Lebesgue inner measure formula 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.
 A: 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
$$m(E) = m_\ast(E \cap A) + m^\ast(E \smallsetminus A).\tag{$*$}$$


*

*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 $G = \bigcap G_n$.]

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

*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 $m(E) = m(G) + m(F)$.]

*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 
$$
m(E) \geq m_\ast(E \cap A) + m^\ast(E \smallsetminus A).\tag{1}
$$

*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.
