Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I cannot prove this formula: $E$ is measurable and $A$ is any subset of $E$ show that $m(E)=m_*(A) +m^*(E-A)$.

Define inner measure of $A$ by $m_∗(A)=\sup(m(F))$, where the supremum is taken over all closed subsets $F$ of $E$.

$m(E)$ means $E$ is measurable and for outer measure of $E$, cover $E$ by countable collection $S$ of intervals $I_k$. i.e. $m^\ast(E)=\inf \sum \nu(I_k)$

Thanks and regards.

share|improve this question
    
What did you try? Can you recall the definition of inner and outer measure, then try to see how it may solve the problem. –  Davide Giraudo Nov 13 '11 at 13:45
4  
In any case, when you ask it here you should provide the definitions you are using. –  GEdgar Nov 13 '11 at 14:12
add comment

1 Answer

up vote 5 down vote accepted

With the little information you give on what you know, this is bound to be somewhat tricky.

Here's an outline for a proof of the following more general formula: For every measurable set $E$ and an arbitrary set $A$ we have $$m(E) = m_\ast(E \cap A) + m^\ast(E \smallsetminus A).\tag{$*$}$$

  1. Prove that for all $A$ the identity $m^\ast(A) = \inf{\{m(G)\,:\,G \supset A \text{ measurable}\}}$ holds.

    [If $m^\ast(A) = \infty$ then ... If $m^\ast(A) \lt \infty$ take $G_n \supset A$ to be a union of intervals with $m(G_n) \lt m^\ast(A) + \frac{1}{n}$ and put $G = \bigcap G_n$.]

  2. Prove that for all $A$ we have $m_\ast(A) = \sup{\{m(F)\,:\,F \subset A \text{ is measurable}\}}.$

  3. If the measure of $E$ is finite, we have $m^\ast(E \smallsetminus A) \lt \infty$ and we may write $$ \begin{align*} m(E) - m^\ast(E \smallsetminus A) & = m(E) - \inf{\{m(G)\,:\,G \supset E \smallsetminus A \text{ is measurable}\}} \\ & = m(E) - \inf{\{m(G)\,:\,E \supset G \supset E \smallsetminus A \text{ is measurable}\}} \\ & = m(E) - \inf{\{m(E) - m(F)\,:\,F \subset E \cap A \text{ is measurable}\}} \tag{#} \\ & = \sup{\{m(F)\,:\,F \subset E \cap A \text{ is measurable}\}} \\ & = m_\ast(E \cap A) \end{align*} $$ so that $m(E) = m_\ast(E\cap A) + m^\ast(E \smallsetminus A)$, which is the desired formula (for sets $E$ of finite measure).

    [Note that in $(\#)$ we used measurability of $E$ and finiteness of its measure to conclude that for $E \supset G \supset E \cap A$ and $F = E \smallsetminus G$ we have $m(E) = m(G) + m(F)$.]

  4. By definition $m_\ast (A) = \sup{\{m(F)\,:\,F \subset A \text{ is closed}\}}$. For each closed $F \subset E \cap A$ we have $$ m(E) = m(F) + m(E \smallsetminus F) \geq m(F) + m^\ast(E \smallsetminus A) $$ since measurability of $F$ gives $m(E \smallsetminus F) = m^\ast(E \smallsetminus F)$, and since $E \smallsetminus F \supset E \smallsetminus A$ yields $m^\ast(E \smallsetminus F) \geq m^\ast(E \smallsetminus A)$ because $m^\ast$ is monotone.

    Taking the supremum over all closed $F \subset E \cap A$ we get $$ m(E) \geq m_\ast(E \cap A) + m^\ast(E \smallsetminus A).\tag{1} $$

  5. Finally, we can establish $(\ast)$ for arbitrary measurable $E$ by observing that monotonicity of $m_\ast$ and $m^\ast$ give $$ \begin{align*} m_\ast(E \cap A) + m^\ast(E \smallsetminus A) &\geq \sup{\{m_\ast (F \cap A) + m^\ast(F \smallsetminus A)\,:\,F \subset E \text{ measurable, }m(F) \lt \infty\}} \\ &= \sup{\{m(F)\,:\,F \subset E \text{ measurable, } m(F) \lt \infty\}} \\ &= m(E) \end{align*} $$ and combining this with the inequality in $(1)$. In the first equality here we used what we proved in 3.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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