Does the operator $T(f)(t) := f(t) - f(0)$ preserve measurability?

There is a Borel set $E$ in $\mathbb R^2$ such that $F := \{x-y\colon (x,y) \in E\}$ is not a Borel set.

Let $A := \{f \in \mathbf{C}\colon (f(1), f(0)) \in E\}$. Then $A \in \mathcal{B}_{\left[0,\infty\right)}$.

How about $T(A)$? In fact $$ T(A) = \{g \in \mathbf{C}\colon g(0)=0, g(1) \in F\} $$ and is not Borel.

added Mar 10
Why is $T(A)$ not Borel?

First, note that $\mathcal{B}_{\left[0,\infty\right)}$ is the same as the Borel sets for the topology of uniform convergence on bounded sets for $\mathbf{C}_{[0,\infty)}$, a Polish space.

Suppose (for purposes of contracidtion) that $T(A)=\{g \in \mathbf{C}\colon g(0)=0 \text{ and } g(1) \in F\}$ is Borel. Then so is its complement $T(A)^c := \{g \in \mathbf{C}\colon g(0) \ne 0 \text{ or } g(1) \in F^c\}$. For Polish spaces, the continuous image of a Borel set is an analytic set. Now $\pi_{01} \colon \mathbf{C}_{[0,\infty)} \to \mathbb R^2$ defined by $$ \pi_{01}(f) = (f(0),f(1)) $$ is continuous. So $$ G_1:= \{(x,y) \in \mathbb R^2 \colon x=0 \text{ and } y \in F\},\qquad G_2:= \{(x,y) \in \mathbb R^2 \colon x \ne 0 \text{ or } y \in F^c\}, $$ are both analytic sets in $\mathbb R^2$. But then cross-sections $$ \{y\colon (0,y) \in G_1\} = F \qquad\text{and}\qquad \{y\colon (0,y) \in G_2\} = F^c $$ are both analytic sets in $\mathbb R$, and therefore $F$ is Borel. This contradiction shows that our assumption that $T(A)$ is Borel is wrong.