# Measurable sets – Part 2

I am going to start now exactly where I stopped in the last post. I will prove that the set of measurable sets is indeed a $\sigma$-algebra. To do so, I only need to prove that if $\{E_n\}_{n=1}^\infty$ is a collection of measurable sets, then $\bigcup_{n=1}^\infty E_n$ is also a measurable set. First I want to prove a small theorem:

Let $E_1,\dots,E_n$ be some pairwise disjoint measurable sets, and define $E=\bigcup_{k=1}^n E_k$. Then, for every $A\sub \mathbb{R}$:

m^*(A\cap E) = \sum_{k=1}^n m^*(A\cap E_k)

The proof is going to be done with induction. The case $n = 1$ is trivial. For general $n$, since $E_n$ is measurable, we have:

m^*(A\cap E)=m^*((A\cap E)\cap E_n)+m^*((A\cap E)\cap E_n^C)

I am just using the definition of a measurable set here. Now, recall that $E=\bigcup_{k=1}^n E_k$, then: $E\cap E_n = E_n$ and $E\cap En^C = \bigcup_{k=1}^{n-1} E_k$ – this one follows from the fact that the sets are pairwise disjoint. Apply that to the equation in the avobe:

m^*(A\cap E)=m^*(A\cap E_n)+m^*(A\cap \bigcup_{k=1}^{n-1}E_k)=m^*(A\cap E_n)+m^*( \bigcup_{k=1}^{n-1}(A\cap E_k))

By induction, we know that: $m^*( \bigcup_{k=1}^{n-1}(A\cap E_k)) = \sum_{k=1}^{n-1} m^*(A\cap E_k)$, thus we get:

m^*(A\cap E)=m^*(A\cap E_n)+m^*( \bigcup_{k=1}^{n-1}(A\cap E_k))
=m^*(A\cap E_n)+\sum_{k=1}^{n-1} m^*(A\cap E_k)=\sum_{k=1}^{n} m^*(A\cap E_k)

And the proof is now complete.

Ok, so we proved this theorem, what can we do with it? Well, if you notice, this theorem applies for any $A \sub \mathbb{R}$, therefore, we can pick $A=E$ and get:

m^*(\bigcup_{k=1}^n E_k)=m^*(E)=m^*(E\cap E) = \sum_{k=1}^n m^*(E\cap E_k)=\sum_{k=1}^n m^*(E_k)

And that’s the finite version of property 4! If you don’t remember what is property 4, or just too lazy to go the last post, fine, I’ll write it here again:

• If we consider the disjoint union $E=\bigcup_{n=1}^{\infty}E_{n}$ we want that: $m(E)=\sum_{n=1}^{\infty}m(E_{n})$.

That’s actually great, since this is a the property that caused us to narrow down the domain to measurable sets only. However, this is not exactly the whole property, we also need to prove the case where the union is not finite, but it is countable. In order to prove it, I want to prove another theorem first, which is more general and is related to algebra of sets:

Let $X$ be a set, and $S$ algebra of sub-sets of $X$. Suppose that $\{E_n\}_{n=1}^\infty\sub S$, then there is a sequence $\{F_n\}_{n=1}^\infty$ of sets in $S$ such that:

1. $F_n\sub E_n$ for all $n$.
2. $F_n$ are pairwise disjoint
3. $\bigcup_{k=1}^n E_k = \bigcup_{k=1}^n F_k$ for every $n$.
4. $\bigcup_{k=1}^\infty E_k = \bigcup_{k=1}^\infty F_k$

The proof is am implicit construction of $F_n$. $F_n$ is going to be the set of all the ‘new’ elements in the sequence $E_n$. Formally, the set $F_n$ is defined as: $F_n=E_n\cap(\bigcup_{k=1}^n E_k)^C$ – we are taking the ‘new’ elements only!

Look at the illusration carefully – I think it gives a lot of intuition. It’s really not hard to see that this construction satisfies all 4 properties, I won’t show it though.

This theorem gives us a great advantage – we can now replace a sequnce of sets with a sequence of pairwise disjoint sets.

That’s it, we are now ready for out major theorem:

Let $\{E_n\}_{n=1}^\infty$ be a collection of measurable sets. Let $E = \bigcup_{n=1}^\infty E_n$ be thier union. Then $E$ is measurable.

Using the last theorem we can assume that $E_n$ are pairwise disjoint (if not, we can replace them with $F_n$ like in the above). Now, for every $n \in \mathbb{N}$ we define:

G_n=\bigcup_{k=1}^nE_k

Of course, $G_n \sub E$, thus $E^C \sub G_n^C$ . Let $A\sub R$ be some set, since $G_n$ is measurable (we proved in the previous post that finite union of measurable sets, is measurable) we get:

m^*(A)=m^*(A\cap G_n)+m^*(A\cap G_n^C)

since $E^C \sub G_n^C$, we get that $A\cap E^C \sub A\cap G_n^C$ as well, Thus: $m^*(A\cap G_n^C)\geq m^*(A\cap E^C)$. Apply that to get:

m^*(A)=m^*(A\cap G_n)+m^*(A\cap G_n^C)\geq m^*(A\cap G_n)+m^*(A\cap E^C)
=m^*(A\cap \bigcup_{k=1}^nE_k)+m^*(A\cap E^C)=m^*( \bigcup_{k=1}^n (A\cap E_k))+m^*(A\cap E^C)
=\sum_{k=1}^nm^*(A\cap E_k)+m^*(A\cap E^C)

Eventually we have:

m^*(A)\geq\sum_{k=1}^nm^*(A\cap E_k)+m^*(A\cap E^C)

and when $n\to\infty$ we get:

m^*(A)\geq\sum_{k=1}^\infty m^*(A\cap E_k)+m^*(A\cap E^C)

I want you to remember this inequality!

