For $A$ Lebesgue measurable, $\exists H$ Borel s.t. $H$ is a countable intersection of open sets, $A \subseteq H$ and $\bar\lambda(H\setminus A) = 0$

analysisborel-setslebesgue-measuremeasure-theoryreal-analysis

I'm given the following question, which I shall state in full. It's actually a bit more involving than what the title asks for.

Let $A \subseteq \mathbb{R}$ be Lebesgue measurable. Show that there
are Borel sets $H_1$ and $H_2$ such that:

  1. $H_1 \subseteq A \subseteq H_2$,
  2. $H_1$ is a countable union of compact sets and $H_2$ is a countable intersection of open sets,
  3. $\bar\lambda(H_2 \setminus H_1) = \bar\lambda(H_2 \setminus A) = \bar\lambda(A \setminus H_1) = 0$.

Prior exercises have yielded the following lemmas:

If $A \subseteq \mathbb{R}$, then $\lambda^\ast (A) = \inf\left\{\bar\lambda(H) : H \ \mathrm{open}, A \subseteq H\right\}$

and

If $A \subseteq \mathbb{R}$ is Lebesgue measurable, then $\bar\lambda (A) = \sup\left\{\bar\lambda(F) : F \ \mathrm{compact}, F \subseteq A\right\}$

Using these two lemmas, I was able to show (1) and (2), however not (3) in all cases — I'll evaluate on this soon. To get Borel sets satisfying (1) and (2), for any $n \in \mathbb{N}$ one can find an open set $O_n$ and a compact set $F_n$ satisfying, respectively:

  • $A \subseteq O_n$ and $\bar\lambda(O_n) \leq \bar\lambda(A) + \frac{1}{n}$ by properties of infimum.
  • $F_n \subseteq A$ and $\bar\lambda(A) \leq \bar\lambda(F_n) + \frac{1}{n}$ by properties of supremum.

Taking countable intersections of $\left\{O_n\right\}_{n \in \mathbb{N}}$ and countable unions of $\left\{F_n\right\}_{n \in \mathbb{N}}$ yields Borel sets $H_1$ and $H_2$ satisfying (1) and (2). Indeed, $\bar\lambda(H_1) = \bar\lambda(A) = \bar\lambda(H_2)$.

Is (3) satisfied?

Consider the case when $\bar\lambda(A) < +\infty$. Recall that for a content $\mu$ on a ring $\mathcal{R}$, given two sets $B$ and $C$ in $\mathcal{R}$ such that $B \subseteq C$ and $\mu(B) < +\infty$, then $\mu(C\setminus B) = \mu(C) – \mu(B)$. Using this fact, clearly then for example, $\bar\lambda(H_2 \setminus H_1) = \bar\lambda(H_2) – \bar\lambda(H_1) = 0$ since $\bar\lambda(H_1) = \bar\lambda(A) = \bar\lambda(H_2)$. Similarly for the others.

The issue arises with whenever $\bar\lambda(A) = +\infty$. Such a technique as used above cannot be used, so it's out of the window immediately. What follows is part of my attempt, however I'm still stuck. I'll focus on $H_2$, as I believe that I have managed to find an $H_1$ which satisfies (3) in all cases.

For $n \in \mathbb{N}$, define $A_n = A \cap [-n, n]$. Since $A$ and $[-n, n]$ are Lebesgue measurable, then so is $A_n$. For each $A_n$ we can find, noting that $\bar\lambda(A_n) < +\infty$, an $H_2^{(n)}$ satisfying (1) – (3) by the above arguments.

If I take $H_2$ as $\bigcup_{n \in \mathbb{N}} H_2^{(n)}$, I get a Borel set which satisfies (1) and (3) this time, but not (2) — why? Each $H_2^{(n)}$ is the countable intersection of open sets – but $H_2$ is the countable union of the Borel sets $H_2^{(n)}$ i.e. a countable union of countable intersections of open sets, and not simply a countable intersection of open sets as required.

This is also the reason why I believe that I have managed to prove the result for $H_1$ (countable union of countable unions, defining $H_1^{(n)}$ in a similar manner to $H_2^{(n)}$). I have also seen exercises which simply ask one to find a Borel set $G$ such that $A \subseteq G$ and $\bar\lambda(G \setminus A) = 0$ — which the above would yield, no problem (each $H_2^{(n)}$ is indeed Borel).

Am I missing something? Am I over complicating things? Your input will be greatly appreciated! Thanks 🙂

Best Answer

See the equivalence of parts (a), (c), and (f) of Theorem 2.71 of Measure, Integration & Real Analysis, which has an electronic version that is legally available for free at https://measure.axler.net/.

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