# Indicators & simple functions – Lebesgue integral

After we’ve seen what measurable sets, measurable functions and measurable spaces are We are finally ready to define the integral.

Yes, the integral I’ve been talking about since the first post! But, this is not going to be so easy.

The construction of the integral is going to be done in 4 parts.

• The first two are going to be covered in this post.
• The third is going to be kind of tough – It may take me two post to cover it.
• After the hard work in part 3, part 4 is going to be pretty easy.

Throughout the entire posts, I am going to deal with a (positive) measure space $(X,S,\mu)$, so the construction is going to be pretty abstract.

We begin our construction where we already know what are measurable sets. And those are the sets we would like to integrate on.

Moreover, we also know what measurable functions are – so at least at the beginning, we are going to deal with them.

Let’s begin:

## Part 1 – Indicators

#### What is an indicator

Indicators are the most elementary measurable functions:

For $E\in S$, we will define the indicator:

1_E:X\to\{0,1\}

As:

1_{E}(x)=\begin{cases}
1 & x\in E\\
0 & x\notin E
\end{cases}

The name ‘indicator’ is a pretty good name. It indicates wheter $x\in E$ or not.

If you recall, We’ve actully met an indicator before, In the introduction. It was the Dirichlet function. We now know that in fact, this function is just an indicator where $E=\mathbb{Q}$.

#### Indicator as a measurable function

It’s not hard to see that $1_E$ is a measurable function if and only if $E$ is measurable. Let’s prove it:

Suppose that $1_E$ is measurable. Thus:

E=f^{-1}([1,\infty))

is measurable.

Conversely, suppose that $E$ is measurable. Pick some $a\in\mathbb{R}$. There are now three cases:

• $1\in(a,\infty),0\notin(a,infty)$. So $f^{-1}((a,\infty))=E$ which is measurable.
• Both 0 and 1 are in the interval. Thus, $f^{-1}((a,\infty))=X$ which is measurable.
• None of them is in the interval. Then $f^{-1}((a,\infty))=\emptyset$ which is measurable.

In all the cases we got that $f^{-1}(a,\infty)$ is measurable – thus, $f$ is measurable.

#### Defining the integral

Obsereve at the plot of the indicator $I_{(0,1)\cup(2,4)}$.

We would want that the integral will be the area under the graph, which is the sum of the area of the rectangles marked in the plot.

The sum of the areas is $1\cdot 1 + 2\cdot 1 = 1 + 2 =3$. On the other hand:

m((0,1)\cup(2,4))=m((0,1))+m((2,4))=1+2=3

The indicator translates the measure of the set into area under the graph. Therefore, we would want to define the integral as:

\int_X1_Ed\mu=\mu(E)

That’s it. This is so simple but it actually makes sense. It’s not hard to see that the integral is defined if and only if $E$ is measurable.

With this simple definition we can calulate the integral of Dirichlet function in the interval $[0,1]$:

\int_\mathbb{R}1_{\mathbb{Q}\cap[0,1]}dm=m(\mathbb{Q}\cap[0,1])\overset{|\mathbb{Q}\cap[0,1]|=\aleph_0}{=}0

Notice how we just solved the problem from the first post with no effort! According to reimann integral, the integral on this function does not even defined. Here, not only it is defined – we also know it’s value.

Well, that’s it for indicators, let’s move on for something more challenging, but still, simple:

## Part 2 – Simple functions

#### What is a simple function

The definition is really simple – A simple function $\varphi:X\to \mathbb{R}$ is a measurable function such that $\varphi(X)=\text{Im}(\varphi)=\{a_1,\dots,a_n\}\sub\mathbb{R}$. It’s image is finite.

In other words, the output of $\varphi$ can only be one of $n$ different values for $n\in\mathbb{N}$.

For example, the function:

\varphi(x)=\begin{cases}
1 & x\geq0\\
0 & x<0
\end{cases}

Is simple, since $\text{Im}\varphi(X)=\{0,1\}$.

