Can you prove that $\Bbb R$ is uncountable using the Lebesgue measure?

Question

I have been studying measure theory from the ground up, and am quite excited by the seeming power it holds. I thought of this last evening, and I wish to ask if the following proof of uncountability is fully rigorous:

Note: In the lecture notes I have been reading, to develop the measure-theoretic tools I use in this proof, the uncountability of $\Bbb R$ was not used, so I believe this avoids circular reasoning.

Any countable subset $Q$ of $\Bbb R$ admits an enumeration of points $\{q_n:n\in\Bbb N\}$, i.e. that $x\in Q\iff\exists n\in\Bbb N:q_n=x$, and $q_n=q_m\iff n=m$, by definition of countability. The Lebesgue outer measure $\mu^*$ is defined as the infimum of the volume of countable rectangular covers of that set; taking some arbitrary $\varepsilon\gt0$, one can construct the cover of "rectangles" $R_n=[q_n-\varepsilon\cdot2^{-(1+n)},q_n+\varepsilon\cdot2^{-(1+n)}]$. By construction, $\bigcup_{n\in\Bbb N}R_n\supseteq Q$, and $\sum_{n\in\Bbb N}\mu(R_n)=\varepsilon$ from the geometric series. The outer measure is defined with the infimum of covers, so $\mu^*(Q)\le\varepsilon$ as $\{R_n\}$ is a valid countable cover of $Q$. It follows from $\varepsilon$ as arbitrary that $\mu^*(Q)=0$, and also that $\mu(Q)=0$ - null sets w.r.t the outer measure are Lebesgue measurable. Now take any non-trivial closed interval $[a,b]$ in $\Bbb R$; as it is a "rectangle", it has Lebesgue measure $|b-a|\neq0$, (by definition of the outer measure, right?) Thus any closed interval in $\Bbb R$ cannot be countable, as otherwise it would have null measure (furthermore it cannot be reached as a countable limit of countable sets, since the measure is $\sigma$-additive and the set would still have $0$ measure). Thus the reals are uncountable, and this argument generalises easily to $\Bbb R^n,\Bbb C^n$.

Is this rigorously sufficient as a proof? It feels like cheating somehow, but I can't place my finger on why I feel that, and I certainly can't see any mistakes or circular reasoning.

Answer 1

You claim that $\mu^*([a, b]) = b - a$. But this is not obvious from the definition of $\mu^*$: For $S \in 2^{\mathbb{R}}$, $$\mu^*(S) = \inf\{\sum_{j = 1}^{\infty}L(I_j) : I_j \subset \mathbb{R} \text{ are intervals }, \bigcup_{j = 1}^{\infty}I_j \supset S\}.$$ From the above definition we get $\mu^*([a, b]) \leq L([a, b]) = b - a$ since $[a, b] \supset [a, b]$. The hard part is proving $\mu^*([a, b]) \geq b - a$. This requires special property of $\mathbb{R}$ because if you define an analogous outer measure $\mu_{\mathbb{Q}}^*$ on subsets of $\mathbb{Q}$, the result doesn't hold.

For $S \in 2^{\mathbb{Q}}$, define $$\mu_{\mathbb{Q}}^*(S) = \inf\{\sum_{j = 1}^{\infty}L(I_j) : I_j \subset \mathbb{Q} \text{ are intervals }, \bigcup_{j = 1}^{\infty}I_j \supset S\}.$$ The same argument that shows $\mu^*(\mathbb{Q}) = 0$ shows that $\mu_{\mathbb{Q}}^*(\mathbb{Q}) = 0$. Hence $\mu_{\mathbb{Q}}^*(I) = 0$ for any interval $I \subset \mathbb{Q}$.