3
$\begingroup$

I want to prove that for Lebesgue (outer) measure (for the real line)

$$m(A) = \inf \lbrace \sum l( (a_i,b_i] |A \subset \cup_{i} (a_i, b_i] \rbrace$$

where $l( (a_i, b_i ] ) = b_i - a_i $, if set $A$ is Lebesgue measurable then for any real number $x$ and $c$, sets $$A+x, cA$$ are also Lebesgue measurable and $m(A+x) = m(A), m(cA) =|c| m(A) $

It seems that to prove that $A+x$ and $A$ have the same outer measure value is not that difficult but I'm somewhat stuck in verifying $A+x$ is also Lebesgue measurable. To do that we have to show that $$m(E) = m(E \cap (A+x) ) + m(E \cap {(A+x)}^{c} ) $$ for any $E$ given $m(E) = m(E \cap A ) + m(E \cap A^{c} ) $ for any $E$.... but It seems tricky.

Furthermore It seems that in order to prove $m(cA) = |c| m(A)$ I have to verify in advance that two outer measures

$$m_1(A) = \inf \Big\lbrace \sum l_1 \Big( (a_i,b_i] \Big) \Big|A \subset \cup_{i} (a_i, b_i] \Big\rbrace$$ $$m_2(A) = \inf \Big\lbrace \sum l_2 \Big( [a_i,b_i) \Big) \Big |A \subset \cup_{i} (a_i, b_i] \Big\rbrace$$

coincides for Lebesgue sigma algebra where $ l_1 \Big( (a,b] \Big)= b-a = l_2 \Big( [a,b) \Big) $. I can readily see using Caratheodory Extension Theorem that those two coincide on Borel sigma algebra but I have no idea how to check the coincidence on Lebesgue algebra. If a measure on Borel sigma algebra is given, is the extension to Lebesgue sigma algebra unique?

In addition, to prove $cA$ is Lebesgue measurable seems also tricky...

Can anybody give me a help or provide me detailed reference?

Thanks in advance.

$\endgroup$
2
$\begingroup$

You know that $A$ is measurable and want to show that $$ m(E) = m(E \cap (A+x) ) + m(E \setminus {(A+x)} ) $$ where $E$ is arbitrary.

Let $E'=E-x$; then $E'+x=E$ and your goal is now $$ m(E'+x) = m((E'+x) \cap (A+x) ) + m((E'+x) \setminus {(A+x)} ) $$ Now since the intersection and set difference both commute with translating by $x$ this is the same as $$ m(E'+x) = m((E'\cap A)+x) + m((E'\setminus A)+x) $$ and you already know that the outer measure ignores the translation, so this is the same as $$ m(E') = m(E'\cap A) + m(E'\setminus A) $$ which is true because $A$ is measurable.


For the gap in your $cA$ proof, note that if $$A\subseteq \bigcup_i (a_i,b_i]$$ then you also have $$A\subseteq \bigcup_i [a_i,b_i+2^{-i}\varepsilon) $$ for every $\varepsilon>0$, and the total length of the latter is only $\varepsilon$ more than the former. (And vice versa in the other direction, of course).

$\endgroup$
  • $\begingroup$ This solved all my doubtful points. Thanks! $\endgroup$ – Guldam Sep 12 '15 at 12:09

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.