0
$\begingroup$

Theorem: If $A \subset \mathbb R$ and every subset of $A$ is Lebesgue measurable then $m(A)=0$

Corollary: Every set of positive measure has non-measurable subsets

$m$ in here denote Lebesgue measure.

Why such corollary is true?

$\endgroup$
  • 1
    $\begingroup$ What is your question here? Why the second assertion is indeed a corollary of the first one? Or why the first assertion is true? $\endgroup$ – TZakrevskiy Apr 11 '15 at 14:03
  • $\begingroup$ Why the second assertion is indeed a corollary of the first one? $\endgroup$ – SamC Apr 11 '15 at 14:05
  • $\begingroup$ It's the contrapositive of the preceding theorem. $\endgroup$ – MJD Apr 11 '15 at 14:22
3
$\begingroup$

You have the following statements: $$P=\text{"All subsets of $A$ are Lebesgue measurable"}\\Q="m(A) =0".$$ The theorem says that $P$ implies $Q$, in other words, the logical formula $P\to Q$ is always true. On the other hand, we can always deduce from that the logical formula $\neg Q \to \neg P$ is also always true (if presence of $P$ implies presence of $Q$, then absence of $Q$ implies absence of $P$).

Now return to our meanings of $P$ and $Q$: $$\neg P=\text{" $A$ has non-measurable subsets"}\\\neg Q="m(A) >0".$$

Therefore, the corollary: "Sets of positive measure have non-measurable subsets".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.