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.


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).

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

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