Show there is a set $H$ which is the intersection of countable many open sets containing $A$ with $\mu(H \setminus A)=0$

Question

Let $A \subset [0,1]$ be a Lebesgue measurable set and let $\mu$ denote the Lebesgue measure.

I am trying to show that there is a set $H$ which is the intersection of countable many open sets containing $A$ with $\mu(H \setminus A)=0$

This is the proof.

For each $n=1,2,\ldots$ there is an open set $G_n$ containing $A$ such that $\mu(G_n \setminus A) < \frac{1}{n}$

This follows from the fact that the Lebesgue outer measure $\mu(A)$ is the infimum of $\mu(G)$ over open sets $G$ containing $A$, and from measurability of $A$, $\mu(G_n \setminus A) = \mu(G_n)-\mu(A)$.

Why is $\mu(G_n \setminus A) < \frac{1}{n}$? Where does $\frac{1}{n}$ come from? And what is the explanation above trying to say?

Let $H=\bigcap^\infty_{n=1} G_n$. $H$ need not be open but it is the intersection of countably many open sets containing $A$ in particular it is Borel and Lebesgue measurable.

Why does $H=\bigcap^\infty_{n=1} G_n$? And how do we know it is Borel and Lebesgue measurable?

$\mu(H \setminus A) <\frac{1}{n}$ for every $n$ so $\mu(H \setminus A)=0$

In general what is this proof trying to say?

Answer 1

It looks as if you're trying to read a proof without knowing the basic definitions and without knowing some basic conventions.

As for "Why does $H=\bigcap^\infty_{n=1} G_n$?", the answer is that they just said "Let $H=\bigcap^\infty_{n=1} G_n$". That just means they're giving a name to this intersection of sets; they're calling it $H$.

You quoted a definition: Lebesgue outer measure $\mu(A)$ is the infimum of $\mu(G)$ over open sets $G$ containing $A$. That means $\mu(A)$ is the largest number that does not exceed the measure of any open set $G$ "containing" $A$. To say $G$ "contains" $A$ means that $A\subseteq G$, i.e. every member of $A$ is a member of $G$. To say that $\mu(A)$ is the largest number not exceeding the measure of any open set $G$ containing $A$ implies that every number larger than $\mu(A)$ does exceed the measure of some open set $G$ containing $A$. Thus for every number larger than $\mu(A)$ there is an open set $G$ containing $A$ whose measure is less than that number larger than $\mu(A)$. In particular, if $n\in\{1,2,3,\ldots\}$, then there is an open set $G_n$ containing $A$ whose measure is less than $\mu(A)+\frac 1 n$. The reason for concerning ourselves with $1/n$ for $n\in\{1,2,3,\ldots\}$ rather than with arbitrary positive numbers is that we want a set of positive numbers that is countably infinite rather than uncountable. Any for reasons that appear later in the proof, we want a set of such numbers that is bounded below by $0$ but is not bounded below by any positive number. The set $\{1/1, 1/2, 1/3,\ldots\}$ qualifies.

To understand how we know $H$ is Borel measurable, you need to recall what "Borel measurable" means. A Borel-measurable set, or "Borel set", is any set that can be made by starting with open sets and applying complementation, countable unions, and countable intersections. $H$ was defined to be a countable intersection of open sets.

Say the measure of $A$ is $0.8$. Then the outer measure of $A$ and the inner measure of $A$ are both $0.8$. That outer measure is $0.8$ means that for every number $0.8+\varepsilon$ bigger than $0.8$, there is some open set containing $A$ whose measure is less than $0.8+\varepsilon$, but there is no open set containing $A$ whose measure is less than $0.8$. That means in particular

• There is some open set $G$ containing $A$ whose measure is less than $0.8+ \dfrac 1 2$; and
• There is some open set $G$ containing $A$ whose measure is less than $0.8+ \dfrac 1 3$; and
• There is some open set $G$ containing $A$ whose measure is less than $0.8+ \dfrac 1 4$; and
• There is some open set $G$ containing $A$ whose measure is less than $0.8+ \dfrac 1 5$;
• and so on.

The intersection of all of these must have measure at least $0.8$ since it contains $A$. But the measure of the intersection must be less than $0.8 + \frac 1 2$, and less than $0.8+\frac13$ and less than $0.8+\frac14$, and so on. The only number not less than $0.8$ that is less than all of those, is $0.8$. Hence the measure of the intersection must be $0.8$. And the intersection is the intersection of countably many open sets.

Answer 2

Infimum means greatest lower bound. The infimum of a set of real numbers can be approximated arbitrarily closely with elements from that set, in particular, more closely than the reciprocal of arbirarily large natural numbers $n.$

$H$ is defined in that way, it does not appear earlier in the proof. Countable intersections of measurable sets are measurable in any $\sigma$-algebra.

A nonnegative real number that is less than the reciprocal of arbitrarily large natural numbers must be zero.

Answer 3

1. You should think of $\frac1n$ as a "small quantity"; as $n$ increases, $\frac1n$ gets smaller and smaller. So, the first part of the proof is just telling you that your set can be approximated from above in measure with open sets whose measure is closer and closer to the one of $A$ (that is $\mu(G_n\setminus A)<\frac1n$). This is true by the fact that $\mu(A)$ is the infimum of the measures of the open sets containing $A$.

2. The set $H$ is defined to be the intersection of the $G_n$'s. Why? Because we need it to be such a set. $H$ contains $A$ because every $G_n$ does. By definition, $H$ is a $G_{\delta}$ set, i.e. a countable intersection of open sets. $H$ needs not to be open, but it is a Borel set (as open sets are Borel sets) and hence is Borel measurable. It is automatically Lebesgue measurable, as all Borel measurable set are Lebesgue measurable. Since the intersection os over all $n$, by taking the limit you get that $\mu(H\setminus A)=0$.

3. The proof is "trying to tell you" that such a set $H$ exists. Indeed, the proof is a construction of such set. Sometimes in measure theory it might be handy to have a Borel set containing your set with the same measure.