Are $\overline{X}$, $\overline{Y}$ and $\overline{Z}$ independent? I have
$$
X_i, i=1,...,n_1,
$$
$$
Y_j, j=1,...,n_2,
$$
$$
Z_k, k=1,...,n_3,
$$
whereat $X_i, Y_j$ and $Z_k$ are all independent. Does then follow that 
$$
\bar{X}=\frac{1}{n_1}\sum_{i=1}^{n_1}X_i, \bar{Y}=\frac{1}{n_2}\sum_{=1}^{n_2}Y_j, \bar{Z}=\frac{1}{n_3}\sum_{k=1}^{n_3}Z_k
$$
independent, too?
 A: Yes. Use the definition of independence in terms of functions: random variables $X_i$'s are all mutually independent means that for every choice of measurable functions $f_i$'s,
$$
\mathbb{E}\left[ \prod_{i=1}^{n} f_i(X_i) \right] = \prod_{i=1}^{n} \mathbb{E} f_i(X_i)
$$
as soon as all the expectations are defined.
Try to use the same definition for the 3 r.v.'s $\bar{X},\bar{Y},\bar{Z}$ (for all choices of $f,g,h$) and see how it is implied by the independence of the $X_i,Y_j,Z_k$'s.
A: Yes.
So, $\bar{X},\bar{Y},\bar{Z}$ will be independent iff the corresponding $\sigma$-fields $\sigma(\bar{X}),\sigma(\bar{Y}),\sigma(\bar{Z})$ are independent. 
Now, $\sigma(\bar{X}) \subset \sigma(X_1,\ldots,X_{n1})$. The same goes for $\sigma(\bar{Y})$ and $\sigma(\bar{Z})$. Now, since all the $X_i,Y_j,Z_k$ are independent, $\sigma(X_1,\ldots,X_{n1}),\sigma(Y_1,\ldots,Y_{n2}),\sigma(Z_1,\ldots,Z_{n3})$ are independent. Hence their respective subsets $\sigma(\bar{X}),\sigma(\bar{Y}),\sigma(\bar{Z})$ are also independent. So you are done.
In fact, using the same logic, you can claim a much stronger result:
1. Take $K$ independent subsets of random variables (In your case $K=3$).
2. Apply any $K$ measurable functions to each subset separately. (sums, averages, products, and many more. You took the average function for each subset)
3. The resulting $K$ random variables will be independent.
I recently gave the measure theoretic definition of independence of random variables in this post:https://stats.stackexchange.com/questions/105102/if-a-random-variable-v-is-independent-of-two-independent-random-variables-x-and/105104#105104
Check it out.
