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Exercise from Royden's Real Analysis:

Show that if $E\subset \Bbb R$ has a finite measure and $\epsilon>0$ then $E$ is the disjoint union of a finite number of measurable sets each of which has measure at most $\epsilon$.

Surely if $E$ is expressed as a finite union of disjoint measurable sets then $E=E_1\cup E_2\cup....\cup E_n$ where $E_i\cap E_j=\emptyset \implies m^*(E)=\sum_{i=1}^n m^*(E_i)$ then if any of the $E_i's$ have outer measure greater then $\epsilon$ then $m^*(E)>\epsilon$ which is false .

Hence $m^*(E_i)<\epsilon\forall i$.

But I am unable to prove that $E$ is the disjoint union of a finite number of measurable sets.Please help or give some hints .

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  • $\begingroup$ Is this true? If $X=\{1,\ldots, n\}$ and $\mu(X=i)=\mu_i>0$ then if we take $E = \{i_j\}_{j=1}^{q}\subseteq X$, then the finest partitioning for it will be $E=\cup_j \{i_j\}$, but it will not necessarily be $\mu_{i_j}<\epsilon$. I believe that it should be assumed that the space is not discrete and the measure does not have any atoms (i.e., meas. sets $A$ with $\mu(A)>0$ so that $B\subset A$ implies $\mu(B)=0$. $\endgroup$ – Pantelis Sopasakis Feb 17 '16 at 3:32
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You mentioned $m$, so I am assuming you are talking about the Lebesgue measure $m$ on $\mathbb{R}^k$.

Given $\epsilon > 0$, the entire space can be written as the union of countably many disjoint $k$-cells $W_n$ such that $m(W_n) < \epsilon$ for all $n$. Since $m(E) < \infty$, we can choose compact $K\subset E$ such that $m(E-K) < \epsilon$. $K$ is bounded, hence covered by finitely many $W_n$, thus the proof.

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