Inequalities in Caratheodary measurability criterion The following is a multiple choice question:
Q. Let $m^*$ be the Lebesgue outer measure over $\mathbb{R}$. Then a subset E of $\mathbb{R}$ is Lebesgue measurable if and only if, for any subset A of $\mathbb{R}$

*

*$m^*(A)\geq m^*(A\cap E)+m^*(A^C\cap E)$

*$m^*(A)= m^*(A\cap E)+m^*(A\cap E^C)$

*$m^*(A)\leq m^*(A\cap E)+m^*(A\cap E^C)$

*$m^*(A)\geq m^*(A\cap E)+m^*(A\cap E^C)$
Options $2$ and $3$ are clear to me since the subadditivity of $m^*$ can be applied to $A=(A\cap E) \cup (A \cap E^c)$. Can we find the strict inequalities in options 1 and 4 by some non-measurable (Lebesgue) sets?
 A: $E\subset \Bbb{R}$ measurable if $\forall A\subset \Bbb{R}$
$m^*(A)= m^*(A\cap E)+m^*(A\cap E^C)$

As $A=(A\cap E) \cup (A \cap E^c)$
By monotonicity of outer measure
$m^*(A)=m^*( (A\cap E) \cup (A \cap E^c))\le m^*(A\cap E)+m^*(A\cap E^C)$
Hence it is enough to show $ m^*(A)\geq m^*(A\cap E)+m^*(A\cap E^C)$

Hence $2, 4$ are equivalent and equivalently imply measurability.

$3 ) $ is true for every set $E$ as shown above. But not all sets are measurable ( one need Axiom of Choice to construct such sets like Vitali set, Bernstein set, in fact any set with positive outer measure  contains a non measurable set.)
So $3$ is not equivalent to measurability.

Suppose $1$ is true i.e $m^*(A)\geq m^*(A\cap E)+m^*(A^C\cap E)$
Since $E=(A\cap E) \cup (A^c \cap E)$
$m^*(E)\le  m^*(A\cap E)+m^*(A^C\cap E)\le m^*(A)$
Hence $0\le m^*(E) \le m^*(A) $ and since this is true for every $A\subset \Bbb{R}$ . In particular it is true for null sets i.e sets with $m^*(A) =0$.
Hence $m^*(E) =0$ implies $E$ is measurable.

Converse is not true i.e not every measurable set satisfy
$m^*(A)\geq m^*(A\cap E)+m^*(A^C\cap E)$ for every $A\subset \Bbb{R}$.
As by previous argument any set satisfying $3$ is a null set.
Example :
Let us choose the Cantor set $\mathcal{C}$ , then for measurable set $E=[0, 1]$
$\begin{align}0=m^*(\mathcal{C}) &\ge  m^*(\mathcal{C}\cap [0,1])+m^*(\mathcal{C}^c \cap [0,1])\\&=0+1\end{align}$ Contradiction(!)
Hence $1$ implies measurability but not the converse.

Lastly recall $a>b$ also implies $a\ge b$ i.e any set $E$ which satisfy strictly inequality in $1$ and $4$ also satisfy $1$ and $4$ respectively. Both $1$ and $4$ implies measurability. Hence what you have tried to find( non measurable set satisfying stict inequality in $1$ and $4$ )is impossible.
