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I know that similar question has been asked and answered here, here, and here. But I am looking for a different proof based on different definitions.

We have the following definition of an outer and inner measure indeuced by a measure:

Definition$\quad$ Let $(X,\mathcal{A})$ be a measurable space, let $\mu$ be a measure on $\mathcal{A}$, and let $A$ be an arbitrary subset of $X$. Then $\mu^*(A)$, the outer measure of $A$, is defined by \begin{align*} \mu^*(A) = \inf\{\mu(B):A \subseteq B\ \text{and}\ B \in \mathcal{A}\}, \end{align*} and $\mu_*(A)$, the inner measure of $A$, is defined by \begin{align*} \mu_*(A) = \sup\{\mu(B):B \subseteq A\ \text{and}\ B \in \mathcal{A}\}. \end{align*}

I need to prove that $\mu_*(A) \leq \mu^*(A)$ holds for each subset $A$ of $X$.

This is how I would like to proceed and where I got stuck:

Assume to the contrary that $\mu_*(A) > \mu^*(A)$ for some $A \in X$. Then there is a $B \in \mathcal{A}$ such that $A \subseteq B$ and $\mu(B) < \mu_*(A)$.

From here, I wanted to say that there is a set $C \in \mathcal{A}$ such that $C \subseteq A$ and $\mu(C) > \mu(B)$, which leads to a contradiction because $C \subseteq A \subseteq B$ implies that $\mu(C) \leq \mu(B)$. However, I do not know how to state rigorously that such a set $C$ exists.

Could anyone please help me out?

Reference: Measure Theory by Donald Cohn Page 33.

Beerus
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4 Answers4

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Using the definitions you provided, I think you can make your argument work by just using the definition of the infimum and the supremum.

Suppose towards a contradiction that $\mu_*(A) - \mu^*(A) = \varepsilon > 0$. By the definition of infimum, there must exist $B_1 \in \mathcal{A}$ such that $A \subseteq B_1$, and

$$ \mu^*(A) > \mu(B_1) - \frac{\varepsilon}{2}. $$

Similarly, by the definition of supremum there is $B_2 \in \mathcal{A}$ such that $B_2 \subseteq A$, and

$$ \mu_*(A) < \mu(B_2) + \frac{\varepsilon}{2}. $$

Using these two inequalities, we can now estimate

$$ \varepsilon = \mu_*(A) - \mu^*(A) < \mu(B_2) + \frac{\varepsilon}{2} - \mu(B_1) + \frac{\varepsilon}{2} = \mu(B_2) - \mu(B_1) + \varepsilon, $$

from which we deduce that $\mu(B_2) > \mu(B_1)$. But since $B_2 \subseteq A \subseteq B_1$, by monotonicity of the measure $\mu$ we must have $\mu(B_2) \leq \mu(B_1)$, which gives a contradiction.

