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.)
 A: 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.
A: Let's try this for the tail field.
Notation
$\mathcal{B}_1$ the Borel sets for $\mathbb R$, a Polish space,
$\mathcal{B}_2$ the Borel sets for $\mathbb R^2$, a Polish space,
$\mathcal{B}_{\left[0,\infty\right)} = \sigma\big(\pi_t \colon t \in [0,\infty)\big)$ the Borel sets for $\mathbf{C}_{\left[0,\infty\right)}$,  a Polish space,
for $s > 0$, $\mathcal{T}_s = \sigma\big(\pi_t \colon t \ge s\big)$,
$\mathcal{T} = \bigcap_{s>0}\mathcal{T}_s$, the tail field.
There is $E \in \mathcal{B}_2$ such that $F := \{x-y\colon (x,y) \in E\} \notin \mathcal{B}_1$.  
Let $\phi$ be a sawtooth function, $\phi(t) = t$ for $0 \le t \le 1$, $\phi(t) = 2-t$ for $1 < t \le 2$, and periodic with period $2$.  Let $Z = \{a\phi \colon a \in \mathbb R\}$, a closed set in $\mathbf{C}_{\left[0,\infty\right)}$ (one-dimensional subspace), and thus $Z \in \mathcal{B}_{\left[0,\infty\right)}$.  
For $n \in \mathbb N$, let
$$
A_n := \left\{f \in \mathbf{C}_{\left[0,\infty\right)} \colon 
(f(2n+1),f(2n)) \in E, \text{ and for all } t \ge 2n, f(t+2)=f(t)\right\} .
$$
Because we can check the last part (periodic from $2n$ on) just using rational $t$, we have $A_n \in \mathcal{T}_{2n}$.  Also note $A_1 \subseteq A_2 \subseteq A_3 \subseteq \cdots$.  Let $A = \bigcup_n A_n$.  Then $A$ is a tail event, $A \in \mathcal{T}$.  
We claim $T(A) \notin \mathcal{B}_{\left[0,\infty\right)}$.  Suppose (for purposes of contradiction) that $T(A) \in \mathcal{B}_{\left[0,\infty\right)}$.  Then also its complement $T(A)^c \in \mathcal{B}_{\left[0,\infty\right)}$.  Therefore $Z \cap T(A) \in \mathcal{B}_{\left[0,\infty\right)}$ and $Z \cap T(A)^c \in \mathcal{B}_{\left[0,\infty\right)}$.  But, in fact,
$$
T(A) \cap Z = \{a\phi\colon a \in F\},\qquad
T(A)^c \cap Z = \{a\phi\colon a \in F^c\} .
$$
Now (in Polish spaces) the continuous image of a Borel set is analytic, so the two sets sets
$$
\pi_1\big(T(A)\cap Z\big) = F\qquad\text{and}\qquad
\pi_1\big(T(A)^c\cap Z\big) = F^c
$$
are both analytic in $\mathbb R$, and therefore $F$ is Borel.  This contradiction completes our proof.
A: The following is an addendum to GEdgar's answer, aimed to clarify some points for my future reference, as well as for the sake of other readers who, like me, do not find these points self evident. The principal results are propositions 6 and 7 below.
Notation
We denote the borel field on the real line by $\mathcal{B}$, the Borel field on $\mathbb{R}^2$ by $\mathcal{B}_2$ and in general for any $n \in \mathbb{N}_1$ the Borel field on $\mathbb{R}^n$ by $\mathcal{B}_n$. When convenient, we take $\mathbb{R}^2$'s elements to be column vectors.

Lemma 1
A plane Borel set exists, whose projection on the $x$ coordinate is not a Borel set.
Proof
This is known as Suslin's Theorem. Q.E.D.

Lemma 2
If $A \in \mathcal{B}_2$ and $y \in \mathbb{R}$, then the section $A_y := \left\{x \in \mathbb{R} \mid: (x,y) \in A\right\}$ is Borel (i.e. $A_y \in \mathcal{B}$).
Proof
See Theorem A (p. 141) in Chapter 34 ("Sections") of Halmos, Paul R., "Measure Theory", Springer, 1974. Q.E.D.

Lemma 3
a. If $T:\mathbb{R}^m \rightarrow \mathbb{R}^n$ for some $m,n \in \mathbb{N}_1$ is linear, then for all $A \in \mathcal{B}_n$, $T^{-1}(A) \in \mathcal{B}_m$.
b. If $T:\mathbb{R}^n \rightarrow \mathbb{R}^n$ for some $n \in \mathbb{N}_1$ is linear and invertible, then for all $A \in \mathbb{R}^n$, $A \in \mathcal{B}_n \iff T(A) \in \mathcal{B}_n$.
Proof
A linear transformation between finite dimensional linear spaces is continuous, and continuous functions are Borel. Q.E.D.

