6
$\begingroup$

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.

$\endgroup$

4 Answers 4

3
$\begingroup$

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.

$\endgroup$
9
  • 2
    $\begingroup$ 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 $B_2$. $\endgroup$
    – Beerus
    Apr 1 at 23:51
  • $\begingroup$ (Cont'd) (plus without specifing what $\mu$ is) $\endgroup$
    – Beerus
    Apr 2 at 2:23
  • 2
    $\begingroup$ That’s just the definition of infinum. If there were no such $B_1$, then the infimum would be higher. $\endgroup$
    – Eric
    Apr 3 at 5:12
  • 1
    $\begingroup$ @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 $A\neq\emptyset$. $\endgroup$
    – Beerus
    Apr 3 at 15:34
  • 2
    $\begingroup$ 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 $\mu_*(A)$. $\endgroup$
    – Eric
    Apr 3 at 19:34
3
$\begingroup$

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 $X$.

$\endgroup$
2
  • 1
    $\begingroup$ I think this is exactly the argument I had in mind, good to see all the details worked out! $\endgroup$ Apr 3 at 6:25
  • $\begingroup$ @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! $\endgroup$
    – Beerus
    Apr 3 at 14:12
3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $\mu_*(A) \leq \mu^*(A)$.

$\endgroup$
2
  • $\begingroup$ +1 This is clearly the "right" proof --- except that, in the last paragraph, "now at least" should be "now at most". $\endgroup$ Apr 4 at 17:47
  • $\begingroup$ No, it’s right, though I tweaked it to be clearer. $\endgroup$
    – Eric
    Apr 4 at 18:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .