Measurable sets

In the last post, we were seeking for an ‘ideal measure function, however, we found out that such a function does not exits. There were four properties we wanted from the function to fulfill:

  1. For every E\subset \mathbb{R}, m(E) is defined and satisfies 0\leq m(E)\leq \infty
  2. For every interval E\subset \mathbb{R}, m(E)=|E|.
  3. If we ‘shift’ / ‘slide’ a set, we don’t want the output of the measure to change.
  4. 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}).

We have seen that the outer measure fulfills properties 1,2,3 and part of property 4, i.e m(E)\leq\sum_{n=1}^{\infty}m(E_{n}). Property 4 is actually an important one, and since the function fulfills all the other three, we know that the second property 4 must not be true – which is sucks. However, maybe we can make one property weaker, such that it is weak ‘enough’ to ‘allow’ property 4 to be true. That’s exactly what we are going to do. We will make the first property weaker, since some sets (like vitali set that we have seen before) are not that interesting, so it wouldn’t be too bad to exclude them out of the function’s domain.

What sets are going to be included though? I will adopt the critirea the mathematician Cartheodory suggested: First, we will call every set in the domain “measurable”. We say that a set E\sub \mathbb{R} is measurable if for every A\sub \mathbb{R}:

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

I don’t know about you, but this definition is really wierd, but i’ll try to find some explanation to it: By definition E\cup E^C=\mathbb{R}, so we are ‘splitting’ the real line into two parts, and calculating the outer measure of A in each part – and if the set / partition is nice enough, then the sum of the measures from each part will be exactly the outer measure of the set, which makes sense.

Let’s see what can we do with this definition: Let E,F be some measurable sets such that E\cap F = \emptyset. Let’s try to figure out what is the value of m^*(E\cup F). First, I’ll denote A=E\cup F. Now, since E is measurable, then by defintion:

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

Notice that: A\cap E= (E\cup F)\cap E =(E\cap E) \cup \overbrace{(E\cap F)}^{\emptyset}=E. And from similar process, we get A\cap E^C = F. Thus we get:

m^*(E\cup F)=m^*(A)=m^*(E) +m^*(F)

And that’s great! cause by induction we can prove that if E_1,\dots,E_n are measurable sets are pairwise disjoint then:

m^*(\bigcup_{i=1}^k E_i)=\sum_{i=1}^km^*(E_i)

And that’s a weaker version of property 4! This example motivates us to stick to this definition and maybe narrow down the domain to be the collection of the measurable sets. So that’s my goal for this post – I want to show that when dealing with measurable sets is enough.

Notice that E is measurable if and only if E^C is measurable, why? Since (E^C)^C=E we get for every set A\sub R:

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

The symmetry of the definiton proves that if one is measurable, so as it’s complement.

So that’s a nice property, I want to present now a short and very helpful statement that I am going to use a lot (I am even going to name it: “The star of the post”):

E\sub \mathbb{R} \text{ is measurable} \iff \text{for every}A\sub \mathbb{R}:m^*(A) \geq m^*(A\cap E)+m^*(A\cap E^C)

It might look like a weaker condition than the condition to being measurable, however, turns out that they are equivalent! So the direction (\Rightarrow) is obvious. We only need to show the other direction.

Suppose that m^* (A) \geq m^*(A\cap E)+m^*(A\cap E^C), we have proved in the last post, that for any two disjoint (wev’e proved it for countable many sets) sets X,Y:

m^*(X\cup Y) \leq m^*(X) + m^*(Y)

We can pick X=A\cap E^C and Y=A\cap E, those are dijoint sets, and their union X\cup Y is exactly A, thus:

m^*(A) \leq m^*(A\cap E^C) + m^*(A\cap E)

Combining it with our assumption to get an equality, which proves that E is measureable.

Fine, so we have a nice criteria to determine whether or not a set is measurable, but we haven’t asked the most imporant question yet… Are there measurable sets? if there are, Who are they? Let’s figure out

What are the measurable sets?

Let’s start with a simple kind of sets – the sets with measure 0, i.e a set E such that m^*(E)=0. For any set A, it’s always true that A\cap E \sub E, thus:

 0= m^*(E)\geq m^*(A\cap E)

(Iv’e proved it in previous post). For E^C we get

m^*(A)\geq m^*(A\cap E^C)

Combining the result to get:

m^*(A)\geq m^*(A\cap E^C)=m^*(A\cap E^C)+0\ge m^*(A\cap E^C)+m^*(A\cap E)

Now we use “The star of the post” to conclude that E is measurable.

Perfect, we now have so many examples for measurable sets! since wev’e seen that m^*(\{x_0\})=0 for every x_0\in R, we now know that all the singletons in P(\mathbb{R}) are all measurable, and there are \aleph singeltons – and I’ll say that it’s quite a few.

Since we have proved that there are measurable sets, we can get back to proving properties of measurable sets – without taking the risk that no set is measurable (trust me, a case like that happened many times to many mathematicians – they started with some sort of structure, proved theorems related to it, and then discovered that it doesn’t even exist! That’s really sucks , however, I thinks it’s hilarious).

Ok, let’s try to prove now that if E is measurable, then for every x_0\in\mathbb{R}, x_0+E is measurable as well.

Suppose that E is measurable, x_0\in \mathbb{R} and pick an arbitraty set A\sub \mathbb{R}, I am now going to focus on the set A-x_0 (which has the same measure as A, as I proved before). Since E is measurable, we know that:

m^*(A)=m^*(A-x_0)=m^*((A-x_0)\cap E) + m^*((A-x_0)\cap E^C)

now, notice that (A-x_0)\cap E + x_0 = A\cap(E+x_0) and (A-x_0)\cap E^C + x_0 = A\cap(E^C+x_0) (try proving it if you are not comfortable with it), therefore we get:

m^*(A)=m^*(A-x_0)=m^*(A\cap (E+x_0)) + m^*(A\cap (E^C+x_0))

which proves that E+x_0 is measurable!

