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I'm reading about Construction of Probability Measures. During discussion of the book, it says:

Suppose $\Omega=[0,1],$ and let us define a set function $P$ on intervals of the form $P((a,b])=b-a,$ where $0\leq a\leq b\leq 1.$ This is a natural "probability measure" that assigns the usual length of an interval as its probability. Suppose we want to extend $P$ in a unique way to $2^{[0,1]},$ all subsets of $[0,1]$ such that

$(i) P(\Omega)=1;$ and $(ii)P(\cup_{n=1}^{\infty}A_{n})=\sum_{n=1}^{\infty}P(A_{n})$ for any disjoint sequence of subsets $(A_{n})_{n\in\mathbb{N}}.$ Then there isn't such $P.$

Why does not such $P$ exist? Why the collection $2^{[0,1]}$ is to big for this work?

I don't get the reasons of such probability doesn't exist. Maybe because each probability $P$ defined on an algebra $\mathcal{A}_{0}$ has a unique extension on $\mathcal{A}=\sigma(\mathcal{A}_{0}),$ but I don't get it clear.

Any kind of help is thanked in advance.

Squird37
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    I believe such an extension can be made (at least under some scary set-theoretic assumption like existence of large cardinals), though this is a very nontrivial result, see here. What's true is that there is no extension which is translation invariant. For that, look up Vitali's set. – Wojowu Dec 28 '17 at 21:35
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    For a good explanation of the issue with a translation invariant extension shown by Vitali's set you might want to look at: https://math.stackexchange.com/a/1869015/256128 – Rhys Steele Dec 28 '17 at 21:36
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    If the extension were unique, it would be translation invariant. – Robert Israel Dec 28 '17 at 21:37

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