sigma field generated by a field Suppose $F$ is a field, a collection of subsets of a set $\Omega$.
Define $\sigma(F)$ as the smallest $\sigma$-field containing the sets of $F$.
My Question:
(1) What is the correct way of thinking about $\sigma(F)$? The way I think about is... by definition of $\sigma$-field, if we replace the closer under finite union with closure under countable union, so this would make $F$ a $\sigma$-field. To make it the smallest, we add the definition of it being in any sigma-field of $F$.
(2) Why do we care about the "smallest"?
(3) I am studying a section which states its goal is to show that a $\sigma$-finite measure on a field $F$ has a unique extension to the minimal $\sigma$-field over $F$, namely $\sigma(F)$. Why is it that we are interested in showing this? For example, why does it have be $\sigma$-finite instead of just some finite measure? In addition, why does it have to be that it extends uniquely to the minimal sigma field instead of just any sigma field?
A pedagogical answer would be appreciated.
 A: Constructions like these appear in every topic of mathematics. Let us start by considering the most elementary example that you know: linear algebra. In linear algebra, we construct the span $\langle S \rangle$ of a set $S$ which is per definition the smallest vector space containing $S$. In much of the same flavour, we are interested in $\sigma(F)$.
Maybe this has already answered your questions. To be more concrete:
(1) Really the best way to think about $\sigma(F)$ is its definition: the smallest sigma-algebra containing $F$. Of course, we can try to think of it via a constructive manner but that is a bit much, needing ordinal numbers of steps as I can recall. For instance, the Borel sets $\mathscr{B}^n$ being the sigma algebra of open sets in $\mathbb{R}^n$ are very complicated to describe and best thought of just what it is: the smallest-sigma algebra containing the open sets of $\mathbb{R}^n$.
(2) Why would we not care about "smallest"? If we didn't care about that, there would be a trivial way to construct a sigma-algebra containing $F$: just take $\mathcal{P}(F)$, the power set. So now you may ask this: why not just always take $\mathcal{P}(F)$? Isn't this much easier? The subtlety is that we want to endow a measure on the sigma-algebra and usually $\mathcal{P}(F)$ is way too large to define a useful measure on it. But we can throw out many of the sets in $\mathcal{P}(F)$ since we are not interested in them. It suffices to look at $\sigma(F)$, we discard extraneous information. Here, we can do much more.
(3) Again, why should it not be interesting to show the existence and uniqueness of $\sigma(F)$? Such procedures show up everywhere in mathematics. Hopefully, I've convinced you in (2) that existence is of much interest. Why uniqueness? Well, it would be annoying if there are many different smallest sigma-algebras, wouldn't it? About the sigma-finiteness: good question. (Maybe I'm recalling it wrong but I think people have tried finiteness first, historically.) One of the goals in measure theory is to ultimately construct the Lebesgue measure. Ultimately, we want to approximate shapes in $\mathbb{R}^n$ with cuboids - with infinitely many to get the right precision. That's one of the reasons finiteness is not enough.