#### representing simple functions

Suppose that $\varphi:X\to\mathbb{R}$ is measurable and $\varphi(X)=\{a_1,\dots,a_n\}$. For each $1\leq k\leq n$ we can define:

E_k=\varphi^{-1}(a_k)

Since $\varphi$ is measurable, $E_k$ is measurable.

We can now represent $\varphi$ as a linear combination of indicators. How exactly? That’s how:

\varphi=\sum_{k=1}^n a_k\cdot1_{E_k}=a_1\cdot 1_{E_1}+\dots+a_n\cdot1_{E_n}

Let’s verify it: Suppose that $x\in X$ and $\varphi(x)=a_k$ for some $k$. Thus, $x\in E_k$. Therefore:

• $1_{E_k}(x)=1$.
• $1_{E_{k_o}}(x)=0$ for $k_0\neq k$.
\sum_{k=1}^n a_k\cdot1_{E_k}(x)=a_1\cdot \overbrace{1_{E_1}(x)}^0+\dots+a_k \overbrace{1_{E_k}(x)}^1+\dots+a_n\cdot \overbrace{1_{E_n}(x)}^0=a_k

The sum $\sum_{k=1}^n a_k\cdot1_{E_k}$ is called the cannonical form of $\varphi$.

#### Some notes

In fact, we can represent a simple function in many ways. For example:

\varphi=1_{[0,1]}

Can be one representation of $\varphi$. However:

\varphi=1_{[0,\frac{1}{2}]}+1_{(\frac{1}{2},1]}

is another way or representing $\varphi$. However, the canonical form is unique!

One more thing: It’s not hard to see that if $f,g$ are simple then:

• $f\pm g$
• $c\cdot f$
• $fg$

are all simple as well.

#### Defining the integral

Just like before, I want to show a plot of a simple function:

Notice how simple functions look like stairs, again, it is kind of clear what we want our integral to be. Try to guess it before I’ll present the definition!

Ok, here is the definition:

Suppose that $\varphi:X\to\mathbb{R}$ is simple non-negative (we’ll treat negative function in part 4) where:

\varphi=\sum_{k=1}^na_k\cdot1_{E_k}

Then we will define the integral as:

\int_X\varphi d\mu=\sum_{k=1}^na_k\mu(E_k)

Where we agree on one thing:

• If $a_k=0$ and $\mu(E_K)=\infty$ then $a_k\mu(E_k)=0$.

This agreement can cause us problem, but I’ll take the risk for now. We will see in the future that it turns out to be harmless.

#### Integral on a set

If $E\sub X$ is measurable we can define the integral over $E$ as :

\int_E\varphi d\mu=\int_X\varphi\cdot 1_Ed\mu

The function $\varphi\cdot 1_E$ is the same as $\varphi$ in $E$. However, it is 0 elsewhere.

In fact, we can also define it as:

\int_E\varphi d\mu=\int_X\varphi|_E

Where $\varphi|_E= \sum_{k=1}^n a_k1\cdot 1_{E_k\cap E}$.

Either way, the value of the integral is:

\sum_{k=1}^na_k\mu(E_k\cap E)

#### Working with non-canonical forms

Sometimes we will have to deal with simple functions in their non-canonical form. It’s essential to know how to integrate in those situations.

Turns out that if $\varphi=\sum_{j=1}^n a_j\cdot 1_{E_j}$ is a simple function, and $E_j$ are pairwise disjoint then:

\int_X\varphi d\mu=\sum_{j=1}^{n}a_j\mu(E_j)

And that’s great, it allows us to deal with all non-canonical forms and it is hard at all to bring a present a simple function that way (Why?).

