This question comes from Bass, Ex 4.15. Given a finite measure space $(X, \mathcal{A}, \mu)$, we can define an outer measure $\mu^*$ as $$ \mu^*(A) := \inf\{ \mu(B) : A \subset B , B \in \mathcal{A} \} \:. $$ (One can check that this is an outer measure and that $\mu^*$ agrees with $\mu$ on $\mathcal{A}$).

I want to show now that if $\{A_n\}_{n \geq 1}$ is a sequence of increasing sets (not necessarily in $\mathcal{A}$) such that $A_n \nearrow A$, then $\mu^*(A_n) \nearrow \mu^*(A)$.

I have this so far. By monotonicity of outer measure, $\mu^*(A_n) \leq \mu^*(A)$ for all $n$, hence $\lim_{n \rightarrow \infty} \mu^*(A_n) \leq \mu^*(A)$. It remains to show that $\mu^*(A) \leq \lim_{n \rightarrow \infty} \mu^*(A_n)$.

My rough idea is as follows. Fix any $\epsilon > 0$. We can pick a sequence of sets $\{G_i\}_{i \geq 1}$ in $\mathcal{A}$ such that $\mu^*(A_i) \geq \mu(G_i) - \epsilon/2^{i}$. If the $\{G_i\}$ were an increasing set then we'd be done by using the continuity of measure from below. However, they are not. But, we can define $H_n := \bigcup_{i=1}^{n} G_i$ which is an increasing sequence by definition. Since $A \subset \bigcup_{n \geq 1} H_n$, we have that $$ \mu^*(A) \leq \mu( \bigcup_{n \geq 1} H_n ) = \lim_{n \rightarrow \infty} \mu(H_n) \:. $$ Now I am not quite sure how to finish off. I want to say something to the effect of $\mu(H_n) \leq \mu^*(A_n) + \epsilon$. However, since the $A_n$'s are not measurable, I am having trouble making such a comparison.

  • $\begingroup$ Your idea is essentially correct. You just need to define $H_n=\bigcap_{i=n}^\infty G_i$, not as $H_n=\bigcup_{i=1}^{n} G_i$. See my answer below for details. $\endgroup$
    – Ramiro
    Commented Dec 16, 2015 at 11:38

2 Answers 2


As you wrote, you have already proved $\lim_{n \rightarrow \infty} \mu^*(A_n) \leq \mu^*(A)$. It remains to show that $\mu^*(A) \leq \lim_{n \rightarrow \infty} \mu^*(A_n)$.

Your idea to complete the proof is essentially correct, all it needs is a small adjustment.

Given any $\epsilon > 0$. We can pick a sequence of sets $\{G_i\}_{i \geq 1}$ in $\mathcal{A}$ such that $A_i\subseteq G_i$ and $\mu^*(A_i) \leq \mu(G_i) \leq \mu^*(A_i) + \epsilon/2^{i}$.

Define $H_n=\bigcap_{i=n}^\infty G_i$. Then $\{H_n\}_{n\geq 1}$ is an increasing sequence of sets in in $\mathcal{A}$.

For any $n \geq 1$, if $i \geq n$ then $A_n\subseteq A_i \subseteq G_i$. So, for any $n \geq 1$, $A_n \subseteq \bigcap_{i=n}^\infty G_i=H_n$. So we have $A_n \subseteq H_n\subseteq G_n$ and we get $$ \mu^*(A_n) \leq \mu(H_n) \leq \mu(G_n) \leq \mu^*(A_n) + \epsilon/2^{n}$$

So we have $\lim_{n \to +\infty} \mu^*(A_n) = \lim_{n \to +\infty} \mu(H_n)$.

Since $A_n \nearrow A$, it is easy to see that that $A\subseteq \bigcup_{n\geq 1} H_n$. But $H_n \nearrow \bigcup_{n\geq 1} H_n$, so using the the continuity of measure from below, we have
$$\mu^*(A)\leq \mu\left(\bigcup_{n\geq 1} H_n\right)=\lim_{n \to +\infty} \mu(H_n)= \lim_{n \to +\infty} \mu^*(A_n)$$

  • 1
    $\begingroup$ Une jolie preuve ! $\endgroup$ Commented Nov 14, 2016 at 13:52
  • 2
    $\begingroup$ I cannot see why do you need $\frac{ε}{2^{i}}$ and not just $\frac{1}{i}$. Is there a reason? $\endgroup$ Commented Mar 18, 2019 at 14:50


Your reasoning is correct. You just need a lemma that estimates the closeness of the approximation of a union by a union of approximants. Details follow:

Let $A\subset X$ and $C \in \mathcal{X}$, $C\supset A$. We have $$\mu^*(A) = \inf \{\mu(B) \ | \ B \in \mathcal{X}, B \supset A\} = \inf \{\mu(B) \ | \ B \in \mathcal{X}, C \supset B \supset A\} $$ Therefore we have $$\mu(C) - \mu^{*}(A) = \sup \{ \mu(C) - \mu(B) \ | \ C \supset B \supset A\} = \sup\{ \mu(D) \ | \ D \in \mathcal{X},D \subset C \backslash A\} $$

Let's define the innner measure of a set $M$ to be $$\mu_*(M)=\sup\{ \mu(D) \ | D \in \mathcal{X},\ D \subset M\} $$ Therefore, for every $A \subset X$ and $C \in \mathcal{X}$, $C \supset A$ $$\mu(C) - \mu^{*}(A) = \mu_*(C \backslash A)$$

Let us prove the following: for $A_1$, $A_2 \subset X$, $C_i \in \mathcal{X}$ and $C_i \supset A_i$ we have $$\mu_*((C_1\cup C_2) \backslash (A_1 \cup A_2)) \le \mu_*(C_1 \backslash A_1) + \mu_*(C_2 \backslash A_2) $$

Indeed, let $D \subset (C_1\cup C_2) \backslash (A_1 \cup A_2)$, $D \in \mathcal{X}$. Then $(D \cap C_i) \subset C_i \backslash A_i$ and therefore $\mu(D) \le \mu(D\cap C_1) + \mu(D\cap C_2) \le \mu_*(C_1 \backslash A_1) + \mu_*(C_2 \backslash A_2)$. Now take the sup over $D$ and get the inequality.

From the above we get

$$\mu(C_1\cup C_2) - \mu^{*} (A_1 \cup A_2) \le (\mu(C_1) - \mu^*(A_1))+ (\mu(C_2) -\mu^*(A_2))$$

Your argument runs like this it seems: let $A_1 \subset A_2 \subset \ldots$. Take $\epsilon > 0$, approximate $A_n$ from the outside with $B_n$ better than $\epsilon/2^{n}$, and take $C_n = B_1 \cup \ldots B_n$ that approximates $A_n$ by better than $\epsilon ( 1/2 + \cdots 1/2^2) < \epsilon$. Now we got an increasing sequence of approximants. Let's show that $\mu(\cup B_n) - \mu^{*}(A) \le \epsilon$. Indeed: the difference is $\mu_* ( \cup B_n \backslash A)$. Let $D \subset \cup B_n \backslash A$, $D \in \mathcal{X}$. Then $D$ is the increasing union of $D \cap B_n$ and each has measure $< \epsilon$, so take the limit. We are about done.


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