noam.szyfer
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    Thanks a lot for your help! I do have some concern about your answer though. We assumed that $\mu_(A)-\mu^(A)=\epsilon>0$. But for a general space $X$, not necessarily $\mathbb{R}$ or $\mathbb{R}^d$, how can we make sure the existence of a $B_1\in\mathcal{A}$ which is a superset of $A$ such that $\mu^(A)>\mu(B_1)-\frac{\epsilon}{2}$. I think the best we can get is $\mu^(A)\leq\mu(B_1)<\mu_*(A)$. A similar concern is also raised for your $B_2$. – Beerus Apr 01 '24 at 23:51
  • (Cont'd) (plus without specifing what $\mu$ is) – Beerus Apr 02 '24 at 02:23
  • (Cont'd) For example, let $X$ be a nonempty set, and let $\mathcal{A}$ be a $\sigma$-algebra on $X$. Let $x$ be a member of $X$. Define a function $\mu_x:\mathcal{A}\to[0,+\infty]$ be letting $\mu_x(A)$ be $1$ if $x\in A$ and letting $\mu_x$ be $0$ if $x\notin A$. Then $\mu_x$ is a measure. But such desired $B_1$ may not exist. – Beerus Apr 02 '24 at 13:30
  • In your counterexample, if $x \in A$, then the outer measure of $A$ is 1 and every measurable superset of $A$ will have measure 1 as well, so my claimed inequality should be fine. If $ x\notin A$ and its outer measure is 0, then there must be some superset not containing $x$ (otherwise the outer measure would be 1) so the inequality still works. If the outer measure is again 1, we are in a similar case as above. I might be overlooking something, can you clarify? – noam.szyfer Apr 02 '24 at 15:13
  • Well, that indeed may not be a solid example to explain my concern. And after a second thought, I think your idea is valid but need some adjustment in expression. I added something to your answer to illustrate my point. Please let me know your thought. I really appreciate it! – Beerus Apr 02 '24 at 16:04
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    That’s just the definition of infinum. If there were no such $B_1$, then the infimum would be higher. – Eric Apr 03 '24 at 05:12
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    @Eric I think I was worried about $\infty$. If $\mu(B)=+\infty$ for all $B\in\mathcal{A}$ such that $A\subseteq B$, then $\mu^(A)=+\infty$, and so we cannot write the strict inequality $\mu(B_1)<\mu^(A)+\frac{\epsilon}{2}$. Moreover, such a senario is certainly possible. For instance, we can define a measure $\mu:\mathcal{A}\to[0,+\infty]$ on $(X,\mathcal{A})$ by letting $\mu(B)=+\infty$ if $B\neq\emptyset$ and $\mu(\emptyset)=0$, and then let $A\neq\emptyset$. – Beerus Apr 03 '24 at 15:34
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    Ahh, sure - but that doesn’t matter. In that case the original theorem is obviously true (everything $\leq \infty$). In particular, we assumed for the sake of contradiction that $\mu^(A)$ is strictly lower than $\mu_(A)$. – Eric Apr 03 '24 at 19:34
  • @Eric Yes, you're indeed correct! Honestly, sometimes I am not sure whether I should state detail like this in a proof. Because if I don't, I feel the proof is not rigorous; but if I do, it may make the proof prolix. – Beerus Apr 04 '24 at 01:13
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Based on @noam.szyfer's answer and our discussion in the comment section under his post, I would like to propose the following solution to improve the rigorousness. I really appreciate @noam.szyfer's help!

We want to prove \begin{align*} \mu_*(A)\leq\mu^*(A)\ \text{holds for each subset $A$ of $X$}. \end{align*}

Assume to the contrary that $\mu_*(A)>\mu^*(A)$ for some $A\in X$. If $\mu^*(A)=+\infty$, then we have already reached a contradiction. So assume that $\mu^*(A)<+\infty$. Note that it follows from this assumption that there exists a $B_1 \in \mathcal{A}$ such that $A \subseteq B_1$ and $\mu(B_1)<+\infty$.

Without loss of generality, suppose that $\mu_*(A)=\mu^*(A)+\epsilon$ for some $A\in X$ and $\epsilon>0$. Since $\mu^*(A)$ is the greatest lower bound of the set $\{\mu(B):A \subseteq B\ \text{and}\ B\in\mathcal{A}\}$, it follows that there is a $B_1\in\mathcal{A}$ such that $A \subseteq B_1$ and \begin{align*} \mu(B_1)<\mu^*(A)+\frac{\epsilon}{2}.\tag1 \end{align*} For otherwise, if $\mu(B_1)\geq\mu^*(A)+\frac{\epsilon}{2}$ for all $B_1 \in \mathcal{A}$ such that $A \subseteq B_1$, then $\mu^*(A)$ would not be the greatest lower bound of the set $\{\mu(B):A \subseteq B\ \text{and}\ B\in\mathcal{A}\}$, because $\mu^*(A)+\frac{\epsilon}{2}$ would be a lower bound bigger than $\mu^*(A)$.

Similarly, $\mu_*(A)$ is the least upper bound of the set $\{\mu(B):B \subseteq A\ \text{and}\ B \in \mathcal{A}\}$. If $\mu_*(A)=+\infty$, then there exists a $B_2\in\mathcal{A}$ such that $B_2 \subseteq A$ and $\mu(B_2)=+\infty$. But $B_2 \subseteq A \subseteq B_1$ and $B_1,B_2\in\mathcal{A}$ would imply that $\mu(B_2)\leq\mu(B_1)$, which is impossible because $\mu(B_1)<+\infty=\mu(B_2)$. So we obtain that $\mu_*(A)<+\infty$. Note that the finiteness of $\mu_*(A)$ implies the finiteness of $\epsilon$. It then follows that there is a $B_2\in\mathcal{A}$ such that $B_2 \subseteq A$ and \begin{align*} \mu_*(A) < \mu(B_2) + \frac{\epsilon}{2},\tag2 \end{align*} which (since $\epsilon<+\infty$) implies \begin{align*} \mu_*(A) - \frac{\epsilon}{2} < \mu(B_2). \end{align*} For otherwise, if $\mu_*(A)-\frac{\epsilon}{2}\geq\mu(B_2)$ for all $B_2\in\mathcal{A}$ such that $B_2 \subseteq A$, then $\mu_*(A)$ would not be the least upper bound of the set $\{\mu(B):B \subseteq A\ \text{and}\ B \in \mathcal{A}\}$, because $\mu_*(A)-\frac{\epsilon}{2}$ would be an upper bound smaller than $\mu_*(A)$.

