Definition of multivariate random variable Let $(\Omega,\mathcal E,P)$ be  a probability space and $X_1,\dots,X_n$ random variables on $(\Omega,\mathcal E, P)$.
Then I can define the vector $X=(X_1,\dots,X_n)\colon \Omega \to \mathbb R ^n$ and call it multivariate random variable (and this comes from wikipedia).
Now, let us denote as $\mathcal B^1$ the borel sets of $\mathbb R$, so that $X_i^{-1}(B)\in \mathcal E$ for each $B\in \mathcal B^1$ for each $1\leq i \leq n$, since $X_i$ is a RV for each $i$. 
My question is: does something similar happen for $X$ defined as above? I mean, is it true that if I take a borel set $I\in \mathcal B^n$ then $X^{-1}(I)\in \mathcal E$, provided that $X$ is simply defined as a vector of one-dimensional random variables?
I guess it is, but I can't say exactly why. Can someone help figuring out this?
 A: If $X=(X_1,\ldots,X_n)$, then $X:\Omega\to \mathbb{R}^n$ is $(\mathcal{F},\mathcal{B}(\mathbb{R}^n))$-measurable  if and only if $X_i:\Omega\to \mathbb{R}$ is $(\mathcal{F},\mathcal{B}(\mathbb{R}))$-measurable for $i=1,\ldots,n$. 
To see why this is true, just note that if $X$ is measurable
$$
X_i^{-1}(B)=X^{-1}(\mathbb{R}\times\cdots\times B\times\cdots\times\mathbb{R})\in\mathcal{F}
$$
for all $B\in\mathcal{B}(\mathbb{R})$. Here $B$ is on the $i$th position. This shows that $X_i$ is measurable for all $i$.
On the other hand, if all $X_i$'s are measurable, then
$$
X^{-1}(B_1\times \cdots \times B_n)=\bigcap_{i=1}^n X_i^{-1}(B_i)\in\mathcal{F}
$$
for all $B_1,\ldots,B_n\in\mathcal{B}(\mathbb{R})$ showing that $X$ is measurable.
A: Sorry I misread the question.
The answer is still true; in fact it is clear that if $I \in \mathcal B^n$ is in the form $I = \prod (-\infty, a_i]$, $X^{-1}(I) \in \mathcal E^n$
Plus, you know that the sets in that form generate $\mathcal B^n$ (written: $\mathcal B^n = \sigma \left(\prod(-\infty, a_i]\right)$ )
And you also know that the fact that $X$ is measurable with respct to a subset of $\mathcal B^n$ that generates $\mathcal B^n$ is equivalent to the fact that $X$ is measurable with respect to $\mathcal B^n$, hence you are done.
If it is not clear tell me and I'll try to fix it!
