Folland, Chapter 1 Problem 17

Solution 1:

Let $\{A_j\}_{1}^{\infty}$ be a sequence of disjoint $\mu^*$-measurable sets and $E\subset X$. By countable subadditivity, $$\mu^*\left(E\cap\left(\bigcup_{j=1}^{\infty}A_j\right)\right) = \mu^*\left(\bigcup_{1}^{\infty}E\cap A_j\right) \leq \sum_{1}^{\infty}\mu^*(E\cap A_j)$$ Let $B_n = \bigcup_{1}^{n}A_j$. For each $n\geq 2$, since $A_n$ is $\mu^*$-measurable we have \begin{align*} \mu^*(E\cap B_n) &= \mu((E\cap B_n)\cap A_n) + \mu^*((E\cap B_n)\cap A_n^c)\\ &= \mu^*(E\cap A_n) + \mu^*(E\cap B_{n-1}) \end{align*} By induction, \begin{align*} \mu^*(E\cap B_n) &= \mu^*\left(E\cap \bigcup_{1}^{n}A_j\right) + \mu^*\left(\bigcup_{1}^{n}E\cap A_j\right)\\ &= \sum_{1}^{n}\mu^*(E\cap A_j) \ \forall n\geq 1 \end{align*} So then by monotonicty, $$\mu^*\left(E\cap \bigcup_{1}^{\infty}A_j\right) \geq \mu(E\cap B_n) = \sum_{1}^{n}\mu^*(E\cap A_j) \ \forall n\geq 1$$ thus as $n\rightarrow \infty$ we have that $$\mu^*\left(E\cap \bigcup_{1}^{\infty}A_j\right) \geq \sum_{1}^{\infty}\mu^*(E\cap A_j)$$ therefore $$\mu^*\left(E\cap \bigcup_{1}^{\infty}A_j\right) = \sum_{1}^{\infty}\mu^*(E\cap A_j)$$

Solution 2:

Hint: The cases I would distinguish are $\mu^*(E\cap \bigcup_{i=1}^\infty A_i) <+\infty$ and $\mu^*(E\cap \bigcup_{i=1}^\infty A_i) =+\infty$, rather than $\mu^*(E) = +\infty$ and $\mu^*(E) < +\infty$.

Or said differently, you might just as well substitute $E \to E\cap \bigcup_{i=1}^\infty A_i$, since the part of $E$ outside of $ \bigcup_{i=1}^\infty A_i$ has no relevance.

Solution 3:

Here's a cute alternative solution:

From the outer measure $\mu^*$, we get an outer measure $\mu^*|_{E}$ on $\mathcal{P}(E)$ defined as $$ \mu^*|_E(B) := \mu^*(E \cap B). $$ We claim then that each $E \cap A_j$ is $\mu^*|_E$-measurable, which is to say for any $B \subseteq E$ one has $$ \mu^*|_E(B) = \mu^*|_E(B \cap (E \cap A_j)) + \mu^*|_E(B \setminus (E \cap A_j)). $$ This of course follows from $\mu^*$-measurability of $A_j$, as \begin{align*} \mu^*|_E(B) = \mu^*(B) &= \mu^*(B \cap A_j) + \mu^*(B \setminus A_j), \\ &= \mu^*(B \cap A_j \cap E) + \mu^*(B \setminus (A_j \cap E)), \\ &= \mu^*|_E(B \cap (E \cap A_j)) + \mu^*|_E(B \setminus (E \cap A_j)). \end{align*} Using this, by Caratheodory's theorem, $\mu^*|_E$ is a measure on the collection of $\mu^*|_E$-measurable sets and therefore, \begin{align*} \mu^*\bigg( E \cap \Big(\bigcup_{j=1}^\infty A_j\Big)\bigg) &= \mu^*|_E\bigg(\bigcup_{j=1}^\infty E \cap A_j\bigg), \\ &= \sum_{j=1}^\infty \mu^*|_E(E \cap A_j), \\ &= \sum_{j=1}^\infty \mu^*(E \cap A_j). \end{align*}