William C. answered 03/22/23
PhD Student at UCLA
As you have stated, the equation m*(A) = m*(A^E) + m*(A^(Ec)) has to hold for any subset A of R, but you only proved it for A = E. However, your approach is definitely on the right track.
In particular, to fill in the details for you, I'd argue the following.
I'm going to show two things:
(1) m*(A^E) = 0
(2) m*(A^(Ec)) = m*(A)
And these two will be enough to show that right-hand side of the equation above is equal to m*(A)
For (1) we note that m*(A^E) <= m^*(E) = 0 and since m*(A^E) >= 0 (since measures are always nonnegative), it follows that m*(A^E) = 0
For (2) we immediately note that m*(A^(Ec)) <= m*(A), and thus if we can show the reverse inequality we will be done. However, the reverse inequality follows from the following: m*(A^(Ec)) = m*(A\E) >= m*(A) - m*(E), and since m*(E) = 0, it follows that m*(A^(Ec)) >= m*(A) as desired.
As far as "other ways to prove it" I'm not aware of another method although the method you proposed seems to be standard (which is a good thing because it makes it applicable to a lot of other questions).
I know this is a lot so let me know if you have any follow-up questions!
Ashley P.
Thank you so for much for the informative explanation!03/23/23