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I want to prove the equality of inner and outer measures for measurable sets. The hypothesis are these:

Let be $R$ an algebra over $X$, $\mu$ a quasi-measure over $R$ and $R^*$ the set of the measurable subsets of $X$, so we can build the outer and inner measures respectively: $$\mu^*(E)=\text{inf}\{\mu(A): A\in S(R),E\subseteq A\}$$ $$\mu_{*}(E)=\text{sup}\{\mu(A): A\in S(R),A\subseteq E\}.$$ Then: $E\in R^*$ iff $\mu_{*}(E)=\mu^*(E)$. We can suppose $\mu^*(E)<\infty.$

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  • $\begingroup$ I think you need to explain what a "putero" space is...and your question isn't understandable. You can write in your own language and perhaps someone an understand here. $\endgroup$ – DonAntonio Dec 5 '16 at 7:50
  • $\begingroup$ The hypothesis "ate" these. $\endgroup$ – астон вілла олоф мэллбэрг Dec 5 '16 at 7:54
  • $\begingroup$ Sorry, I've already corrected the statement, i had some troubles writing in my cell phone. $\endgroup$ – Alex Rosas Dec 5 '16 at 8:04
  • $\begingroup$ What is your definition of measurable set? Also, what is $S(R)$? Is this the $\sigma$-algebra generated by $R$? $\endgroup$ – Bungo Dec 5 '16 at 8:16
  • $\begingroup$ S(R) is $\sigma$-algebra generated by $R$ and we say that $E\subseteq X$ is measurable iff for all $A\subseteq X$ $$\mu^*(A)=\mu^*(A\cap E)+\mu^*(A-E)$$ $\endgroup$ – Alex Rosas Dec 5 '16 at 8:21
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Step 1: Let $E \subseteq X$ be an arbitrary set. Show that there exists $B \in S(R)$ such that $E \subseteq B$ and $\mu(B) = \mu^*(E)$.

Step 2: Let $E \subseteq X$ be such that $\mu^*(E) < \infty$. Choose $B$ be as in step 1. Then

$$\begin{align*} \mu_*(E) &= \sup \{\mu(A); A \subseteq E, A \in S(R)\} \\ &= \sup\{\mu(A); B \cap A^c \supseteq B \cap E^c, A^c \in S(R)\} \\ &= \sup\{\mu(B)-\mu(B \cap A^c); B \cap A^c \supseteq B \cap E^c, B \cap A^c \in S(R)\} \\ &= \mu(B) - \inf\{\mu(C); C \supseteq B \cap E^c, C \in S(R)\} \\ &= \mu(B)-\mu^*(B \backslash E). \end{align*}$$

Step 3: "$\Rightarrow$" Let $E \in R^*$ be such that $\mu^*(E) < \infty$. By the measurability of $E$, we have

$$\mu^*(E) = \mu^*(B) = \mu^*(B \cap E) + \mu^*(B \backslash E) = \mu^*(E) + \mu^*(B \backslash E).$$

As $\mu^*(E)<\infty$, this implies $\mu^*(B \backslash E)=0$. Combining this with step 2, we get

$$\mu_*(E) = \mu(B) = \mu^*(E).$$

Step 4: "$\Leftarrow$": Suppose that $E \subseteq X$ is such that $\mu^*(E) = \mu_*(E)$. Choose $B$ as in step 1. It follows from step 2 that $\mu^*(B \backslash E)=0$. For an arbitrary set $A \subseteq X$, we have

$$\begin{align*} \mu^*(A \cap E) + \mu^*(A \cap E^c)&\leq \mu^*(A \cap B) + \mu^*\big( (A \cap B^c) \cup (B \backslash E) \big) \\ &\leq \mu^*(A \cap B) + \mu^*(A \cap B^c) + \mu^*( B \backslash E) \end{align*}$$

which implies by the measurability of $B$

$$\mu^*(A \cap E) + \mu^*(A \cap E^c) \leq \mu^*(A) + \mu^*(B \backslash E) = \mu^*(A).$$

Since the inequality "$\geq$" is trivially satisfied (because of the subadditivity of the outer measure), this shows that $E$ is measurable.

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  • $\begingroup$ Thank you, the step 2 was truly useful, I was thinking in such a set like in step 1 but I had no idea how to use it. $\endgroup$ – Alex Rosas Dec 6 '16 at 4:00
  • $\begingroup$ @AlexRosas You are welcome. If $\mu$ is a finite quasi-measure, then it is very common to define the inner measure via $$\mu_*(E) = \mu(X) - \mu^*(X \backslash E).$$ $\endgroup$ – saz Dec 6 '16 at 6:38

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