1
$\begingroup$

I have a kind of tricky question here that I would like to discuss with you.

Define $\mu_0 : \mathbb{I} \to \overline{\mathbb{R}}_{+}$ by $\mu_0((a,b]) = F(b)-F(a)$ where $F$ is a weakly increasing right-continuous function. Define the set function $\mu^*(E) = \inf \sum_{i=1}^{\infty} \mu_0(I_i)$, where the infimum is taken over all countable coverings $I_i \in \mathbb{I}$ of $E$. Show that $\mu^*$ is an outer measure.

So $\mu^*(\emptyset)=0$ since $\mu_0(\emptyset)=0$.

Next we need to show that for all $A \subset B \subset \mathbb{R}$, $\mu^*(A) \leq \mu^*(B)$. Suppose that $\mu^*(A) > \mu^*(B)$. But since $A \subseteq B$, for every $i_0$ there exists a $j_0$ such that $I_{i_0} \subseteq J_{j_0}$, whence $\mu_0(J_{j_0}) \geq \mu_0(I_{i_0})$ by definition of $F$. But this contradicts the fact that $\inf \sum_{i=1}^{\infty} \mu_0(I_i) > \inf \sum_{i=1}^{\infty} \mu_0(J_i)$. Thus, $\mu^*(A) \leq \mu^*(B)$.

I marked the shaky party above with boldface, is it correct?

Lastly, we need to show that for any sequence $(A_i)_{i \in \mathbb{N}} \subset \mathbb{R}$, we have

\begin{equation} \mu^* \left( \bigcup_{i=1}^{\infty} A_i \right) \leq \sum_{i=1}^{\infty} \mu^*(A_i). \end{equation}

I'm a little at a loss with this last one. How do I relate the countable coverings $I_i$ to the countable union of $A_i$'s?

$\endgroup$
2
$\begingroup$

Let me offer a better, direct proof. Since $A\subseteq B$, every cover of $B$ is also a cover of $A$. This means that $$\left\{\sum_{C\in \mathscr C}\mu(C):\mathscr C \text{is a countable cover of } B\right\}\subseteq \left\{\sum_{C\in \mathscr C}\mu(C):\mathscr C \text{is a countable cover of } A\right\}$$

What happens to $\inf S$ and $\inf T$ when $S\subseteq T$?

$\endgroup$
  • $\begingroup$ Ahhh, I think I see where you're getting at! :) Then $\inf S \leq \inf T$? Shouldn't $A$ and $B$ be switched in the $\{ \} \subseteq \{ \}$ expression you just wrote though? Since $A \subseteq B$, not $B \subseteq A$ $\endgroup$ – Numbersandsoon Feb 24 '14 at 7:26
  • $\begingroup$ @BoSchmidt You're not reading what I wrote. Read it again. Also, your infimum inequality is wrong, for example, $[0,1]\subset [-1,1]$, and $\inf [0,1]=0> -1=\inf[-1,1]$. $\endgroup$ – Pedro Tamaroff Feb 24 '14 at 7:30
  • $\begingroup$ Ah, because $A$ has at least as many coverings than $B$ (every covering of $B$ is a covering of $A$), that inclusion follows, I get it! And $\inf S \geq \inf T$ hence $\inf \{ \sum_{C \in \mathcal{C}} \mu(C) : \mathcal{C} \text{ is a countable cover of } $B$ \}$ $\geq \inf \{ \sum_{C \in \mathcal{C}} \mu(C) : \mathcal{C} \text{ is a countable cover of } $A$ \}$ $\endgroup$ – Numbersandsoon Feb 24 '14 at 7:33

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.