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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

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    $\begingroup$ No, your reasoning is not correct. A set of positive measure does not necessarily contain a (non-trivial) open interval (consider e.g. $[0,1] \backslash \mathbb{Q}$). $\endgroup$ – saz Jan 11 at 15:24
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    $\begingroup$ See this. $\endgroup$ – David Mitra Jan 11 at 15:33
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    $\begingroup$ No, a closed set need not contain a rational number. $\endgroup$ – David C. Ullrich Jan 11 at 15:35
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    $\begingroup$ 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 $m(V)<\infty$. $\endgroup$ – David C. Ullrich Jan 11 at 15:38
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    $\begingroup$ 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. $\endgroup$ – Dave L. Renfro Jan 11 at 16:12
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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...)

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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.
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