Outer Measure limit equality

Question

Suppose M is the class of measurable sets with respect to an outer measure $\mu^*$ defined on the subsets of $\Omega$. Take {$E_n$} to be a monotone increasing sequence of sets in M and A any set in $\Omega$. Prove that:

lim $\mu^*( A \cap E_n)$ = $\mu^*($lim $A \cap E_n)$

(the limits are taken with $n$ going to $\infty$)

I could prove "less or equal" using monotonicity of the outer measure, but I'm having a bit of trouble with the other side.
This is exercise 8 from section 4.1 of S. J. Taylor's Introduction to Measure and Integration

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Edit: $\mu^*$ here is any outer measure, not necessarily an induced one. In such manner, we can't say that for every set $A$ there exists a measurable set that covers it and has the exact same measure. When this property holds, Taylor refers to $\mu^*$ as a "regular outer measure".

Answer 1

Lemma: Let $A \subseteq \Omega$ and $E, F \in M$ such that $E \cap F=\emptyset$, then $$\mu^*(A \cap (E \cup F)) = \mu^*(A \cap E) + \mu^*(A \cap F)$$

Proof: Since $E \in M$ and $E \cap F=\emptyset$, we have \begin{align} \mu^*(A \cap (E \cup F)) &= \mu^*(A \cap (E \cup F) \cap E) + \mu^*(A \cap (E \cup F) \cap E^c) =\\ &= \mu^*(A \cap E) + \mu^*(A \cap F) \end{align} $\square$

Now for the main result:

Since for all $n$, $A \cap E_n \subseteq \lim_k A \cap E_k$, we have $\lim_n \mu^*( A \cap E_n) \leqslant \mu^*(\lim_n A \cap E_n)$.

Now, let $E= \lim_n E_n = \bigcup_n E_n$. Let us define $F_0 = E_0$ and, for each $n$, $F_{n+1} =E_{n+1} \setminus \bigcup_{k=0}^n E_k$. Then the sets $F_n$ are disjoint set and for all $n$, $F_n \in M$. Moreover $E_n= \bigcup_{k=0}^n F_k$ and $E= \lim_n E_n = \bigcup_n E_n= \bigcup_n F_n$.

So we have $$\lim_n \mu^*( A \cap E_n) \leqslant \mu^*(\lim_n A \cap E_n)= \mu^* \left (\bigcup_n ( A \cap F_n) \right ) \leqslant \sum_n\mu^*(A\cap F_n)$$

where the last step is $\sigma$-sub-additivity of $\mu^*$. So we have:

$$\lim_n \mu^*( A \cap E_n) \leqslant \mu^*(\lim_n A \cap E_n)\leqslant \lim_k\sum_{n=0}^k\mu^*(A\cap F_n) \tag{1}$$

Using the Lemma and induction, we have that for all $k$, $$\sum_{n=0}^k\mu^*(A\cap F_n)= \mu^*\left (A \cap \bigcup_{n=0}^k F_n \right ) = \mu^*(A \cap E_k) \tag{2}$$

Combining $(1)$ and $(2)$:

$$\lim_n \mu^*( A \cap E_n) \leqslant \mu^*(\lim_n A \cap E_n)\leqslant \lim_k\sum_{n=0}^k\mu^*(A\cap F_n) = \lim_k\mu^*(A \cap E_k) $$

So $\lim_n \mu^*( A \cap E_n) = \mu^*(\lim_n A \cap E_n)$

Answer 2

The function $\mu^*_A(B):=\mu^*(A\cap B)$ defines a measure on the $\sigma$-algebra $\mathcal{A}_A:=\{B\cap A: B\in\mathcal{M}_{\mu^*}\}$. One can reconstruct the steps of Caratheodory's construction (postings of @Ramiro and @Snoop) or apply the Caratheodory extension theorem directly.

Notice that $\mu^*_A$ is finitely additive on $\mathcal{A}_A$. Suppose $B,C\in \mathcal{M}_{\mu^*}$ and such that $A\cap B\cap C=\emptyset$, then \begin{align} \mu^*((A\cap B)\cup(A\cap C))&=\mu^*\Big(\big((A\cap B)\cup(A\cap C)\big)\cap B\Big)\\ &\qquad+\mu^*\Big(\big((A\cap B)\cup(A\cap C)\big)\setminus B)\\ &=\mu^*(A\cap B) +\mu^*(A\cap C) \end{align} Clearly $\mu^*$ is subadditive; hence, by Caratheodory's theorem $\mu^*_A$ extends to a measure $\nu$ for which each $B\in\mathcal{A}_A$ is $\nu$-measurable and $\nu(B)=\mu^*(B)$ for all $B\in\mathcal{A}_A$.

