# Why every set of positive measure has non-measurable subsets

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?

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

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 $Q$: $$\neg P=\text{" A has non-measurable subsets"}\\\neg Q="m(A) >0".$$