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 -algebra. To do so, I only need to prove that if is a collection of measurable sets, then is also a measurable set. First I want to prove a small theorem:

Let be some ** pairwise disjoint** measurable sets, and define . Then, for every :

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

The proof is going to be done with **induction**. The case is trivial. For general , since 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 , then: and – 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: , 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 , therefore, we can pick 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**we want that: .

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 be a set, and **algebra of sub-sets** of . Suppose that , then there is a sequence of sets in such that:

- for all .
- are pairwise disjoint
- for every .

The proof is am implicit construction of . is going to be the set of all the ‘new’ elements in the sequence . Formally, the set is defined as: – 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 be a collection of measurable sets. Let be thier union. Then is measurable.

Using the last theorem we can assume that are pairwise disjoint (if not, we can replace them with like in the above). Now, for every we define:

G_n=\bigcup_{k=1}^nE_k

Of course, , thus . Let be some set, since 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 , we get that as well, Thus: . 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 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 , we have 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 is measurable as required!

Yes! we have finally proved that the set of measurable sets is indeed a -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 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: **-additive**.

## Conclutions

Notice how this proof telling us that the intersection is measurable as well(Why? Hint: de-morgan).

We can also conclude that if is **countable**, we can write , and that’s a measurable set with **zero **measure!

Great, we have seen how powerful the reduction of to measurable sets is, so we are going to call it from now **Lebesgue’s measure** and denote it by (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
- Our set 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 is measurable, thus, each 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 -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

And Wev’e already seen why this is a contradiction.

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

***Topology alert***

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

Suppose that is open. we know that open inetrvals are measurable. In addition, has a **countable **sub-basis:

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

We have proved that intervals of the form are measurable, thus, is made up of measurable sets.

Since is a sub-basis, we can express as:

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

Where . is a finite intesection of measurable sets, so it’s measurable.

Therefore, 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!

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