# Measurable functions & Measure Spaces

So far, all of what I’ve been talking about was concerning the real line. The measurable sets that I’ve defined were subsets of $\mathbb{R}$, there is a good for reason for that though: If you remember, our main goal since the first post was to give a better definition to an Integral.

However, in math we like to generalize things and get a wider prespective on them. That’s what I’m going to do in the first part of this post, I am going to generilize what we have done so far in order to prove statements that are not specific to the real line with the lebesgue measure (which is the outer measure reduced to the measurable sets).

## Abstraction beginning

There are 2 main things that I can generalize here. If you recall a $\sigma$-algebra $S$ on a set $X$ is a collection of subset with the following properties:

• $E\in S \Rightarrow E^C\in S$. This means that for any set in $S$ , it’s complement is in $S$ as well – $S$ is closed under complement
• If $\{E_n\}_{n=1}^\infty$ is a sequence of sets in $S$ then their union $\bigcup_{n=1}^\infty E_n$ is an element of $S$ as well. $S$ is closed under union.
• $\emptyset\in S$.

In the previous posts, we worked really hard in order to prove that the set of all (lebesgue) measurable sets is indeed a $\sigma$-algebra.

So that’s going to be our first abstraction, a pair of a set $X$ and a $\sigma$-algebra on $X$$(X,S)$ is called a Measurable space.

Great, so we now have a definition of a measurable space, however, we are not done since this is a really boring an not so useful definition. Recall that the definition of a measurable set didn’t came out of nowhere – We had a function we worked with. That brings me to next definition:

Let $(X,S)$ be a measurable space. A measure on $(X,S)$ is a function:

## The abstract measure

\mu:S\to[0,\infty]

such that:

• $\mu(\emptyset) = 0$.
• $\mu$ is $\sigma$-additive: If $\{E_n\}_{n=1}^\infty$ in $S$ are pairwise disjoint then $\mu(\bigcup_n E_n)=\sum_n\mu(E_n)$.

As you probably remember, we we worked very hard in order to prove the second axiom, which is now a given.

Finally we are ready for the definition we were looking for: the trio $(X,S,\mu)$ is called a measure space.

That is the axiomatic approach of a measure space, and we can start proving some properties using the definitions. Before that, I would like to give two examples for measure space first:

• $(\mathbb{R},\mathcal{L}(\mathbb{R}),m)$, where $\mathcal{L}(\mathbb{R})$ is the set of the lebesgue measurable sets. This is what we have done so far.
• The counting measure – If $X$ is a set, we can define on all the subset of $X$, the measure $\mu(E)=|E|$ (if $E$ is infinite, then it’s measure is $\infty$). What this function does is just counting the number of elements in a set.

There are a lot of other measure spaces we can define which are really useful but I want to present some properties of a general measure here, so I’ll stop with the examples for now.

## Some properties

First, Suppose that $(X,S,\mu)$ is a measure space and $E,F\in S$ such that $E\subseteq F$. Notice that we can present $S$ as a disjoint union: $F=E\cup(F\setminus E)$. With the help of $\mu$ being $\sigma$-additive, we get:

\mu(F)=\mu(E\cup(F\setminus E))=\mu(E)+\mu(F\setminus E)\geq\mu(E)

Thus, in any measure space, we know that if $E\subseteq F$, then $\mu(E)\leq \mu(F)$.

Moreover, we can use the last equation to get that $\mu(F\setminus E)=\mu(F)-\mu(E)$, which is a pretty nice property.

How about one more property? Suppose that $\{E_n\}_{n=1}^\infty$ are sets in $S$ and $E=\bigcup_n E_n$ Then:

\mu(E)\leq \sum_{n=1}^\infty \mu(E_n)

If you recall, we have proved in this post that we can pick a collection of sets $\{F_n\}_{n=1}^\infty$ which are pairwise disjoint, $F_n\subseteq E_n$ and $E=\bigcup_n F_n$. We can now use the $\sigma$-additive property to get:

\mu(E)=\sum_n\mu(F_n)\leq \sum_n\mu(E_n)

As we wanted.

There are many more properties we can prove and we can discuss measurable spaces forever, but we have a different goal though – integrals. We already dealt with sets, now it is time to deal with functions.

## Looking for desired functions

Before I’ll start talking about functions, I want to give a little ‘spoiler’ – I am going to show the idea behind our new definition to an integral.

If you recall, Riemann integral is based on a partition of the $x$-axis, we are going to take a different approach here, we are going to use a partition on the $y$-axis.