Let’s prove it: Suppose that $\text{Im}(\varphi)=\{b_1,\dots,b_m\}$. Since $E_j$ are pairwaise disjoint, for every $x\in E_j$:

\varphi(x)=a_j

Therefore, each $a_j$ is actually equals to some $b_k$. Now, for every $k=1,\dots,m$ we define:

S_k=\{x\in X:\varphi(x)=b_k\}

Notice that $\varphi = \sum_{k=1}^m b_k 1_{S_k}$ is the canonical form of $\varphi$. Moreover:

S_k=\bigcup_{j:\ a_j=b_k}E_j

And this is a union of pairwise disjoint sets. Thus:

\mu(S_k)=\mu(\bigcup_{j:\ a_j=b_k}E_j)=\sum_{j:\ a_j=b_k}\mu (E_j)

So let’s calculate the integral:

\int_X\varphi d\mu=\sum_{k=1}^nb_k\mu(S_k)=\sum_{k=1}^nb_k\sum_{j:\ a_j=b_k}\mu (E_j)=\sum_{k=1}^n\sum_{j:\ a_j=b_k}b_k\mu (E_j)
=\sum_{k=1}^n\sum_{j:\ a_j=b_k}a_j\mu (E_j)=\sum_{j=1}^na_j\mu(E_j)

As we wanted!

#### How ‘good’ is the definition?

So we now have a definition for a respectful amount of functions. Not all the functions, not even ‘most’ of the functions. However, it is a good place to stop and ponder:

Is my definition good? Is this definitionn satisfies properties that I would like it to satistfy?

So we are going to do just that. I’ll name 5 properties I want an integral to satisfy and we’ll check if it is indeed satisfies them.

##### property 1 – homogeneity

Suppose that $\varphi \geq 0$ is simple and $\alpha\geq 0$. I want the integral to satisfy:

\int_X\alpha\cdot\varphi d\mu=\alpha\int_X\varphi d\mu

Let’s see if this is true:

Suppose that $\varphi = \sum_{k=1}^n a_k\cdot 1_{E_k}$ is the canonical form of $\varphi$. Then $\alpha\cdot\varphi = \sum_{k=1}^n \alpha\cdot a_kE_k$ is the canonical form of $\alpha\cdot\varphi$. Thus:

\int_X\alpha\cdot\varphi d\mu=\sum_{k=1}^n \alpha\cdot a_k\mu(E_k)=\alpha\cdot \sum_{k=1}^n a_k\mu(E_k)=\alpha\int_X\varphi d\mu

Indeed, the integral satisfies the property.

Suppose that $\varphi,\psi \geq 0$ are simple. Is:

\int_X\varphi+\psi=\int_X\varphi+\int_X\psi

true? Let’s find out:

The canonical forms of $\varphi,\psi$ are:

\varphi=\sum_{k=1}^na_k1_{E_k}\ \ \ \ , \ \ \ \ \psi=\sum_{j=1}^mb_j1_{F_j}

Since $E_k$ are pairwise disjoint and $F_j$ are pairwise dijoint as well – The sets $E_k\cap F_j$ are pairwise dijoint.

For every $x\in E_k\cap F_j$ we know that:

\varphi(x)=a_k\ \ \ ,\ \ \ \psi(x)=b_j

Thus, we can use the theorem we proved to get:

\int_X(\varphi+\psi)d\mu=\sum_{j,k}(a_k+b_j)\mu({E_k\cap F_j})=\sum_{j,k}a_k\mu({E_k\cap F_j})+\sum_{j,k}b_j\mu({E_k\cap F_j})

Again, by the same theorem, we know that:

\sum_{j,k}a_k\mu({E_k\cap F_j})=\int_X\varphi d\mu\ \  \ ,\ \ \ \sum_{j,k}b_j\mu({E_k\cap F_j})=\int_X \psi d\mu

Thus:

\int_X(\varphi+\psi)d\mu=\sum_{j,k}a_k\mu({E_k\cap F_j})+\sum_{j,k}b_j\mu({E_k\cap F_j})=\int_X\varphi d\mu +\int_X \psi d\mu

As we wanted!