Now, since $E=\bigcup_{n=1}^\infty E_n$, we have $A\cap E=A\cap \bigcup_{n=1}^\infty E_n = \bigcup_{n=1}^\infty A\cap E_n$ We have:

m^*(A\cap E)\leq \sum_{k=1}^\infty m^*(A\cap E_k)

(I proved this statement at the end of this post ) We can now apply it to get:

m^*(A)\geq\sum_{k=1}^\infty m^*(A\cap E_k)+m^*(A\cap E^C)\geq m^*(A\cap E)+m^*(A\cap E^C)

We now use the  “The star of the post” to conclude that $E$ is measurable as required!

Yes! we have finally proved that the set of measurable sets $\mathcal{M}$ is indeed a $\sigma$-algebra, as we wanted. However, we haven’t proved that the property 4 is valid, or have we? Remember the inequality I wanted you to remember? No? Fine I am talking about this one. Let’s plug in $A=E$ in it:

m^*(E)\geq\sum_{k=1}^\infty m^*(E\cap E_k)+m^*(E\cap E^C)=\sum_{k=1}^\infty m^*( E_k)+m^*(\empty)=\sum_{k=1}^\infty m^*( E_k)

And in this post, I proved the other direction of the inequality, and together, they complete the proof!

From now on, I am going to call property 4 by it’s real name: $\sigma$-additive.

## Conclutions

Notice how this proof telling us that the intersection $\bigcap_{n=1}^\infty E_k$ is measurable as well(Why? Hint: de-morgan).

We can also conclude that if $S\sub \mathbb{R}$ is countable, we can write $S=\bigcup_{n=1}^\infty\{x_n\}$, and that’s a measurable set with zero measure!

Great, we have seen how powerful the reduction of $m^*$ to measurable sets is, so we are going to call it from now Lebesgue’s measure and denote it by $m$ (finally I can stop typing this little star).

## Filling Holes

If you noticed, I haven’t even shown a non-measurable set yet (to be honest, I have, but I haven’t proved that it is not measurable. In order to wrap everything up, I would like to give an example for such a set, Let’s begin:

We’ve actually seen a non-measurable set, this was exactly the set that motivated us to come up with the idea of measurable set. This is the set I used in this post. I’ll remind you how to construct it:

• define an equivalence relation on \mathbb{R}R: $x\sim y \iff x-y\in\mathbb{Q}$
• Our set $E\sub(0,1)$ will be made from exactly one elements from each equivalence class (I am using the axiom of choice in order to construct this set).

In this post, we found out that:

(0,1)\sub \bigcup_{r\in\mathbb{Q}\cap(-1,1)}{E+r}\sub(-1,2)

Suppose now that $E$ is measurable, thus, each $E+r$ is measurable (we have also seen that the union is disjoint), apply the measure to get:

m((0,1))\leq m(\bigcup_{r\in\mathbb{Q}\cap(-1,1)}{E+r})\leq m((-1,2))

Since the measure is $\sigma$-additive (for those who missed it, this is the real name of property 4) we get:

1\leq \sum_{r\in\mathbb{Q}\cap(-1,1)}m({E+r})\leq 3

Let’s improve it a bit:

1\leq \sum_{r\in\mathbb{Q}\cap(-1,1)}m(E)\leq 3

Great, we now have an example of a non-measurable set, I don’t know how about you, but I thing this is a really not that big of a deal – I mean, why would we want to measure such a set?

## Open sets

As you probably know, we really like working with open sets (in the euclidian topology). Are they measurable? Let’s find out:

Suppose that $U\in\mathbb{R}$ is open. we know that open inetrvals are measurable. In addition, $\mathbb{R}$ has a countable sub-basis:

\mathcal{B}=\{(a,\infty):a\in\mathbb{Q}\}

We have proved that intervals of the form $(a,\infty)$ are measurable, thus, $\mathcal{B}$ is made up of measurable sets.

Since $\mathcal{B}$ is a sub-basis, we can express $U$ as:

U=\bigcup_{n=1}^\infty(\bigcap_{k=1}^{m_n}O_{k,n})

Where $O_{k,n}\in \mathcal{B}$. $\bigcap_{k=1}^{m_n}O_{k,n}$ is a finite intesection of measurable sets, so it’s measurable.

Therefore, $U$ is measurable as a union of measurable sets.

That’s great! Notice that since open sets are measurable, then so as closed sets – it follows from the simple fact that closed sets are complements of open sets.

Those facts are going to be quite helpful in the future, so make sure that you understand them.

## Summary

So wev’e seen our definition for a measurable set works great, and wev’e also seen an example for a non-measurable set. As a result, we are now one step closer to the ‘fixed’ definition for integral – An integral involves two thing: A set and a function. Wev’e talked about sets, so we still have to talk about functions. However, This will be in later posts, for now, I think that’s enough. See you in the next one!