Let $X$ be a nonempty set and $\mu^*:P(X)\rightarrow [0,\infty]$ be an outer measure on $X.$ Consider a class of the subsets of $X$ by
$$m_{\mu}=\{A \in P(X):\mu^*(E)\ge \mu^*(E\cap A)+\mu^*(E/A)\;\;\forall E\in P(X)\}.$$
Then $m_{_\mu}$ is a $\sigma$-algebra on subsets of X and the restriction of $\mu^*$ to $m_{\mu},$ that is $\mu:m_{_\mu}\rightarrow[0,\infty]$ given by $\mu(A):=\mu^*(A)$ is a measure on $X$.
In addition, the measure space $(X,m_{\mu},\mu^*)$ is complete, that is, if $\mu^*(A)=0$,then $A\in m_{\mu}.$
My doubt is why my course instructor has chosen the set $m_{\mu} =\{A$ $ \in$ $ P(X):\mu^*(E)\ge \mu^*(E\cap A)+\mu^*(E/A)$ ,$\forall E\in P(X) \}$ with greater and equal to rather than equal.
I think by subadditivity it will be just $\mu^*(E)= \mu^*(E\cap A)+\mu^*(E/A),\;\forall E\in P(X).$
2)$\mu^*(E)\ge \mu^*(E\cap A)+\mu^*(E/A),\;\;\forall E\in P(X)$
similarly has define $L(R^k)=\{A$ $ \subset$ $ R^k:\mu^*(E)\ge \mu^*(E\cap A)+\mu^*(E/A)$ ,$\forall E\subset R^k \}$
he says outer measure means measuring from out side.
He has given this example and told about $L(R)$ with strict inequality and says this is why strict inequality is there