The statement of the OP now follows immediately: \begin{align} \lim_n\mu^*(A\cap E_n)&=\lim_n\nu(A\cap E_n)=\nu(\bigcup_n(A\cap E_n))\\ &=\nu(A\cap\bigcup_nE_n)= \mu^*(A\cap\bigcup_nE_n) \end{align}

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Notes:

1. If the assumption that $(E_n)\subset\mathcal{M}_{\mu^*}$ is dropped, the the statement still hols for for outer measures $\mu^*$ for which $$\mu^*(E)=\inf\{\mu^*(A): A\in\mathcal{M}_{\mu^*},\, E\subset A\}$$ Here is a proof: Let $C_n=A\cap E_n$. $C_n$ is a monotone sequence and $C_n\nearrow C=A\cap\bigcup_nE_n=\bigcup_n A\cap E_n$. Clearly $\mu^*(C_n)\leq \mu^*(C)$. If $\lim_n\mu^*(C_n)=\infty$, there is nothing to prove.
   Suppose $\lim_n\mu^*(C_n)<\infty$. For each $n$, there is a measurable set $B_n$ such that $C_n\subset B_n$ such that $\mu^*(C_n)=\mu(B_n)$ (why?). Let $B'_n=\bigcap^\infty_{k=n}B_n$. Then $C_n\subset B'_n$ for each $n$ and $\mu^*(C_n)=\mu(B'_n)$. The sequence $B'_n$ monotone nondecreasing and so $$\lim_n\mu^*(C_n)=\lim_n\mu(B'_n)=\mu(B), \quad\text{where}\quad B=\bigcup_nB'_n$$ Since $C\subset B$, $\lim_n\mu^*(C_n)\leq \mu^*(C)\leq\mu(B)$; thus $\mu^*(C)=\mu(B)$.

• For the why? notice that of for any set $A$, if $\mu^*(A)<\infty$, then by definition of outer measure, for any $n$ there is a measurable set $B_n$ such that $A\subset B_n$ and $$\mu(B_n)<\mu(A)+2^{-n}$$ Set $B=\bigcap_nB_n$.

• For a general outer measure, the result does not hold unless the sequence $(E_n)\subset\mathcal{M}_{\mu^*}$. Here is an example. Let $\Omega=\mathbb{N}$, define $\mathcal{T}=\{\{n\}, \Omega, \emptyset: n\in\Omega\}$ and let $\tau:\mathcal{T}\rightarrow[0,\infty]$ by $\tau(\emptyset)=0$, $\tau(\{n\})=2^{-n+1}$, and $\tau(\Omega)=\infty$. Define doe any $E\subset \mathbb{N}$ $$\mu^*(E)=\inf\{\sum_n\tau(A_n):A_n\in\mathcal{T}, E\subset\bigcup_nA_n\}$$ Then, for $E\neq\Omega$, $\mu^*(E)=\sum_{n\in E}2^{-n+1}$, and $\mu^*(\Omega)=\infty$.
  $E_n=\{1,\ldots,n\}\nearrow\Omega$ and $$\lim_n\mu^*(E_n)=2<\lim_n\mu^*(\Omega)$$

Answer 3

I will be using this resource. Feedback appreciated. We start by defining the outer measure $\mu^*:2^X\to [0,\infty]$ with $\mu^*(\emptyset)=0$ and $\mu^*(A)\leq \sum_{n}\mu^*(A_n)$ for all $A\subseteq \cup_nA_n$, inducing the property $\mu^*(A)\subseteq \mu^*(B),\,\forall A\subseteq B$. Then, we define (Definition 1.2) the usual $\mu^*$-measurable sets $$\mathscr{M}:=\{A\subseteq X:\mu^*(S)=\mu^*(S/A)+\mu^*(S\cap A),\,\forall S \subseteq X\}$$ Further, we can show that for $\mu^*$-measurable $(A_n)_n$ pairwise disjoint, we have (Remark 1.7) $$\mu^*(S\cap (\cup_nA_n))=\sum_{n \in \mathbb{N}}\mu^*(S\cap A_n),\,\forall S\subseteq X$$ We then obtain similarly to (1.8) that for $\mu^*$-measurable $(E_n)_n$ s.t. $E_n\subseteq E_{n+1}$ and $A\subseteq X$ we get $$\begin{aligned}\mu^*((\cup_nE_n)\cap A)&=\mu^*((\cup_n(E_n/E_{n-1})\cap A)=\\ &=\lim_{n\to \infty}\sum_{k\leq n}\mu^*((E_k/E_{k-1})\cap A)=\\ &=\lim_{n\to \infty}\mu^*((\cup_{k\leq n}(E_k/E_{k-1}))\cap A)=\\ &=\lim_{n\to \infty}\mu^*(E_n\cap A)\end{aligned}$$