# If $\{X_\alpha\}$ is a collection of mutually disjoint meas. subsets of $\mathbb{R} \rightarrow$ then at most countable of them has positive measure.

Let $$\{X_\alpha\}_{\alpha \in I}$$ be a collection of mutually disjoint measurable subsets of $\mathbb{R}. Show that at most countable of them has positive measure. I want to see if my proof is correct. Proof: Let $$\{X_\alpha\}_{\alpha \in \Gamma \subset I}$$ be the subcollection such that $$m(X_\alpha) > 0, \forall \alpha \in \Gamma$$. Since each set is measurable and has positive measure, $$\forall X_{\alpha}, \alpha \in \Gamma, \exists O_{\alpha} = (a_\alpha,b_\alpha) \mbox{ such that } O_\alpha \subset X_{\alpha}$$ Now we must show that $$|\Gamma| \leq |\mathbb{N}|$$ Let $$f: \{O_{\alpha}\}_{\alpha \in \Gamma} \to \mathbb{Q}$$ be the function that $$f(O_\alpha) = q_{\alpha}, \mbox{ where }q_\alpha \in (a_\alpha,b_\alpha)$$ This is possible because $$\mathbb{Q}$$ is dense in $$\mathbb{R}$$ This function is clearly injective since $$\{O_\alpha\}$$ is a disjoint collection of open sets. Hence, the same rational can't be in two different open intervals from this collection. Since $$f$$ is injective, $$\{O_\alpha\}_{\alpha \in \Gamma}| = |\Gamma| \leq |\mathbb{Q}| = |\mathbb{N}| = \aleph_0$$ Q.E.D • No, your reasoning is not correct. A set of positive measure does not necessarily contain a (non-trivial) open interval (consider e.g.$[0,1] \backslash \mathbb{Q}$). – saz Jan 11 at 15:24 • See this. – David Mitra Jan 11 at 15:33 • No, a closed set need not contain a rational number. – David C. Ullrich Jan 11 at 15:35 • For example, say the ratinoals are$q_1,\dots$. Let$V=\bigcup(q_n-1/n^2,q_n+1/n^2)$,$F=\Bbb R\setminus V$. Then$V$is open so$F$is closed,$V$contains every rational so$F$contains no rational, and$F\ne\emptyset$since$m(V)<\infty\$. – David C. Ullrich Jan 11 at 15:38
• FYI, you might find it interesting to observe how the main idea used in the proofs by David C. Ullrich and saz can also be used to show that: (a) a monotone function can have at most countably many jump discontinuities; (b) in a "convergent uncountable sum" of non-negative real numbers, at most countably many of the numbers can be nonzero. – Dave L. Renfro Jan 11 at 16:12

Hints: (I take it that you are considering $$\mathbb{R}$$ equipped with the Lebesgue measure $$\lambda$$.)
1. Let $$(\Omega,\mathcal{A},\mu)$$ be a probability space. Show that if $$(A_{\alpha})_{\alpha \in I}$$ is a collection of measurable mutually disjoint sets, then $$\{\alpha; \mu(A_{\alpha})>0\}$$ is countable. Hint: Check that $$\{\alpha; \mu(A_{\alpha}) > 1/k\}$$ is a finite set for any $$k \in \mathbb{N}$$.
2. Now let $$(X_{\alpha})_{\alpha \in I}$$ be a collection of measurable mutually disjoints subsets of $$\mathbb{R}$$. Define $$A_{\ell,\alpha} := X_{\alpha} \cap [\ell,\ell+1], \qquad \ell \in \mathbb{Z}.$$ Use Step 1 to prove that $$\{\alpha; \lambda(A_{\ell,\alpha})>0\}$$ is countable for each $$\ell$$. Conclude that $$\{\alpha; \lambda(X_{\alpha})>0\}$$ is countable.
No, that's totally wrong. Saying $$X$$ has positive measure does not imply $$X$$ contains an open interval.
Hint: For $$n=1,2\dots$$ let $$\Gamma_n=\{\alpha:m([-n,n]\cap X_\alpha)>0\}.$$Show that each $$\Gamma_n$$ is countable and that $$\{\alpha:m(X_\alpha)>0\}=\bigcup_{n=1}^\infty\Gamma_n.$$
(To show $$\Gamma_n$$ is countable: Show that $$\{\alpha:m([-n,n]\cap X_\alpha)>1/k\}$$ is finite...)