Proving a mapping T from d-dimensional R to d-dimensional R is Borel measurable. Let $(i_1,...,i_d)$ be a permutation of $(1,...,d)$ and define $T:\mathbb{R}^d\rightarrow\mathbb{R}^d$ by $T(x_1,...,x_d)=(x_{i_1},...,x_{i_d}).$ Prove that $T$ is $\mathscr{B}(\mathbb{R}^d)/\mathscr{B}(\mathbb{R}^d)$-measurable.
I need to show that $T^{-1}(\mathscr{B}(\mathbb{R}^d))\subseteq\mathscr{B}(\mathbb{R}^d).$ Let $A=(a_1,....,a_d)\in\mathscr{B}(\mathbb{R}^d)$.
Doesn't the result just follow from the fact that any permutation of $A$ is always going to be another d-dimensional open set in $\mathscr{B}(\mathbb{R}^d)$? In other words, since $\mathscr{B}(\mathbb{R}^d)$ contains every open d-dimensional subset of $\mathbb{R}^d$ and any permutation of A is also in $\mathscr{B}(\mathbb{R}^d)$, shouldn't the inverse inclusion listed above be obvious?
What am I missing here? 
 A: You only have to prove that if $A\subseteq\mathbb{R}^d$ is open then $T^{-1}(A)\in\mathscr{B}(\mathbb{R}^d)$ (since $\mathscr{B}(\mathbb{R}^d)$ is the $\sigma$-algebra generated by the topology of $\mathbb{R}^d$).
Now, $T$ is a linear isomorphism and therefore is a homeomorphism.
Thus $T^{-1}(A)$ is open, which obviously implies $T^{-1}(A)\in\mathscr{B}(\mathbb{R}^d)$.

In fact, it can be more general.
If $(X, \mathscr{F})$ and $(Y,\mathscr{G})$ are measurable spaces in which $\mathscr{G}$ is the $\sigma$-algebra generated by some set $\mathcal{S}\subseteq 2^Y$, then a function $f\colon(X,\mathscr{F})\to(Y,\mathscr{G})$ is measurable if and only if $f^{-1}(S)\in\mathscr{F}$ for all $S\in\mathcal{S}$.
In particular, if $(X,\tau_X)$ and $(Y,\tau_Y)$ are topological spaces, $\sigma(\tau_X)$ (the $\sigma$-algebra generated by $\tau_X$) is by definition the Borel $\sigma$-algebra on $(X,\tau_X)$, respectively for $Y$. Therefore, a continuous function $f\colon(X,\tau_X)\to(Y,\tau_Y)$ is always a measurable function $f\colon(X,\sigma(\tau_X))\to(Y,\sigma(\tau_Y))$ since $f^{-1}(U)\in\tau_X\subseteq\sigma(\tau_X)$ for all $U\in\tau_Y$.
