Does the operator $T(f)(t) := f(t) - f(0)$ preserve measurability? Denote by $\mathcal{B}$ the Borel field on $\mathbb{R}$, denote by $\mathbf{C}_{\left[0,\infty\right)}$ the set of continuous, real-valued functions over the domain $\left[0,\infty\right)$ and denote by $\mathcal{B}_{\left[0,\infty\right)}$ the minimal $\sigma$-algebra that renders each projection function $\pi_t:\mathbf{C}_{\left[0,\infty\right)}\rightarrow\mathbb{R}$
$$
\pi_t\left(f\right) := f\left(t\right)
$$
$\mathcal{B}_{\left[0,\infty\right)}/\mathcal{B}$-measurable.

Consider the following operator $T: \mathbf{C}_{\left[0,\infty\right)} \rightarrow \mathbf{C}_{\left[0,\infty\right)}$
$$
T\left(f\right)\left(t\right) := f\left(t\right) - f\left(0\right)
$$

Let $A \in \mathcal{B}_{\left[0,\infty\right)}$.


Is $T\left(A\right) \in \mathcal{B}_{\left[0,\infty\right)}$?
If the answer to the first question is: "No", is it still "No" if $A$ is a tail event? ($B \in \mathcal{B}_{\left[0,\infty\right)}$ is a tail event iff $B \in \bigcap_{t \in \left[0,\infty\right)}\sigma\left(\pi_s :\mid s \in \left[t, \infty\right)\right)$.)




My attempts at solving this problem


Attempt #1

If $T$ were defined instead like this:
$$
 T\left(f\right) := f + c
 $$
for some constant $c \in \mathbb{R}$, in other words, if $T$ were a rigid translation of $f$, then $T$ would be measurable and invertible, its inverse being itself a translation, and therefore $T\left(A\right)$ would indeed belong to $\mathcal{B}_{\left[0,\infty\right)}$. Unfortunately, $f\left(0\right)$ is not constant, so this approach fails.
Attempt #2

Suppose we know that for all $f \in A$, $f\left(0\right) \in \mathbb{Q}$. Then
$$
 A = \bigcup_{q \in \mathbb{Q}}\underbrace{A \cap \left\{\pi_0^{-1} \in \left\{q\right\}\right\}}_{=: A_q}
 $$
and therefore
$$
 T\left(A\right) = \bigcup_{q \in \mathbb{Q}}T\left(A_q\right)
 $$
For each $q \in \mathbb{Q}$, $A_q \in \mathcal{B}_{\left[0\infty\right)}$ and $T\left(A_q\right) = A_q - q$, so by attempt #1, $T\left(A_q\right) \in \mathcal{B}_{\left[0,\infty\right)}$ and hence $T\left(A\right) \in \mathcal{B}_{\left[0,\infty\right)}$. Unfortunately, it may not be the case that for every $f \in A$, $f\left(0\right) \in \mathbb{Q}$, in which case this attempt fails.

However, the rationals are dense in $\mathbb{R}$ and in addition, every $f \in \mathbf{C}_{\left[0,\infty\right)}$ is uniquely determined by the values $f$ assumes at the non-negative rationals. Can we take advantage of this additional structure, together with the result of attempt #2, to solve question 1 in the affirmative? (In which case question 2 is automatically resolved as well.)
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.
Hi I think it is true that the operator $T$ that you defined is such that the image of any measurable set of $\mathcal{B}([0,+\infty))$ is a measurable set or as claimed that $T$ preserves measurability. 

The idea is to view things in a "topological" way first. If you "examine" carefully $T$ I think that you would shortly realize that it is a continuous linear application from one Banach spaces to another and that moreover it is surjective, from this, every ingredients are here for the Banach-Schauder theorem to apply unless mistaken. 

This proves that the image of every open sets of the space of continuous functions equipped with the sup norm (on every compacts) are open sets of the space at arrival which is naturally embedded in the very same space (you could refine the argument by defining the space of continuous functions that start at 0 but this doesn't really matter). 

Now let's reconcile topological and meausre theory, it is proven in Karatzas and Shreve "Brownian Motion and Stochastic Calculus" (may be as an exercise left to the reader I can't remember) that the $\sigma$-algebra defined the way you did by projection restricted to the space of continuous functions is the same $\sigma$-algebra that the Borelian $\sigma$-algebra of the space of continuous functions equipped with the sup norm (on every compacts). 

Hence I think we are done because $T$ preserves open sets which is a collection of sets the generated our $\sigma$-algebra $\mathcal{B}([0,+\infty))$, an application of the Monotone Class Theorem should be enough to conclude I think and I leave this part to you. 

Best regards