This question was inspired by the well-known fact that there's no uniform probability distribution on a countably infinite set, because this would contradict $\sigma$-additivity of the probability measure.
I can't have an countably infinite number of equal positive numbers add up to $1$, and the sum of a countably infinite number of zeroes, is still zero.
But $\sigma$-additivity would be no obstacle to a probability measure on $\omega_1$, the first uncountable ordinal, which gave points probability zero, since $\omega_1$ isn't a union of countably many points. I had naively thought, in fact, that I could pull back the uniform distribution on $[0, 1]$ to a continuous distribution on $ω_1$ as follows:
List off all the dyadic intervals in the following order: $$I_0 := [0, 1];$$ $$I_1 := [0, 0.5], I_2 := [0.5, 1];$$ $$I_3 := [0, 0.25], I_4 := [0.25, 0.5], I_5 := [0.5, 0.75], I_6 := [0.75, 1];$$ ...etcetera, following the same pattern.
Define an injection of $\omega_1$ into $[0, 1]$ using the Axiom of Choice as follows. For each nonnegative integer $n$, pick an arbitrary element $x_n$ of the $n$th interval, different from any that came before.
Then, for each ordinal of the form $ω + n$, $n \geq 0$, pick an element $x_{\omega + n}$ of the $n$th interval different than any previous choice.
Then the same for each ordinal of the form $ω \cdot 2 + n$, $n \geq 0$.
...
Then the same for each ordinal of the form $ω^2 + n = \omega \cdot \omega + n$, $n \geq 0$.
Then the same for each ordinal of the form $ω^2 + ω + n = (ω + 1) \cdot ω + n$, $n \geq 0$.
...
And so on by transfinite induction; for every non-finite countable ordinal $\alpha$ and all nonnegative integers $n$, we are choosing a new element $x_{\alpha \cdot ω + n}$ of the $n$th interval to correspond to the ordinal $\alpha \cdot ω + n$, which is always possible since all of our original intervals are uncountable.
Eventually our construction yields an injection $$\iota: \omega_1 \to [0, 1],$$ whose image $\mathcal{S} := \iota( \omega_1 )$ is an uncountable dense set which has the property that $\mathcal{S} \cap (a, b)$ is also uncountable (i.e. $\iota^{-1}((a, b))$ is cofinal in $\omega_1$) for all $0 \leq a < b \leq 1$.
My intuition was that I could just define a probability distribution $\Bbb{P}: \omega_1 \to [0, 1]$ using the $\sigma$-algebra on $\omega_1$ generated by the preimages $\iota^{-1}((a, b)) \subset \omega_1$, with the obvious probability assignment $$\Bbb{P}(\iota^{-1}((a, b))) = b - a,$$ regardless of the particular injection I constructed, or the individual choices of $x_{\alpha \cdot \omega + n}$.
My reasoning was, once I have an uncountable, dense subset of $[0, 1]$, any cover of that set by a countable set of open intervals (which seemingly, would combine to form $[0, 1]$ minus the countably many endpoints of those intervals, a set of full measure), would have to have Lebesgue outer measure $1$, regardless of whether $\mathcal{S}$ itself was measurable or not. (I know this wouldn't work if $\mathcal{S}$ was countable because I could just list the points of $\mathcal{S}$ off, cover the $n$th point of $\mathcal{S}$ by an open interval of length $\epsilon \cdot 2^{-n}$, and get an outer measure of $\epsilon$.)
However, the Strong Law of Large Numbers/Borel's normal number theorem furnishes multiple simple examples of uncountable dense sets whose intersection with any subinterval of $[0, 1]$ is also uncountable dense, yet which have Lebesgue measure $0$.
For instance, if I accidentally constructed $\mathcal{S}$ so that all $x \in \mathcal{S}$ had the digit $1$ appear twice as frequently as $0$ in their binary expansion (asymptotically speaking), then this clearly couldn't work because my "pullback" probability would have to be zero on every subset of $\omega_1$.
My question is, what am I missing? Clearly my intuition on the outer measure is leading me astray here. If someone can clarify the mistake in my thinking, I'd be much obliged.
Edit: To clarify, I did not expect that every subset of $\omega_1$ would have a probability/be measurable under the pullback, as explicitly stated:
My intuition was that I could just define a probability distribution $\Bbb{P}: \omega_1 \to [0, 1]$ using the $\sigma$-algebra on $\omega_1$ generated by the preimages $\iota^{-1}((a, b)) \subset \omega_1$
By referencing a particular $\sigma$-algebra I intended to clarify which subsets of $\omega_1$ would be measurable under $\Bbb{P}$. I hope that clears up any further confusion.