# Lebesgue integral of non-negative functions

In the last post we’ve defined the lebesgue integral for indicators – functions of the form:

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

for some measurable set $E$. We’ve defined the integral to be the measure of the set $E$. After doing so, we’ve defined the integral for a bit more comlicated functions, simple functions. However, simple functions are just a linear combination of indicators. So if:

\varphi=\sum_{k=1}^n a_k1_{E_k}

Where $E_j$ are pairwise disjoint, Then:

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

Recall that this represantation of $\varphi$ is not unique, however, we’ve proved that the integral turns out to be the same for any represantation.

Finally, we’ve seen that this definition has some great properties that we would definitely want an integral to fulfill: it’s additive, we can bound it…

Now it’s time to move on to the third part of our journey – defining the integral on non-negative (measurable) functions. As I said, this is where thing are getting a little complicated but we’ll get over it, so let’s begin!

## The idea

Before I present the definition, I want to describe the idea behind it. So far we only know how to find integrals of simple functions. However, we now want to take advantage of that.

The basic idea is to approxiamte non-negative fuctions with simple functions from below. For example, consider this function and it’s approximation with a simple function:

If we would calculate the integral of the simple function, we will get a value that is pretty close to the integral of function (in blue).

Umm.. it seems pretty similar to reimann integral, isn’t it? If you recall, I’ve also said that we are going to define the integral based on partition of the $Y$-axis, and that’s not what I am doing here, so what’s going on here?

Well, even though it seems like the reimann integral, it’s actually already a lot better:

• In reimann integral we are splitting the interval to smaller intervals – we can only approximate the integrals using the length of the intrevals scaled by som number. This is not the case here, we are not restricted to intervals. We can approximate the integral with scaled measures of sets.
• This fact allows us define the integral on a wider class of functions, such as the dirichlet function that I’ve already mentioned in previous posts.

For the second question we will give an answer later, however for now I’ll just say that we can define the sets on the $x$ axis, to be pre-images of sets on the y-axis. However, to do that, we need the function to be measurable! This is just something to think about untill I’ll be able to fully describe it. (I will be able to describe it once we’ll prove the monotone convergence theorem)

Ok, now that we understand the idea, we’re ready to define the integral:

## The definition

Let $(X,S,\mu)$ be a measure space. Moreover, suppose that $f:X\to[0,\infty]$ is a measurable function. We define the integral of $f$ to be:

\int_Xfd\mu=\sup_{0\leq\varphi\leq f}\int_X\varphi d\mu

Where $\varphi$ are measurable functions.

This definition implements exactly our idea. The integral the ‘best’ bound of integrals of simple function, the supremum.

In addition, we have two equivalent ways to define the integral on a set:

\int_Ef d\mu=\sup_{0\leq\varphi(x)\leq f(x)\ :\ x\in E}\int_E\varphi d\mu=\int_Xf\cdot1_E d\mu

Great, the definition fits perfectly with our intuition from the beginning, so let’s try to find some properties of it.

## Properties of the integral

The first property that came to your was probably additivity, but that’s not trivial at all! We will need the monotone convergence theorem (yes, I did mention it before…) in order to do so.

Ok, what other properties can we learn about it? After additivity, I started to wonder, maybe if I’ll find properties of the definition instead of thinking about ‘desired’ properties, it will be easier to find properties.

Well, that will be my approach from now – I am will try to get as much infomation as I can from the definition.

First, let’s try to understand why it’s hard to prove additivity. If $f,g$ are two measuables functions, then by definition:

\int_Xfd\mu=\sup_{0\leq\varphi\leq f}\int_X\varphi d\mu\\ \int_Xgd\mu=\sup_{0\leq\varphi\leq g}\int_X\varphi d\mu

Then their sum is:

\int_Xfd\mu+\int_Xgd\mu=\sup_{0\leq\varphi\leq f}\int_X\varphi d\mu+\sup_{0\leq\varphi\leq g}\int_X\varphi d\mu

On the other, the integral of the sum is:

\int_Xf+gd\mu=\sup_{0\leq\varphi\leq f+g}\int_X\varphi d\mu

So we want to prove that:

\sup_{0\leq\varphi\leq f+g}\int_X\varphi d\mu=\sup_{0\leq\varphi\leq f}\int_X\varphi d\mu+\sup_{0\leq\varphi\leq g}\int_X\varphi d\mu

And that’s pretty hard. Sum of supremums it’s kind of a hard thing to deal with – how can we transform two supremums into one?