Notice how this property and the previous one allows us to prove (with simple induction) that if $\varphi_1,\dots,\varphi_n\geq 0$ are simple and $a_1,\dots,a_n\geq 0$. Then:

\int_X\sum_{k=1}^na_k\varphi_k d\mu=\sum_{k=1}^n a_k\int_X\varphi_k d\mu

And even more than that: if $\varphi=\sum_{k=1}^n a_k 1_{E_k}$ is some representation of $\varphi$ then, we can use our latest result to get:

\int_X\varphi d\mu=\int_X\sum_{k=1}^na_k 1_{E_k} d\mu=\sum_{k=1}^n a_k\int_X 1_{E_k}\mu=\sum_{k=1}^n a_k\mu(E_k)

And that’s a wonderful conclusion! We don’t need the canonical form at all! It doesn’t matter what form the function is currently represented with – the integral will be the same.

##### Property 3 – additivity of the domain

If $E,F$ are measurable and $E\cap F = \emptyset$ I would like that:

\int_{E\cup F}\varphi d\mu =\int_{E}\varphi d\mu + \int_F\varphi d\mu

Let’s try to prove it:

First of all, since $E\cap F = \emptyset$, it’s’ not hard to check that:

1_{E\cup F}=1_{E}+1_{F}

(check it!) Therefore:

\int_{E\cup F}\varphi d\mu =\int_X\varphi 1_{E\cup F}d\mu=\int_X\varphi (1_{E}+1_{F})d\mu=\int_X\varphi 1_{E}+\varphi1_{F}d\mu
=\int_X\varphi 1_{E} d\mu+\int_X\varphi1_{F}d\mu=\int_E\varphi d\mu+\int_F\varphi d\mu

As desired!

Notice that we can easlly conclude by induction that if $E_1,\dots,E_n$ are pairwise disjoint, then:

\int_{\bigcup_{k=1}^nE_k}\varphi d\mu=\sum_{k=1}^n\int_{E_k}\varphi d\mu
##### Property 4 – comparing integrals

If $0\leq \varphi\leq \psi$. I would probably want that:

0\leq\int_X\varphi d\mu +\int_X \psi d\mu

That’s not hard to check though:

If $\varphi = \sum_{k=1}^n a_k 1_{E_k}$ then:

\int_X\varphi d\mu=\sum_{k=1}^n a_k\mu(E_k)\geq0

Moreover, if $0\leq\varphi \leq \psi$ then $\psi-\varphi \geq 0$, and:

\int_X\psi d\mu=\int_X\varphi+(\psi-\varphi) d\mu=\int_X\varphi d\mu+\int_X(\psi-\varphi) d\mu\geq\int_X\varphi d\mu
##### Property 5 – bounding the integral

If $E\in S$ and for every $x\in E$ : $m\leq \varphi \leq M$. I would probably want to bound the integral using those bounds of the function. Thus, I would like that:

m\mu(E)\leq\int_E\varphi d\mu\leq M\mu(F)

But that’s an immediate corollary from property 4: We can define

\varphi_m(x)=m\ \ \ ,\ \ \ \varphi_M(x)=M

Of course, $\varphi_m\leq \varphi \leq \varphi_M$. So by property 4 we get:

\int_Emd\mu=\int_E\varphi _md\mu\leq \int_E\varphi d\mu\leq \int_E\varphi_M d\mu=\int_EMd\mu

And we know that:

\int_Emd\mu=m\mu(E)\ \ \ , \ \ \ \int_EMd\mu=M\mu(E)

Thus:

m\mu(E)\leq \int_E\varphi d\mu\leq M\mu(E)

## Summary

So we are done with the first two parts! So far, it seems like we are doing something that works well – all the properties I wanted the integral to satisfy are indeed properties of our new definition of an integral.

In the next post we will face part 3. And as I said, this is not going to be a walk in the park!