Question: Let $X\_1,...,X\_n$ be mutually independent RVs.  

Suppose $X\_i \perp \mathcal F $ for $1\ \leq \forall i \leq n$. 

How can I show that:

$$\sigma (X\_1,...,X\_n) \perp \mathcal F$$ ?

What I have tried:

$\sigma (X\_1...X\_n) = \sigma(\cup\_{i=1} ^n \sigma(X\_i))$ holds. I think I need some Dynkin's lemma-ish argument but can't see how.

Answer: 
I doubt the conclusion. I remember that there is something like "pairwise independent does not imply independent". 
More precisely, let $X\_1$, $X\_2$, $X\_3$ be random variables such that for any $i\neq j$, $X\_i$ and $X\_j$ are independent. However, $X\_1$, $X\_2$, $X\_3$ need not be independent. If my memory is correct, we put $\mathcal{F}=\sigma(X\_3)$, then we will arrive a contradiction. You may consult any textbook or search "pairwise independent does not imply independent".

I remember that there is such a counter-example:

There is a probability space $(\Omega,\mathcal{F},P)$ and $A\_{i}\in\mathcal{F}$,
for $i=1,2,3$ such that $P(A\_{i}A\_{j})=P(A\_{i})P(A\_{j})$ whenever
$i\neq j$. However, $P(A\_{1}A\_{2}A\_{3})\neq P(A\_{1})P(A\_{2})P(A\_{3})$.

Then we define $X\_{i}=1\_{A\_{i}}$, for $i=1,2,$ and define $\mathcal{G}=\sigma(X\_{3})$.
Obviously, $X\_{1}$ and $X\_{2}$ are independent and $X\_{i}\perp\mathcal{G}$.
However, it is false that $\sigma(X\_{1},X\_{2})\perp\mathcal{G}$.
For, if it is true, then $A\_{1}A\_{2}\in\sigma(X\_{1},X\_{2})$ and $A\_{3}\in\mathcal{G}$,
then $P(A\_{1}A\_{2}A\_{3})=P(A\_{1}A\_{2})P(A\_{3})=P(A\_{1})P(A\_{2})P(A\_{3})$.
I'll answer for $n=2$. This should be sufficient to show that if it's not true for $n>2$.

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> 
> If $\sigma(X),\sigma(Y)$ and $\mathscr {F}$ are independent, then $f(X,Y)$ is independent from $\mathscr{F}$.
> 
> 
> 

In this respect, your question is similar to asking about pairwise independence. If $\sigma(X)$ and $\mathscr {F}$ are independent and $\sigma(Y)$ and $\mathscr {F}$ are independent, then it is not sufficient to conclude $\sigma(X,Y)$ is independent from $\mathscr{F}$. Of course $\sigma(f(X,Y))$ merely $\subseteq \sigma(X,Y)$. That's gotta be some $f$ to achieve independence.

> 
> If $\sigma(X,Y) = \sigma(\sigma(X) \cup \sigma(Y))$ and $\mathscr {F}$ are independent, then $f(X,Y)$ is independent from $\mathscr{F}$.
> 
> 
> 

In this respect, your question is similar to asking if the conclusion holds simply for $\sigma(X) \cup \sigma(Y)$ and $\mathscr {F}$ being independent: $\sigma(X) \cup \sigma(Y)$ isn't even a $\sigma$-algebra!