Measure Theory/Outer Measure Subadditivity

Outer Measure Subadditivity
Recall that countable additivity is the property that, for disjoint sets $$A_1,A_2,\dots$$ in the domain of $$\lambda^*$$,


 * $$\lambda^*\left(\bigsqcup_{i=1}^\infty A_i\right) = \sum_{i=1}^\infty \lambda^*(A_i) $$

Also recall that we have a proof that no function defined on $$\mathcal P(\Bbb R)$$ satisfies all four properties: Nonnegativity, Interval length, Translation invariance, and Countable additivity.

Prove that $$\lambda^*$$ cannot be countably additive.

Although $$\lambda^*$$ may not be countably additive, there is a weaker property which we very often find useful, called subadditivity.

Note that we could state subadditivity for disjoint sets. However, overlap tends to only make the measure of the union smaller than the sum. For example,
 * $$\lambda^*((0,2)\cup(1,3)) = \lambda^*((0,3)) = 3$$

whereas
 * $$\lambda^*((0,2))+\lambda^*((1,3)) = 2+2=4$$

Therefore if there is overlap, the inequality still remains true, as we shall prove below.

Proof
We now state and will prove the following theorem.

$$\lambda^*$$ is subadditive.

Suppose that any $$A_i$$ in the sequence $$A_1,A_2,\dots\subseteq\Bbb R$$ has infinite outer measure. Prove that therefore $$\lambda^*\left(\bigcup_{i=1}^\infty A_i\right) = \infty$$ and that $$\sum_{i=1}^\infty\lambda^*(A_i)=\infty$$.

In the exercise above, you have handled the case where at least one set has infinite outer measure. Therefore, for the rest of this proof, we only need to show that the inequality is true when every set in the sequence has finite measure.

Pairwise Subadditivity
First we prove that if we have just two sets of finite outer measure, $$A,B\subseteq\Bbb R$$, then these satisfy subadditivity.

That is to say, we will show


 * $$\lambda^*(A\cup B)\le\lambda^*(A)+\lambda^*(B)$$

To get started, let $$\varepsilon\in\Bbb R^+$$ and then let $$\mathfrak O,\mathfrak P$$ be any open interval over-approximations of A and B respectively, such that the over-estimates satisfy


 * $$ \sum_{I\in \mathfrak O}\ell(I)< \lambda^*(A)+\varepsilon, \quad \text{ and } \sum_{I\in\mathfrak P}\ell(I) < \lambda^*(B)+\varepsilon $$

Prove that $$\mathfrak O\cup\mathfrak P$$ is an open interval over-approximation of $$A\cup B$$.

Next, prove that


 * $$\sum_{I\in\mathfrak O\cup\mathfrak P}\ell(I) < \lambda^*(A)+\lambda^*(B)+2\varepsilon $$

Next, state the reason why $$\lambda^*(A\cup B)\le\sum_{I\in\mathfrak O\cup\mathfrak P}\ell(I)$$.

Next, complete the rest of the proof for pairwise subadditivity.

Now use the same basic proof that you used for pairwise subadditivity, to prove countable subadditivity. That is to say, start by considering an open interval over-approximation for each set, such that the corresponding over-estimate is underneath "some bound".

Note that this proof will likely require that you encounter a sum of sums, and rearrange the terms. We have said before that the rearrangement of nonnegative terms is not an issue, as proved in elementary analysis. However, this is a kind of rearrangement of terms which is unlike the theorems encountered anywhere in the usual elementary analysis textbooks.

The proof that this kind of rearrangement of terms is valid, is quite tricky and therefore we do not want to get bogged down in the issue. Rest assured that any rearrangement is valid, and therefore use it freely here.

The trick here is to pick the right bound. If you picked the bound in exactly the same way as with pairwise subadditivity, you'd say that it is $$\lambda^*(A_i)+\varepsilon$$. However, that choice will cause a failure later on in the proof, when you have to take an infinite sum of $$\varepsilon$$, which then goes to infinity.

Hint: Use the $$\varepsilon/2^i$$ trick.

Null Adding and Subtracting
In this section we will prove that, if $$A\subseteq\Bbb R$$ is any set of real numbers and $$E\subseteq\Bbb R$$ is a null set, then the outer measure of A is not changed by either adding or removing E. More formally, we will show

Let $$A,E\subseteq\Bbb R$$ be two sets of real numbers, and E a null set. Then


 * $$\lambda^*(A)=\lambda^*(A\cup E)=\lambda^*(A\smallsetminus E)$$

Use subadditivity to show $$\lambda^*(A\smallsetminus E)\ge \lambda^*(A)$$.

Use monotonicity to show the reverse inequality.

Due to ''Exercise 5. Minus Null'', we know that subtracting a null set from a set does not change its measure.

Use this to show that $$\lambda^*(A\cup E) = \lambda(A\smallsetminus E)$$.