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.

  • $\begingroup$ What did you try? Can you recall the definition of inner and outer measure, then try to see how it may solve the problem. $\endgroup$ – Davide Giraudo Nov 13 '11 at 13:45
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    $\begingroup$ In any case, when you ask it here you should provide the definitions you are using. $\endgroup$ – GEdgar Nov 13 '11 at 14:12

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.

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