Lemma 4
Let $A \subseteq \mathbb{R}$. Then $A \in \mathcal{B}$ iff $\left\{(x,0) :\mid x \in A\right\} \in \mathcal{B}_2$.
Proof
Suppose $A \in \mathcal{B}$. Denote by $T:\mathbb{R}^2 \rightarrow \mathbb{R}$ the projection of $\mathbb{R}^2$ on the $x$-coordinate: $T(v) := Pv$ with
$$
P := \left[\begin{array}{cc}
1 & 0
\end{array}\right]
$$ 
Then $T^{-1}(A) \in \mathcal{B}_2$ (Lemma 3a). Then $\left\{(x,0) :\mid x \in A\right\} \in \mathcal{B}_2$, as the intersection of $T^{-1}(A)$ with the $x$-axis, which is $\in \mathcal{B}_2$.
Conversely, suppose that $\left\{(x,0) :\mid x \in A\right\} \in \mathcal{B}_2$. Define the linear transformation $F:\mathbb{R}\rightarrow\mathbb{R}^2$ thus:
$$
F(x) := (x, 0)
$$
Then $A = F^{-1}\left(\left\{(x,0) :\mid x \in A\right\}\right) \in \mathcal{B}$ (Lemma 3a). Q.E.D.

Lemma 5
If $A \in \mathcal{B}$, then $\left\{(x,-x) :\mid x \in A\right\} \in \mathcal{B}_2$.
Proof
Define $B := \left\{(x, 0) :\mid x \in A\right\}$. Then $B \in \mathcal{B}_2$ (Lemma 4). Denote by $T:\mathbb{R}^2\rightarrow\mathbb{R}^2$ the linear transformation $T(v) := Rv$ with
$$
R := \left[\begin{array}{cc}
1 & 0 \\
0 & -1
\end{array}\right]
$$
$T$ is invertible. Hence $\left\{\left(x,-x\right) :\ x \in A\right\} = T(A) \in \mathcal{B}_2$ (Lemma 3b). Q.E.D.

Proposition 6
There is a Borel set $E$ in $\mathbb{R}^2$ such that its difference set $F := \{x-y :\mid \left(x,y\right) \in E\}$ is not a Borel set.
Proof
Let $B \in \mathcal{B}_2$ be such that $B$'s projection on the $x$-coordinate is not Borel (Lemma 1). Consider the projection of $B$ on the $x$-axis, $PB$ with
$$
P = \left[\begin{array}{cc}
1 & 0 \\
0 & 0
\end{array}\right]
$$
Then $PB \notin \mathcal{B}_2$ (Lemma 4).
Define $E$ to be the rotation of $B$ $45^\circ$ clockwise, i.e. $E := AB$, where
$$
A := \left[\begin{array}{cc}
\sqrt{2} & \sqrt{2} \\
-\sqrt{2} & \sqrt{2}
\end{array}\right]
$$
We have $E \in \mathcal{B}_2$ and $\frac{1}{2}APA^{-1}E \notin \mathcal{B}_2$ (Lemma 3b). But
$$
\frac{1}{2}APA^{-1} = \left[\begin{array}{cc}
1 & -1 \\
-1 & 1
\end{array}\right]
$$
Hence
$$
\frac{1}{2}APA^{-1}E = \left\{\left(x - y, -(x - y)\right) :\mid \left(x,y\right) \in E\right\} = \left\{(x,-x) :\mid x \in F\right\}
$$
So $F \notin \mathcal{B}$ (Lemma 5). Q.E.D.

Proposition 7
Using the definitions used in GEdgar's answer, $T(A) \notin \mathcal{B}_{[0,\infty)}$.
Proof
Define a function $g:\mathbb{R}\rightarrow\mathbf{C}_{[0,\infty)}$ thus: for every $t \in \mathbb{R}$ and every $x \in \left[0,\infty\right)$, $g(t)(x) := tx$.
$g$ is $\mathcal{B}/\mathcal{B}_{[0,\infty)}$-measurable. Therefore, had $T(A)$ been in $\mathcal{B}_{[0,\infty)}$, we would have $g^{-1}\left(T(A)\right) \in \mathcal{B}$. But $g^{-1}\left(T(A)\right) = F$. Q.E.D.
A: 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
