$A,\ B\subseteq R$ and $m*(A)<\infty$, show that $m^*(B\cap A^c)\geq m^*(B)-m^*(A)$

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1.$A,\ B\subseteq R$ and $m^*(A)<\infty$, show that $ m^*(B\cap A^c)\geq m^*(B)-m^*(A)$; here $m^*$ is the outer measure.

My work- we can write,

$m^*(B)\leq m^*((B\cap A^c)\cup A)\leq m^*(B\cap A^c)+m^*(A)$, by sigma additivity of outer measure.

then, $m^*(B)\leq m^*(B\cap A^c)+m^*(A)$

since, $m^*(A)$ is finite, we can substract it from both side. hence, $m^*(B)-m^*(A)\leq m^*(B\cup A^c)$

Is this correct proof?

2.In the second part, I'm asked to prove $m^*((a,b)\cup(c,d))=m^*(a,b)+m^*(c,d)$ where (a,b) and (c,d) are disjoint intervals. Should I use the above results for the proof?

Since outer measure is not additive , I don't think we can use additivity here.