prove that if M*(E) is finite if E is finite then M* is an outer measure define M*(E) as the number of points in E if E is finite and M*(E)=infinity if E is infinite. Show that M* is an outer measure. determine the measurable sets.??
 A: This is called the counting measure. See the formal definition of an outer measure in this wikipedia link. Note that since you can define this function $M^*$ in all the subsets of your given set, say $X$, then it will be not only an outer measure but also a measure (your $\sigma$-algebra is the power set $2^X$ of your set).The proof that it actually defines a measure is given here. In this particular case since it is defined in every subset outer measure and measure in the same and the proof of this link will be enough (a simple glance at the definition axioms of outer measure and measure will make this evident).
EDIT: Let me be more precise. You have two options: first, prove directrly the outer measure axioms which are stated in the link above it should be a fairly easy excercise; second, prove it is actually a measure following the other link I gave to you. In this particular case, the prove is valid for the reasons I will just state:

*

*An outer measure requires three things: the outer measure of the empty set is 0, it is countably subadditive, it is monotonic.

*For a measure you have to prove: the measure of emptyset is zero, it is non negative ant it is countably additive.

*A measure is only defined in the measurable sets, and an outer
measure in the whole power set. In this case, however,  every set is
measurable, hence there is no difference here.

*A measure is countably addtive whereas an outer measure need only be
countably subadditive, but by proving the first one you prove the
latter trivially.

*Proving that $M^*(\emptyset)=0$ is trivial since $\emptyset$ is a finite set with $0$ points.

*If you have a measure $M$ then $A \subseteq B$ then $M(B) = M(A\cup(B\setminus A))$ which by countable additivity is equal to $M(A)+M(B\setminus A)$ and since $M(B\setminus A)\geq 0$ by the non-negativity axiom then you can conclude that $M(A)\leq M(B)$, showing that a measure is monotonic.

*All of these can be use to prove that a measure defined in the whole power set is an outer measure.

While this discussion may be unnecesary for your fairly easy excercise I hope this is useful for you to see the relationship of outer measure and measure axioms and how "basic" is the counting measure.
