If $E$ is an measureble set in the real line , with $m(E)=\infty$, then $\forall A \subset E$ is true that: $$ m_*(A)+m^*(E-A) = \infty \quad? $$ I think this statement is false!!!
Mi attempt:
Let $E_n=[n,n+1),\forall n\in\mathbb{Z}_+$,observe that how $E_n\cap E_m=\emptyset , d(E_n,E_m)>0$, since $$E=\bigcup E_n$$ then $$m(E)=\sum_{n=0}^{\infty} m(E_n)=\infty$$(Ok $m(E)=\infty$).
we know that $\exists A\subset E_0\subset [0,1],$such that, $m_*(A)=0$ and $m^*(A)=1$.
Repeat this construction in every $E_n$ , so $A_n\subset E_n:m_*(A_n)=0$ and $m^*(A_n)=1$.
How the sets $A_n$ they are dijoints and with positive distance ( by construction), since $$ A=\bigcup_{n=0}^{\infty} A_n$$ then follow that $$m_*(A)=\sum_{n=0}^{\infty}m_*(A_n)=0$$
Finally , how $A_n\subset E_n$ and $m(E_n)<\infty$, then $m*(E_n-A_n)=m^*(E_n)+m^*(A_n)=1-1=0$.
But $$m^*(E-A)\leq\sum_{n=0}^{\infty}m^*(E_n-A_n)=0$$
Can you help me to verify if this mi prove is correct.
Thanks !!!
Since $m_*(A)=0$, we have $m^*(E-A)=1$. The same holds for each $E_n$ and $A_n$. So the last equation in your proof is not correct. The $0$ at the end should be $\infty$, and that destroys the proof.
A minor point: The distance between $E_n$ and $E_{n+1}$ is $0$. The reason this correction is minor is that your argument didn't use any information about distance.