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If $\Omega$ is an uncountable sample space, why are we not allowed to take the sigma-algebra of events $\cal{F}$ to be the power set of $\Omega$?

The Wikipedia page devoted to Probability spaces states that this power set may be too large so that there may be events for which it is impossible to assign a unique probability measure.

(1) Why can't we assign probabilities to all of the events?

(2) Can't we just set most of the probabilities to be $0$ and deal with the probabilities that we can assign?

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    You are correct. For example, choose $\omega \in \Omega$. For every $A \in 2^{\Omega}$, let $P(A)=1$ if $\omega \in A$ and $P(A) = 0$ if $\omega \notin A$. This is a probability defined over $2^\Omega$. The problem is defining a probability over $2^\Omega$ with an uncountable support. – 2017-01-15

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The problem is not that you can not define a measure on power set of $\Omega$ (example by madprob above shows that you can), but that measures you can define can't have some nice properties you might need.

For example, to model "picking a random number from an interval $[0,1]$", you might need a measure $\mu$ such that

  1. $\mu([a,b])=b-a \; $ for $\; a,b \in [0,1] \;$ and,
  2. translation of subset $A \subset [0,1]$ by $x \in [0,1]$ defined as $$A \oplus x = \{ a+x \mid a \in A, a+x \le 1 \} \cup \{ a+x-1 \mid a \in A, a+x>1\}$$ does not change it's measure, i.e. $\mu(A)=\mu(A \oplus x)$.

It turns out that no such measure can be defined on power set of $[0,1]$, and you will have to contend yourself with such measure defined on some smaller $\sigma$-algebra, usually the $\sigma$-algebra of Borel sets.

The proof of this can be found in every measure theory textbook. It goes something like this: suppose $\mu$ is such measure, and observe equivalence classes of relation defined by $x \sim y$ iff $x-y \in \mathbb Q$ for $x,y \in [0,1]$. By Axiom of Choice, there is set $H$ consisting of precisely one representative of each equivalence class. We can write interval $[0,1]$ as disjoint union $$[0,1] = \bigcup_{x \in [0,1] \cap \mathbb Q} H \oplus x$$. By $\sigma$-additivity of $\mu$, $$\mu([0,1])=\sum_{x \in [0,1] \cap \mathbb Q} \mu(H \oplus x)$$ and by shift invariance all summands are equal. But this infinite sum can only equal 0 and $\pm \infty$, not 1.

ADDED: As madprob mentioned in a comment below, we could drop 2. (translation invariance) and there still would be no measure on the power set of $[a,b]$ satisfying 1. For example, Ulam proved in 1930 that if a measure is defined on all subsets of uncountable set, and all singletons have measure zero, then every set must also have measure zero (thus the only measures defined on power set of $\Omega$ are like the one in madprob's example).

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    I think that the problem is even more fundamental than the one you describe. That is, the lack of shift invariance is not the main reason why a measure is usually not defined over $2^{\mathbb{R}}$. Stronger than that, one can prove that every measure over $2^{\mathbb{R}}$ is discrete. – 2017-01-15
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    @madprob Thanks. – 2017-01-16