Our sum is going to look like:

\sum_{k=1}^n y_k\cdot m(E_k)

Where $E_k = \{x\in[a,b]: y_{k-1}\leq f(x) \leq y_k$ (where $f$ is the function we are integrating on). By this definition, we see that it is kind of crucial that $E_k$ will be measurable. But what is this set? This is exactly the inverse image $f^{-1}((y_{k-1},y_k))$. Therefore, we need that the inverse image of intervals under $f$ will be measurable. This leads us to this new term:

## Measurable functions

As usual, $(X,S)$ is a measurable space, and $f:X\to\mathbb{R}\cup\{-\infty,\infty\}$. Then $f$ is said to be $S$ measurable if it satisfies one of the following:

1. for every $\alpha\in\mathbb{R}$: $f^{-1}((\alpha,\infty))=\{x\in X:f(x)>\alpha\}\in S$
2. for every $\alpha\in\mathbb{R}$: $f^{-1}([\alpha,\infty))=\{x\in X:f(x)\geq\alpha\}\in S$.
3. for every $\alpha\in\mathbb{R}$: $f^{-1}((-\infty, \alpha))=\{x\in X:f(x)<\alpha\}\in S$
4. for every $\alpha\in\mathbb{R}$: $f^{-1}((-\infty, \alpha])=\{x\in X:f(x)\leq\alpha\}\in S$

Notice that you only need to show one property, and then all the other properties are automatically valid to $f$ as well. Let’s prove it:

First notice that for every set and for every function: $f^{-1}(A^C)=(f^{-1}(A))^C$ (try prove it yourself, it’s a well-known fact from set theory). Apply it to our case to get:

(f^{-1}((\alpha,\infty))^C=f^{-1}((\alpha,\infty)^C)=f^{-1}((-\infty,\alpha])

Since $S$ is a $\sigma$-algebra, we conclude that if one on of the sets $f^{-1}((\alpha,\infty),f^{-1}((-\infty,\alpha])$ is in $S$, then so as the other.

So we just proved: $1\iff 4$ and (with similar process) $2\iff 3$.

How about $1\iff 2$? Assume that 1 is true and we will try to prove 3. I’ll use a cool trick here:

f^{-1}((-\infty, \alpha))=(f^{-1}((-\infty, \alpha)^C))^C=(f^{-1}([\alpha,\infty)))^C
=(f^{-1}(\bigcap_n(\alpha-\frac{1}{n},\infty)))^C=(\bigcap_nf^{-1}((\alpha-\frac{1}{n},\infty)))^C

By our assumption, $f^{-1}((\alpha-\frac{1}{n},\infty))$ are all in $S$, which is a $\sigma$-algebra, thus $(\bigcap_nf^{-1}((\alpha-\frac{1}{n},\infty)))^C\in S$ (convince yourself!). The other direction is very similar and so as the directions $3\iff 4$, you can try them as a nice exercise that will also show you how well you understand the definiton.

#### Immediate collolary

Notice that if $f$ is measurable and $x_0\in\mathbb{R}$, then:

f^{-1}(x_0)=f^{-1}(\{x_0\})=f^{-1}((-\infty,x_0]\cap[x_0,\infty))=f^{-1}((-\infty,x_0])\cap f^{-1}([x_0,\infty))

Therefore, a pre-image of a single – point is a measurable set!

Recall that a function $f$ is continuous if and only if a pre-image of an open set is open.

Suppose that $f:\mathbb{R}\to\mathbb{R}$ is continuous. Is it measurable? Indeed! Since $(a,\infty)$ is open then $f^{-1}((a,\infty))\in\mathbb{R}$ is open as well. We proved that open sets in $\mathbb{R}$ are measurable. Therefore, $f^{-1}((a,\infty))$ is measurable.

## Sum of measurable functions

After we’ve defined what measurable functions are, the next step is to find out what the set of measurable function satisfies.

First thing we want to ask is: If $f,g$ are both measurable, then what about their sum? Does it measurable as well?

Obviously, we would expect a positive answer to this question, and luckily, the answer is indeed positive, however, the proof won’t be as easy as you might think!

I’ll try to show that $f+g$ satisfies the third property, i.e. I’ll show that for every $\alpha \in \mathbb{R}$, the set:

\{x\in\mathbb{R}:f(x)+g(x)<\alpha\}

is indeed measurable.

To do so, pick some $x\in\{x\in\mathbb{R}:f(x)+g(x)<\alpha\}$. Thus $f(x)+g(x)<\alpha\Rightarrow f(x)<\alpha-g(x)$. We need to find a way to ‘break’ the connection between $f$ and $g$, since then we will be able to talk about each one separately.

We are going to use the fact that the rationals are dense in $\mathbb{R}$. Thus, there exists some $r\in\mathbb{Q}$ such that:

f(x) < r < \alpha-g(x)

Recall that this is true only for our specific $x$ that we picked! Therefore, a good direction would be to prove the following statement:

\{x\in\mathbb{R}:f(x)+g(x)<\alpha\}=\bigcup_{r\in\mathbb{Q}}(\{x\in\mathbb{R}:f(x)< r\}\cap\{x\in\mathbb{R}:g(x)<\alpha- r\})

The union is over all the rationals since we don’t want to ‘miss’ elements, let’s prove this statement. If we will prove it though, we are basically done – The right side of the equation is a measurable set as a countable union of measurable sets.

if $x\in\mathbb{R}:f(x)+g(x)<\alpha$ then $f(x)<\alpha-g(x)$, thus, there exists some $r\in\mathbb{Q}$ such that $f(x) < r < \alpha-g(x)$, therefore $f(x) and $g(x)<\alpha - r$, which implies that $x\in\{x\in\mathbb{R}:f(x)< r\}\cap\{x\in\mathbb{R}:g(x)<\alpha- r\}$, and in particular belongs to the union.

