Definition of $\sigma(X_{1},X_{2},...,X_{n})$ I am looking at the definition of the sigma algebra generated by a collection of random variables. I do not see why a particular part of the definition is necessary. The definition I am using is as follows:
Suppose $X_{1},X_{2},...,X_{n}$ are RVs on $(\Omega,A,P)$. Set $C=\sigma (X_{1})\cup\sigma(X_{2})\cup ,...\cup\sigma(X_{n})$. Then $\sigma(C)$ is denoted $\sigma(X_{1},X_{2},...,X_{n})$ which is the smallest $\sigma$-algebra that makes the random variables $X_{1},X_{2},...,X_{n}:\Omega\rightarrow R$ measurable.
My question is, why is it necessary to use $\sigma (C)$ rather than just $C$ as our $\sigma$-algebra? It seems like any event $(X_{i})^{-1}(A)\in\sigma(X_{i})\subset C$ for $A\in B(R)$, making all $X_{i}$ measurable with respect to $C$.
EDIT: After thinking more about this I suspect the above does not work since $C$ does not have to be a $\sigma$-algebra, however if this is the case I'm not sure how to show this.
 A: $C$ itself is not a sigma algebra in general.  For example $X_1^{-1}(A)$ and $X_2^{-1}(B)$ are in it (if $A$ and $B$ are Borel sets) but there no reason why their intersection is in $C$. 
Example. Let $n=2,X=I_E$ and $Y=I_F$. Then $E$ and $F$ are in $C$ but $E\cap F$ need not be in $C$ since $\sigma (X) \cup \sigma (Y)=\{\emptyset, \Omega, E,E^{c}F,F^{c}\}$.
Let $D$ be any sigma algebra which makes each $X_i$ measurable. Then $\sigma(X_i) \subset D$ for all $i$ so $C \subset D$. This also implies  $\sigma (C) \subset D$.  Since $\sigma (C)$ is one sigma algebra which makes each $X_i$ measurable it follows that it is the smallest such sigma algebra. 
A: Let me add that it can be proved that:$$\sigma(X_1,\dots,X_n)=X^{-1}(\mathcal B(\mathbb R^n)):=\{X^{-1}(B)\mid B\in\mathcal B(\mathbb R^n)\}$$where $X:\Omega\to\mathbb R^n$ is prescribed by: $$\omega\mapsto (X_1(\omega),\dots,X_n(\omega))$$Here $\mathcal B(\mathbb R^n)$ denotes the Borel $\sigma$-algebra on $\mathbb R^n$ i.e. the smallest $\sigma$-algebra that contains all open sets if $\mathbb R^n$ is equipped with its usual topology.
In words: $\sigma(X_1,\dots,X_n)$ is the set of preimages of Borel-measurable subsets of $\mathbb R^n$ under $X$.
Also it is handsome to know that:$$\mathcal B(\mathbb R^n)=\mathcal B(\mathbb R)\otimes\cdots\otimes\mathcal B(\mathbb R)$$ 
which also needs a proof.
