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
 A: 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...)
A: Hints: (I take it that you are considering $\mathbb{R}$ equipped with the Lebesgue measure $\lambda$.)


*

*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 $k \in \mathbb{N}$.

*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.

