I'm trying to solve the Folland Real analysis p.32 problem 19 and it was easy to show that the inner measure of a measurable set equals the outer measure. However I'm stuck at the converse. Could anyone help me how to prove that if the inner measure equals the outer meausure, then the set is measurable?

This is the definition of measurability in Folland.enter image description here

And the problem 19 is the one that I'm stuck at.enter image description here

  • 1
    $\begingroup$ For those who don't have Folland's book at hand, it would help if you gave the definition of "measurable" that is to be used. Some books define measurable to be when inner measure equals outer measure, which makes your question quite easy. :) $\endgroup$ – Dave L. Renfro Jan 14 '15 at 19:18

Suppose $\mu^*(E) = \mu_*(E) = \mu_0(X) - \mu^*(E)$. Then, because $\mu^*(X) = \mu_0(X) < \infty$, $$\mu^*(X) = \mu_0(X) = \mu^*(E) + \mu^*(E^c).$$

Now, we can use Exercise 18 (a) to show that for each $n\in \mathcal{N}$ there is an $A_n \in \mathcal{A}_\sigma$ such that $A_n \supset E$ and $\mu^*(E) \leq \mu^*(A_n) \leq \mu^*(E) + \frac{1}{n}$. Setting $B = \bigcap A_n$, we have that $$\mu^*(E) \leq \mu^*(B) \leq \mu^*(A_n) \leq \mu^*(E) + \frac{1}{n}, $$ for each $n$. Hence, $\mu^*(B) = \mu^*(E)$. Since each $A\in \mathcal{A}$ is measurable and the measurable sets form a $\sigma$-algebra, $B$ is also measurable.

Thus, we have that $$\mu^*(X) = \mu^*(E) + \mu^*(E^c) = \mu^*(B) + \mu^*(B^c).$$ Since $\mu^*(B) = \mu^*(E)$, we also have $\mu^*(B^c) = \mu^*(E^c)$. Using the measurability of $B$ again we have $$ \begin{align*} \mu^*(E) + \mu^*(E^c) &= \mu^*(E) + \mu^*(B\cap E^c) + \mu^*(B^c\cap E^c) \\ &= \mu^*(E) + \mu^*(B\cap E^c) + \mu^*(B^c)\\ & = \mu^*(E) + \mu^*(B\cap E^c) + \mu^*(E^c). \end{align*} $$

Hence, $\mu^*(B\setminus E) = 0$, so we can apply Exercise 18 (b) to show that $E$ is measurable.

  • $\begingroup$ Why is it true that $\mu^{\ast}(B) = \mu^{\ast}(E) \implies \mu^{\ast}(B^{C}) = \mu^{\ast}(E^{C}) $ $\endgroup$ – Muno Feb 1 '18 at 6:59
  • $\begingroup$ @Muno, because $B$ is measurable and $\mu^*(E) + \mu^*(E^c) = \mu^*(X)$. Note that $E\cup E^c = B\cup B^c$ and, therefore, $\mu^*(E) + \mu^*(E^c) = \mu^*(B) + \mu^*(B^c).$ $\endgroup$ – L.F. Cavenaghi Apr 19 '18 at 19:52

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