On the other hand, if $x$ is in the union, then there exist some $r\in\mathbb{Q}$ such that $f(x), from that we can conclude: $f(x)+g(x) < \alpha$ which implues that $x$ belong to the left side as well.

## Sum makes our life easier

After proving that sum of two measurable functions is measurable as well, we can easlly prove that product of two measurable functions is measurable. To do so. First I’ll show that if $f$ is measurable, then $c\cdot f$ is also measurable.

This is really easy by definition. Pick $\alpha\in\mathbb{R}$ and consider the set $\{x\in\mathbb{R}:c\cdot f(x)<\alpha\}$. This is exactly the set $\{x\in\mathbb{R}:f(x)<\frac{\alpha}{c}\}$ which is measurable since $f$ is measurable (If $c=0$ then $c\cdot f = 0$ which is measurable).

Now for the product, first notice that if $f$ is measurable, then $f^2$ is also measurable. Indeed, pick some $\alpha\in \mathbb{R}$. If $\alpha \geq 0$ then:

\{x:f^2(x)<\alpha\}=\{x:-\sqrt{\alpha}< f(x) < \sqrt{\alpha}\}=\{x: f(x)>-\sqrt{\alpha}\}\cap\{x:f(x)<\sqrt{\alpha}\}

And as an intersection of measurable, the set on the left is measurable as well. If $\alpha <0$, that’s even easier:

\{x:f^2(x)<\alpha\}=\empty\text{ (which is measurable)}

For any two measurable functions $f,g$. We can represent thier product as a measurable function:

f\cdot g=\frac{1}{4}((f+g)^2-(f-g)^2)

And that’s it – convince yourself why the right part of the equation is indeed measurable!

## Sequence of measurable functions

So far we’ve seen some great properties that the set of measurable function satisfies, turns out that measurable functions also behave pretty good when it comes to sequences of functions.

If you recall from your calculus class, the ‘nice’ things we wanted from the sequence to satisfy (such as a continuous limit function, changing the order for the limit- $\lim_{x\to a}\lim_{n\to\infty}f(x)=\lim_{n\to\infty}\lim_{x\to a}f(x)=$, and so on) were only valid when the sequence was converges uniformly. It didn’t work so well if the sequence were only converges pointwise.

As it turns out, measurable function are actually much more forgivable when it comes to pointwise convergence. Let’s see why:

Suppose that $(X,S)$ is a measurable space and $\{f_n\}_{n=1}^\infty$ is a sequnce of $S$measurable functions. Then: $\sup_n f_n(x), \inf_n f_n(x), \overline{\lim_{n\to\infty}}f_{n}, \underline{\lim_{n\to\infty}}f_n$ are all measurable, when the functuion are defined pointwise.

Ok, we have 4 things we need to prove, let’s begin with the suprimum. Denote $f(x)=\sup f_n(x)$. By the definition of supremum we have:

f(x)=\sup_nf_n(x)\leq\alpha\iff f_n(x)\leq\alpha\text{ for every }  n\in\mathbb{N}

Therefore, we conclude that:

\{x\in X: f(x)\leq\alpha\}=\{x\in X| \forall n\in\mathbb{N}: f_n(x)\leq\alpha\}=\bigcap_{n=1}^\infty\{x\in X: f_n(x)\leq\alpha\}

The sets ${x\in X: f_n(x)\leq\alpha}$ are all measurable, then so as thier countable intersection, therefore, ${x\in X: f(x)\leq\alpha}$ is measurable, as we wanted.

For the infimum, notice that:

\inf_nf_n(x)\geq\alpha\iff f_n(x)\geq\alpha\text{ for every }  n\in\mathbb{N}

and the rest is similar to the previous case.

How about the upper and lower limits? what even are they? If you remember from calculus the upper limit is defined as:

\overline{\lim}f_n(x)=\inf_k(\sup_{n\geq k}f_n(x))

From this definition, we can easily see why it is measurable – define $g_k(x)=\sup_{n\geq k}f_n(x)$, then $\overline{\lim}f_n(x)=\inf_kg_k(x)$ and we have just proved that it is measurable.

On the other hand, the lower limit is defined as:

\underline{\lim}f_n(x)=\sup_k(\inf_{n\geq k}f_n(x))

Again, we can define $g_k(x)=\inf_{n\geq k}f_n(x)$ to get $overline{\lim}f_n(x)=\sup_kg_k(x)$.

## Summary

We just proved that if $f_n$ converges pointwise to a function $f$, then both of the upper and lower limit exists and they both equals to $f$. In other words, pointwise convergence preserves the property of being a measurable function, this is not trivial at all! If you recall, back in calculus class, we didn’t really ‘liked’ pointwise convergence since it doesn’t preserve the property of being continuous.

This gives us kind of a clue to how great measurable functions are, and we will see more good properties in the future. In the next post, we will start constructing our new definition for an integral and we will finally define the lebesgue integral!