Background: Let $E = C[0,1]$ be the space of continuous functions on $[0,1]$ and let $\mu$ be the classical Wiener measure on $E$, i.e. the distribution of a Brownian motion and let $H$ be the Cameron-Martin space associated to $\mu$. By the Cameron-Martin theorem, the measure defined by $\mu_x(A) := \mu( A + x)$ (i.e. the push-forward under the shift operator) is singular with respect to $\mu$ precisely when $x \not\in H$.
In particular, this implies that for $x \not\in H$, the expression
$$ I_x(f) = \int_E f(y + x) d \mu(y) $$
is not well-defined for a $\mu$-a.e. equivalence class $f$, i.e. for two functionals $f,g: E \rightarrow \mathbb{R}$ which agree $\mu$-a.e., we may have $I_x(f) \neq I_x(g)$.
An explicit example of this is $f = 1_{A}$ and $g \equiv 1$, where $A := \{ x \in E: \langle x \rangle_1 < \infty \}$ and $\langle \cdot \rangle_1$ denotes the quadratic variation evaluated at $1$. The two agree $\mu$-a.e. since for $B = \{B_t : 0 \leq t \leq 1\}$ a Brownian motion we have $\langle B \rangle_1 = 1 < \infty$ almost surely. However, for any function $x \in E$ with $\langle x \rangle_1 = \infty$ we have
$$ I_x(f) = \mu( y \in E: \langle y \rangle_1 = \infty) = 0 $$
while $I_x(g) = \mu(E) = 1 $.
Question: Where is the mistake in the following argument, claiming that the situation above cannot happen?
Since $f$ and $g$ represent the same element in $L^2(E, \mu)$, they both have a Wiener-Ito-Chaos decomposition with the same coefficients, i.e. there are real coefficients $c_a$ and polynomials of the form $H_a(\cdot) = \prod_{a_i \in a} h_{a_i}(e_i(\cdot))$, where $a$ is a multiindex, $h_{j}$ is the $j$-th Hermite polynomial and $e_j$ is the $j$-th element of an ONB of $H$ which is contained in $E^{\ast}$ s.t.
$$ \sum_{\vert a \vert \leq n} c_a H_a \rightarrow f ~~ \text{in} ~~ L^2(E, \mu), $$
where $\vert a \vert$ is the absolute value of the multi-index $a$. By $H_a$ we mean a bona-fide function, not a $\mu$-a.s. equivalence class.
(*) Since the convergence is in $L^2$ we have
$$ I_x(f) = \lim_{n \rightarrow \infty} \sum_{\vert a \vert \leq n} c_a \int_E H_a(y+x) d \mu(y) = \lim_{n \rightarrow \infty} \sum_{\vert a \vert \leq n} c_a \int_E \prod_{a_i \in a} h_{a_i}(e_i(y+x)) d \mu(y) . $$
To work out the integral (and in order to use the independence of the $e_i$) we use the Binomial Theorem for the Hermite polynomials to compute
\begin{align} H_a(x+y) &= \prod_{a_i \in a} h_{a_i}(e_i(y+x)) \\ &= \prod_{a_i \in a} \sum_{l = 0}^{a_i} { a_i \choose l } h_l(e_i(y)) (e_i(x))^{a_i-l} \\ &= \sum_{\sigma \in S} \prod_{\sigma_i \in \sigma} { a_i \choose \sigma_i } h_{\sigma_i}(e_i(y)) (e_i(x))^{a_i-\sigma_i} \\ \end{align}
where $S = \prod_{a_i \in a} \{0, \ldots, a_i\}$. Plugging this back into the integral and using that the family of random variables $\{h_{\sigma_i}(e_i(\cdot))\}_{\sigma_i \in \sigma}$ is independent (because the indices of the $e_i$ differ)
$$ \int_E \prod_{a_i \in a} h_{a_i}(e_i(y+x)) d \mu(y) = \sum_{\sigma \in S} \prod_{\sigma_i \in \sigma} { a_i \choose \sigma_i } \int_E h_{\sigma_i}(e_i(y)) d \mu(y) (e_i(x))^{a_i-\sigma_i} $$
Now, since the expectation of the Hermite polynomials of order $l > 0$ vanishes, this leaves us with
$$ \sum_{\sigma \in S} \prod_{\sigma_i \in \sigma} { a_i \choose \sigma_i } \int_E h_{\sigma_i}(e_i(y)) d \mu(y) (e_i(x))^{a_i-\sigma_i} = \prod_{a_i \in a} (e_i(x))^{a_i} $$
ultimately giving
$$ I_x(f) = \lim_{n \rightarrow \infty} \sum_{\vert a \vert \leq n} c_a \prod_{a_i \in a} (e_i(x))^{a_i}. $$
The same argument can of course be done with $g$ instead of $f$, leading to $I_x(f) = I_x(g)$.
Thank you very much in advance!
The problem lies in the line marked with (*). The problems is that even though the convergence takes place in $L^2$ with respect to the measure $\mu$, it does not take place in $L^2$ with respect to the shifted measure $\mu_x(\cdot) = \mu(\cdot + x)$. If $x$ were a Cameron-Martin path, then this argument would go through.