Basic Measure Theory: Proving E in L(R)

[snipez90]

Homework Statement

Show that if [itex]E \subset B[/itex] and [itex]B \in L(\mathbb{R})[/itex] (where L(R) denotes the family of Lebesgue measurable sets on the reals) with [itex]m(B) < \inf [/itex], then [itex]E \in L(\mathbb{R})[/itex] if and only if [itex]m(B) = m^{*}(E) + m^{*}(B - E)[/itex], where [itex]m^*[/itex] denotes the Lebesgue outer measure.

Homework Equations

Basic set theoretic manipulations.

The Attempt at a Solution

The forward direction follows by the definition of the outer measure. As for the reverse direction, we need to show that for any set A contained in the reals,
[tex]m(A) \geq m^{*}(A \cap E) + m^{*}(A - E)[/tex]
(the less than or equal direction follows from subaddivity).
I'm not really stuck at this point, but I'm wondering about my approach from here. I simply considered the three cases where A contains B, A is contained in B and contains E, and A is contained in E. I'm fairly certain this works out in each case as it should. This seems like a pretty straightforward approach, but is there a better way? I understand this isn't exactly the most interesting of problems but it helps me to know what other tools are available to me at this point. Thanks in advance.

 

[mighty2000]

Your approach seems reasonable and straightforward. Another approach could be to use the fact that the Lebesgue outer measure is countably subadditive, meaning that for any countable collection of sets {A_n}, we have m^*(∪A_n) ≤ ∑m^*(A_n). This property can be used to show that m^*(A) ≥ m^*(A∩E) + m^*(A∩(B-E)) for any set A contained in the reals. From there, you could use basic set theoretic manipulations to show that this is equivalent to m(A) ≥ m^*(A∩E) + m^*(A-E). This approach may require a bit more technical work, but it is another way to approach the problem.