Okay so I'm asked to show (not necessarily prove but that would be helpful too!) that if the outer measure of a subset F of R is equal to zero .i.e m*(F)=0 then each subset E of F is Lebesgue Measurable.
I'm thinking this is going to follow directly from the definition of Lebesgue measurable but I'm having a little trouble conceptualizing that too.
The way I've learned it is that F is Lebesgue Measurable if for another set S,
the outer measure of S is equal to the outer measure of the intersection of S&F and S&F-complement.
Would I let this subset E of F be equal to S in this case? I assume the criteria for S is pretty loose and just has to be a subset of Reals so E could satisfy this naturally.
Therefore by letting m*(F)=0 I would start:
m*(E)=m*(EnF)+m*(EnF^c)
then distribute this out somehow?
Best Answer
Let $S$ be an arbitrary subset of $\mathbb R$.
In order to prove that $E$ is a Lebesgue measurable set it must be shown that:$$m^*(S\cap E)+m^*(S\cap E^c)=m^*(S)$$
From $S=(S\cap E)\cup (S\cap E^c)$ it follows directly that: $$m^*(S\cap E)+m^*(S\cap E^c)\geq m^*(S)$$
So it remains to prove that: $$m^*(S\cap E)+m^*(S\cap E^c)\leq m^*(S)$$
Based on $S\cap E^c\subseteq S$ we find $m^* (S\cap E^c)\leq m^*(S)$
Based on $S\cap E\subseteq F$ we find $m^*(S\cap E)\leq m^*(F)=0$
Now we are ready.