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Let $\Omega$ be the set of all finite sequences of zeros and ones. Find a probability measure $\mathbb{P}$ on $\mathcal{A}=\mathcal{P}(\Omega)$, such that for each finite sequence $\omega \in \Omega$, $\mathbb{P}(\{\omega\}) > 0$

I know that, for a valid probability measure, I need to establish $\sigma$-additivity and make sure that $\mathbb{P}(\Omega)=1$.

My idea so far was to define $\mathbb{P}(\{\omega\})$ as "Probability that $\omega$ starts with $0$" which I thought to be $\mathbb{P}(\{\omega\}) = \frac{1}{2}$, but I don't know if this assumption is correct or how to mathematically argue for it.

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What if you assigned probability mass $2^{-2n}$ to each unique sequence of length $n$? We can define $\ell(\omega)$ to be the length of $\omega$, and let $\mathbb{P}(\{\omega\}) = 2^{-2\ell(\omega)}$ and $\mathbb{P} : 2^\Omega\to [0, \infty)$ be defined by $\mathbb{P}(S) = \sum_{\omega\in S} \mathbb{P}(\{\omega\})$ (this is well-defined, as the sum is unconditionally convergent). Then, the total measure of $\Omega$ would be $$\sum_{n=1}^{\infty} \frac{\text{# of length-}n\text{ sequences}}{2^{2n}} = \sum_{n=1}^{\infty} \frac{2^n}{2^{2n}} = \sum_{n=1}^{\infty} 2^{-n} = 1$$ We have therefore defined $\mathbb{P}$ such that it is $\sigma$-additive and $\mathbb{P}(\Omega) = 1$.

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Your example would not really work, if $\omega$ does not start with a 0 then $\mathbb P(\{\omega\}) = 0$.

The key thing to note is that the set of all finite sequences of ones and zeros, $\Omega$, is a countable set. This makes it much easier to define a probability measure. Can you think of another countable space with a probability measure on the space that give positive mass to singleton sets? If so, then can you see how can you use it to define a probability measure on $\Omega$?

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