$X_1,...,X_n$ be independent RVs and $X_i \perp \mathcal F $ for $1\ \leq \forall i \leq n$. Show that $\sigma (X_1,...,X_n) \perp \mathcal F$ 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.
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 $n&gt;2$.




  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!