1(f, g) = \int_0^1 |f(x)-g(x)| dx
$$
Then is the space $(C([0, 1]), d_1)$ completely metrizable?
Or equivalently, is $C([0, 1])$ a $G_\delta$ subset of $L^1([0, 1])$?
Of course, $d_1$ is not a complete metric but I think another equivalent metric may be complete.
If $C([0, 1])$ were a $G_\delta$ subset of $L^1([0, 1])$, we could write $C([0, 1]) = \bigcap_{n \in \mathbb{N}} U_n$ where each $U_n$ is dense open set of $L^1([0, 1])$.
So, $C([0, 1])$ would be comeager set. But I didn't able to proceed any more.
Actually, I want to prove that $(C([0, 1]), d_1)$ and $(C([0, 1]), d_\infty)$ are not homeomorphic. Here, $d_\infty(f, g) = \sup_{x \in [0, 1]} |f(x) - g(x)|$. But $d_\infty$ is complete, so it is sufficient to solve the above question. 
 A: 
is the space $(C([0, 1]), d_1)$ completely metrizable? Or equivalently, is $C([0, 1])$ a $G_\delta$ subset of $L^1([0, 1])$?

No, I have two distinct proof of this. 
Let $X=(C([0, 1]), d_1)$ and $\hat X$ be the completion of $X$ as a normed space. So $\hat X$ is a complete metrics space. (According to [KF, Ch. VII, §1] the space $L^1([0, 1])$ is complete (by Theorem 1) and the space $ C([0, 1])$ is dense in $L^1([0, 1])$ (by Theorem 2), but we don’t need this fact). Suppose to the contrary that the space $X$ is completely metrizable.
Proof 1. By [Eng, Theorem 4.3.24], $X$ is a $G_\delta$-subset of $\hat X$. Since the space $X$ is not complete, there exists an element $x\in\hat X\setminus X$. Then $X$ and $x+X$ are dense $G_\delta$ subsets of a complete metric space $\hat X$ with the empty intersection, which is impossible since $\hat X$ is Baire.
Proof 2. By [Eng, Theorem 4.3.26] $X$ is Čech-complete. By [AT, Theorem 4.3.7], $X$ is a Raĭkov complete topological group. According to [R], $X$ is closed in any topological group in which it is isomorphically embedded as a  topological subgroup. In particular, $X$ is closed in its completion $\hat X$, so $X$ is a complete normed space, a contradiction. 
References
[AT] Arhangel'skiĭ A, Mikhail G. Tkachenko, Topological groups and related structures, Atlantis Press, Paris; World Sci. Publ., NJ, 2008.
[FK] Fomin S.V., Kolmogorov A.N., Elements of theory of functions and functional analysis, 4th edn., M.: Nauka, 1976 (in Russian)
[R] Raĭkov D.A. On the completion of topological groups, Izv. Akad. Nauk SSSR 10, (1946), 513–528 (in Russian).   
