# Prove that $\mu_*\leq\mu^*$ where the definition of an inner and outer measure is induced by a measure

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.

Reference: Measure Theory by Donald Cohn Page 33.

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.

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

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 $$X$$.

• I think this is exactly the argument I had in mind, good to see all the details worked out! Apr 3 at 6:25
• @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! Apr 3 at 14:12

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.

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

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