Hence, by inequalities (1) and (2), we have \begin{align*} \mu(B_1) + \mu_*(A) < \mu^*(A) + \mu(B_2) + \epsilon, \end{align*} which (by plugging in our assumption $\mu_*(A)=\mu^*(A)+\epsilon$) implies \begin{align*} \mu(B_1)+\mu^*(A)+\epsilon < \mu^*(A)+\mu(B_2)+\epsilon. \end{align*} Since $\mu^*(A)$ and $\epsilon$ are both finite, it follows that \begin{align*} \mu(B_1)<\mu(B_2). \end{align*} However, since $B_2 \subseteq A \subseteq B_1$ and $B_1,B_2\in\mathcal{A}$ imply that $\mu(B_2)\leq\mu(B_1)$, we reached a contradiction. Therefore, we have proved that $\mu_*(A)\leq\mu^*(A)$ holds for each subset $A$ of $X$.

Beerus
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    I think this is exactly the argument I had in mind, good to see all the details worked out! – noam.szyfer Apr 03 '24 at 06:25
  • @noam.szyfer Thanks a lot for checking the work! Honestly, I am not very experienced in formal proof-based math. And as a self-learner, I sometimes get stuck on details or get suspecious on whether a statement is rigorous or not. For example, without disscussing $\infty$, I am not comfortable writing down an inequality like $(1)$. I think I just need to practice more to build intuition. Anyway, I appreciate your help with this question! – Beerus Apr 03 '24 at 14:12
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Here is an answer which carries on from your initial approach.

Suppose for a contradiction that $\mu^{*}(A) < \mu_{*}(A)$ for some $A\subseteq X$.

Step 1: As $\mu^{*}(A) < \mu_{*}(A)$, $\mu_{*}(A)$ is not a lower bound of $\{\mu (B):A\subseteq B\ \text{and}\ B\in\mathcal{A} \}$. So there is some $B\in\mathcal{A}$ such that $A\subseteq B$ and $\mu (B) < \mu_{*}(A)$.

Step 2: As $\mu^{*}(A) < \mu_{*}(A)$, $\mu^{*}(A)$ is not an upper bound of $\{\mu (C) : C\subseteq A\ \text{and}\ C\in\mathcal{A}\}$. So there is some $C\in\mathcal{A}$ such that $C\subseteq A$ and $\mu^{*}(A) < \mu (C)$.

Step 3: Note that $C\subseteq A \subseteq B$ and

$$\mu (B) < \mu_{*}(A) < \mu^{*}(A) < \mu (C).$$

However, by the monotonicity of the measure, $C\subseteq B$ implies $\mu (C) \leq \mu (B)$. This is a contradiction.

Therefore, $\mu_{*}(A) \leq \mu^{*}(A)$ for all $A\subseteq X$ as desired.

Dean Miller
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Here’s a simpler proof without needing proof by contradiction nor needing to delve into $\epsilon$’s. In particular, all you need is that an infinum is the greatest lower bound and the supremum is the lowest upper bound.

Note that for any sets $B_1,B_2$ with $B_1\subseteq A \subseteq B_2$, we have $B_1 \subseteq B_2$ and so $\mu(B_1)\leq \mu(B_2)$.

For any fixed $B_2$, it’s measure is greater than or equal to the measure of all the $B_1$’s within $A$, so it is an upper bound and so by definition of the supremum $\mu_*(A)\leq \mu(B_2)$.

Similarly, $\mu_*(A)$ is now less than or equal to the value of any measures of sets $B_2$ that contain $A$, so it is a lower bound and by the definition of the infimum $\mu_*(A) \leq \mu^*(A)$.

Eric
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