TITLE: What can go wrong if we let sigma algebra to admit the union of uncountable union of elements?
QUESTION [5 upvotes]: By definition we only allow the union of countable infinite of elements to be also include the $\sigma$ field, why not uncountable many? Is there a historical view behind this？

REPLY [2 votes]: You can show, that it's not possible to construct a measure $\lambda$ defined on all the subsets of $\mathbb{R}$ such that the following natural assumptions hold:

For intervals $[a,b]$, you have $\lambda([a,b])=b-a$.
Translation Invariance: $\forall X\subset\mathbb{R}$ and $\forall t\in\mathbb{R}$ you have $$\lambda(X+t)=\lambda(\{x+t | x\in X\})=\lambda(X)$$
Additivity: For any sequence $(X_i)_{i\in I}$ of disjoint sets in $\mathbb{R}$ you have $$\lambda\Bigl(\bigcup_{i\in I} X_i \Bigr)=\sum_{i \in I}\lambda(X_i)$$

And therefore you need to have some restrictions on your sets (at least if you want to keep the assumptions about your measure, which seems like a good idea)
Counter example (vitali sets):
Define the following equivalence relation on $\mathbb{R}$:
$$x \sim y \quad \text{iff} \quad x-y \in \mathbb{Q}$$
Then, by the axiom of choice, there is a map $f\colon \mathbb{R} / \sim \to \mathbb{R}$ which assigns a representant to each equivalence class. You can assume that $X=f(\mathbb{R} / \sim)\subset [0,1]$ (you can just "add integers").
Now look at $\lambda(X)$:

If $\lambda(X)$ would be equal to $0$, you would have $$\lambda(\mathbb{R})=\lambda \Bigl(\bigcup_{\mathbb{Q}}(t+X) \Bigr)=\sum_{\mathbb{Q}}\lambda(X)=0$$ by the translation invariance and the countability of $\mathbb{Q}$.
If $\lambda(X)=c>0$, you get $$\lambda \Bigl(\bigcup_{\mathbb{Q}\cap  [0,1]}(t+X) \Bigr)=\sum_{\mathbb{Q}\cap  [0,1]} c = +\infty$$ But you also have $$\lambda \Bigl(\bigcup_{\mathbb{Q}\cap  [0,1]}(t+X) \Bigr)\leq \lambda([0,2])=2$$ and therefore another contradiction.