Good, everything is running smooth so far, let’s keep going. If we have two measurable functions E_1,E_2, is thier union E_1\cup E_2 is also measurable? I hope so… Let’s check:

Due to “The star of the post”, we only need to show that for every A \sub \mathbb{R}:

m^*(A) \geq m^*(A\cap (E_1\cup E_2))+m^*(A\cap (E_1\cup E_2)^C)

However, :

A\cap(E_1\cup E_2)=(A\cap E_1)\cup (A\cap E_2)\overset{(*)}{=}(A\cap E_1)\cup (A\cap E_1^C\cap E_2)

Try proving (*) yourself – the logic behind that is if you are an element of the set(A\cap E_1)\cup (A\cap E_2) then you are in A for sure, however you may not be an element of E_1, then you must be in E_2 while you are also in E_1^C

A little sketch that may help you understand it

On the other hand, we have;

A\cap(E_1\cup E_2)^C=A\cap(E_1^C\cap E_2^C)=A\cap E_1^C\cap E_2^C

We now have:

m^*(A\cap(E_1\cup E_2))+m^*(A\cap(E_1\cup E_2)^C)=
m^*((A\cap E_1)\cup (A\cap E_1^C\cap E_2))+m^*(A\cap E_1^C\cap E_2^C)
\leq m^*(A\cap E_1)+\overbrace{m^* (A\cap E_1^C\cap E_2)+m^*(A\cap E_1^C\cap E_2^C)}^{m^*(A\cap E_1^C)}
=m^*(A\cap E_1)+m^*(A\cap E_1^C)=m^*(A)

and we got what we wanted (again, using “The star of the post”).

Now, by induction we conclude that if E_1,\dots, E_n are all measurable, then \bigcup_{k=1}^nE_k is also measurable. But we proved even more than that: since, E_i are all measurable, then E_i^C are measurable as well, then bigcup_{i=1}^nE_i^C is also measureable, thus (\bigcup_{i=1}^nE_i^C)^C=\bigcap_{i=1}^nE_i is, again, measurable. Thus, finite union and the intersection of measurable sets are also measure sets!

Let’s summarize what we have found so far:

  • If E is measurable, then E^C is measurable as well.
  • if E_1,\dots,E_n are measurable, then \bigcup_{i=1}^nE_i is measurable.
  • if E_1,\dots,E_n are measurable, then \bigcap_{i=1}^nE_i is measurable.

Let’s denote by \mathcal{M} the set of all measurable sets, then we can write the properties in the above in a different way:

  • if E\in \mathcal{M} then E^C\in \mathcal{M}.
  • if E_1,\dots,E_n\in \mathcal{M}, then \bigcup_{i=1}^nE_i\in \mathcal{M}.
  • if E_1,\dots,E_n\in \mathcal{M}, then \bigcap_{i=1}^nE_i\in \mathcal{M}.

Such a set, which is closed under finite union, finite intersection, and taking the complement is called an Algebra of sets. If the set is closed under countable union / intersection, then it is called a Sigma-Algebra (\sigma-algebra). It turns out that \mathcal{M} is indeed a sigma-algebra, but we have to work hard in order to prove it and I’ll do it in the next post. So to wrap this one up I would like to present another type of sets which are measureable, and stop there:

Every interval of the form E=(a,\infty) for a\in\mathbb{R} is a measurable set. Before I’ll go into the proof, I just want to mention that from this class, we actually get a lot of new measureable sets! for example, suppose we have proved it, then we know that (0,\infty),(1,\infty) are measuable, then we also know that (1,\infty)^C=(-\infty, 1] is measurable, then, (0,\infty)\cap (-\infty, 1] = (0,1], that’s really nice.

Ok, Let’s prove it: I’ll pick A\sub \mathbb{R} and define:

A_1 = A\cap (a,\infty),A_2=A\cap (-\infty,a]

I’ll try to prove that:

m^*(A)\geq m^*(A_1)+m^*(A_2)

and that will be enough thanks to “The star of the post”.

And now pick some \varepsilon > 0 . since m^*(A) is defined as infimum, there is an open cover A\sub \bigcup_{n=1}^\infty I_n such that:

\sum_{n=1}^\infty|I_n|\leq m^*(A)+\varepsilon

Now, I’ll denote:

I_n^\prime = I_n\cap (a,\infty),I_n^{\prime\prime}=I_n\cap (-\infty,a]

Notice that I_n=I_n^\prime\cup I_n^{\prime\prime} and |I_n|=|I_n^\prime| + |I_n^{\prime\prime}|. Moreover:

A_1\sub \bigcup_{n=1}^\infty I_n^\prime,A_2\sub \bigcup_{n=1}^\infty I_n^{\prime\prime}


m^\star(A_1)+m^*(A_2)\leq\sum_{n=1}^\infty|I_n^\prime|+\sum_{n=1}^\infty|I_n^{\prime\prime}|=\sum_{n=1}^\infty|I_n^\prime|+|I_n^{\prime\prime}|=\sum_{n=1}^\infty|I_n|\leq m^*(A)+\varepsilon

and that’s true for every \varepsilon >0 thus m^*(A_1)+m^*(A_2)\leq\sum_{n=1}^\infty|I_n|\leq m^*(A) as required.


So we now have tons of examples of measurables sets, but we still have one debt, I have to prove that the set of measurable sets \mathcal{M} is indeed a \sigma-algebra, but that will be on the next post.

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