Folland provides a proof for the following proposition:
If $E \subset \mathbb{R}$, the following are equivalent:
- $E$ is Lebesgue Measurable (in the domain Lebesgue measure)
- $E = V - N_1$ where $V$ is a $G_{\delta}$ set and $N_1$ has measure zero.
- $E = H - N_2$ where $H$ is a $F_{\sigma}$ set and $N_2$ has measure zero.
He provides the following proof: To show the first implies the second and third, we being by noting that we can choose a compact $K_j \subset E$ and open $U_j \supset E$ such that $$\mu(U_j) - \frac{1}{2^j} \leq \mu(E) \leq \mu(K_j) + \frac{1}{2^j}$$ If we define $V = \bigcap_1^{\infty}U_j$ and $H = \bigcup_1^{\infty}K_j$, then $H \subset E \subset V$ and the measures of $H$,$E$, and $V$ are all equal.
My question: I can see how the inequality arises, then taking limit on the upper and lower bounds, together with the using the upper and lower continuity on measures I can see that $$\mu(V) \leq \mu(E) \leq \mu(H)$$ But by monotonicty this means: $$V \subset E \subset H$$ All my inclusions are going the wrong way! How is Folland concluding that $H \subset E \subset V$?
First of all, $\mu(V) \leq \mu(E) \leq \mu(H)$ definitely does not imply $V \subset E \subset H$. Monotonicity of the measure gives the converse of this implication, not this implication. For instance, $\mu([0,1])<\mu([2,4])$ but $[0,1]\not\subset[2,4]$.
As for where the inclusions $H \subseteq E \subseteq V$ come from, this is just immediate from the definitions of $H$ and $V$. By definition, $H$ is a union of sets $K_j$ which are all contained in $E$, so $H$ is also contained in $E$. Similarly, $V$ is an intersection of sets $U_j$ which all contain $E$, so $V$ also contains $E$.