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?


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 $S\subseteq T$?

  • $\begingroup$ 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

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