0
$\begingroup$

Denote the Lebesgue outer measure $m^*(A) = inf \{\sum_{n=1}^{\infty} \ell (I_n): A \subseteq \cup_{n=1}^{\infty} I_n$ where $I_n$’s open intervals$\}$.

Consider the “measure” (which is, as it turns out, not actually a measure) $m^{**}$ given by $m^{**}(A) = inf \{\sum_{n=1}^{N} \ell (I_n): A \subseteq \cup_{n=1}^{N} I_n$ where $I_n$’s open intervals$\}$. So, basically, a finite version of the Lebesgue measure.

i) Show that $m^*(A) = m^{**}(A)$ for compact sets $A \subseteq \mathbb{R}$.

ii) Show that $m^{**}(A) = m^{**}(\bar{A})$ for bounded sets $A \subseteq \mathbb{R}$.

I am not really sure how to proceed. Obviously, for the first one, I want to try and use the definition of compactness which says that any open cover of A has a finite sub cover. But then how do I “translate” it to open intervals?

I am not sure exactly how to proceed for the second, but I am hoping to get a better idea on how to work with these functions from the first. Also, obviously I have that $(\bar{A})$ is compact for bounded A.

$\endgroup$
1
$\begingroup$

You have $m^* \le m^{**}$ automatically because the $\inf$ is taken over a larger set with $m^*$.

Now let $\epsilon>0$ and suppose $A$ is compact. Then choose $I_n$ such that $\sum_n l(I_n) < m^*(A) + \epsilon$. Since $A$ is compact, there is a finite index set $J$ such that $A \subset \cup_{n \in J} I_n$ and so $m^{**}(A) \le \sum_{n \in J} l(I_n) \le \sum_n l(I_n) < m^*(A) + \epsilon$. Since $\epsilon $ was arbitrary, we have $m^{**}(A) \le m^* (A)$.

We always have $m^{**}(A) \le m^{**}(\overline{A})$. Choose $\epsilon>0$ and find open $I_n$ such that $\sum_n l(I_n) < m^{**}(A) + \epsilon$. Let $I_N' = (\inf I_n -{1 \over 2m} \epsilon, \sup_n I_n+ {1 \over2 m} \epsilon) $, where $m$ is the number of intervals in the cover. (Observe that the closed cover $[\inf I_n , \sup_n I_n]$ contains the closure of $A$.) Note that $m^{**}(\overline{A}) \le \sum_n l(I_n') < m^{**}(A) + 2\epsilon$ and since $\epsilon>0$ was arbitrary, we have $m^{**}(\overline{A}) \le m^{**}(A) $.

As an aside, if we let $A= [0,1] \cap \mathbb{Q}$, we see that $m^*(A) = 0 < 1 = m^{**}(A) = m^{**}(\overline{A}) = m^*(\overline{A})$.

$\endgroup$
0
$\begingroup$

For the second question, first note that $m^{**}(A)\leqslant m^{**}(\overline{A})$ (and more generally, if $B\subset C$ for any $B,C$, then $m^{**}(B)\leqslant m^{**}(C)$).

We want to show the other direction. Fix $I_1, \ldots, I_m$ open intervals such that $A\subset \cup_{i=1}^m I_i$ and $\sum_{i=1}^m l(I_i)<m^{**}(A)+\epsilon/2$. Then $\overline{A}\subset \overline{\cup_{i=1}^m I_i}=\cup_{i=1}^m \overline{I}_i$. Let $J_i$ be an open interval containing the left endpoint of $I_i$ and with $l(J_i)<\epsilon/4m$. Let $K_i$ be an open interval containing the right endpoint of $K_i$ with $l(K_i)<\epsilon/4m$. Then $\overline{I_i}\subset J_i\cup I_i\cup K_i$, and $\overline{A}\subset \cup_{i=1}^m J_i\cup I_i\cup K_i$. Moreover, each $J_i, I_i, K_i$ is an open interval and \begin{align*}m^{**}(\overline{A}) & \leqslant \sum_{i=1}^m l(J_i)+l(I_i)+l(K_i) < m^{**}(A)+\epsilon/2 +\sum_{i=1}^m l(J_i)+l(K_i) \\ & <m^{**}(A)+\epsilon/2+2m(\epsilon/4m)=m^{**}(A)+\epsilon. \end{align*} Since this holds for any $\epsilon>0$, $m^{**}(\overline{A})\leqslant m^{**}(A)$. Since we already know the reverse inequality, we have equality.

Note also that we can identify the point in this proof which makes this work, but why the analogous thing is false if we try it with $m^*$. The point of difference is that $\overline{\cup_{i=1}^m I_i}=\cup_{i=1}^m \overline{I}_i$ because the union is finite (the closure of the union is the union of the closures for any finite union in any topological space). But for an infinite union, the union of the closures $\cup_{i=1}^\infty \overline{I}_i$ can be a proper subset of $\overline{\cup_{i=1}^\infty I_i}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.