Indeed, sum is problematic… However, the definition is bases on the supremum, which we can bound and increase the set that we calculate it’s supremum. This fact leads me to the first property:

#### Comparing integrals

If $f,g$ are measuable functions where $0\leq f(x)\leq g$ then:

\int_Xfd\mu=\sup_{0\leq\varphi\leq f}\int_X\varphi d\mu\overset{f\leq g}{=}\sup_{0\leq\varphi\leq g}\int_X\varphi =\int_Xgd\mu

That’s pretty nice, we can compare integrals of functions based on the the comparison of the functions. And this fact is actually not new for us – we would expect such a property to exist.

Moreover, we can learn even more from that, what if we want to integrate on a set and not on the whole space? By definition;

\int_Efd\mu=\int_X f\cdot1_Ed\mu

My goal is to compare integrals, so let’s consider two sets $A,B$ such that $A\subseteq B$. Thus:

f\cdot1_A\leq f\cdot1_B

So we can use our latest result to get:

\int_Afd\mu=\int_X f\cdot1_Ad\mu\leq\int_X f\cdot1_Bd\mu=\int_Bfd\mu

Cool! That’s another property we would expect an integral to satisfy.

There is another thing we can learn from it. If $m\leq f\leq M$ in $E$, then we can treat $m,M$ as constant functions and conclude:

\int_Emd\mu\leq\int_Efd\mu\leq\int_EMd\mu

But constant functions are simple, and we know how to calculate their integral:

\int_Emd\mu=m\cdot\mu(E)\ \ \ \ \int_Mmd\mu=m\cdot\mu(E)

Thus:

m\cdot\mu(E)\leq\int_Efd\mu=M\cdot\mu(E)

### Scaling a function

Another property we would want an integral to satisfy is:

\int_X\alpha f d\mu=\alpha\int_Xfd\mu

(Where $\alpha > 0$). Let’s try to prove it and see what we get:

\int_X\alpha f d\mu=\sup_{0\leq\varphi\leq \alpha f}\int_X\varphi d\mu=\sup_{0\leq\frac{1}{\alpha}\varphi\leq f}\int_X\varphi d\mu=\sup_{0\leq\frac{1}{\alpha}\varphi\leq f}\int_X\alpha\cdot\frac{1}{\alpha}\varphi d\mu

Now, since $\frac{1}{\alpha}\varphi$ is a simple function, we can “remove” the scalar out of the integral to get:

=\sup_{0\leq\frac{1}{\alpha}\varphi\leq f}\alpha\int_X\frac{1}{\alpha}\varphi d\mu

My first instinct was to “remove” the saclar out of the supremum. And that’s in fact a valid move! The statement:

\sup \alpha E=\alpha\sup E\ \ (\alpha\geq0)

is true! Verify it yourself, it follows immidealty from the definition of a supremum (“$\sup E = M$if and only if for every $\varepsilon >0$ there exists some $x\in E$ such that $x>M-\varepsilon$“, you can multiply the inequality by $\alpha$ to derive the result).

Great, so now we have:

\int_X\alpha f d\mu=\sup_{0\leq\frac{1}{\alpha}\varphi\leq f}\alpha\int_X\frac{1}{\alpha}\varphi d\mu=\alpha\cdot\overbrace{\sup_{0\leq\frac{1}{\alpha}\varphi\leq f}\int_X\frac{1}{\alpha}\varphi d\mu}^{\int_Xfd\mu}=\alpha\int_Xfd\mu

As desired!

#### Integral of zero

What if $f\equiv 0$ on $E$? Well, then we would expect the integral to be zero as well, let’s see what it is:

\int_Efd\mu=\int_E0d\mu=0\cdot\mu(E)

Umm… What happenes if $\mu(E)=\infty$? Recall that we decided to define the product $0\cdot \mu(E)=0$. Now we see exactly why! That’s not a valid argument for the desicion but it shows why we would decide to define it that way.

To summarize, we got that, indeed, the integral of the zero function is zero.

\int_E0d\mu=0

Similarly, we can conclude that if $\mu(E)=0$ then the integral is zero as well, even if $f(x)=\infty$ in $E$.

I’ll say it again, this follows from our desicion, we still need to see if that won’t cause us problems!

## What now?

Right now, I can’t really think of other ‘interesting’ properties that the integral has. The supremum is something really uncomfortable to work with… We need to ‘attack’ this problem from a different point of view, that will allows us to prevent direct interaction with it.

There is actually a way we do so, I’ve already mentioned it twice in this post. We are going to use Lebesgue’s monotone convergence theorem.

This theorem is one of the most important things in lebesgue’s meaure theory. In fact, lebesgue dedicated a lot of his time to this theorem since it’s so important and useful.

This theorem will allow us, under specific conditions, interchange order of limit of functions that pointwise converges and integration.

This is so great since that is one of the major problem reimann integral has – It doesn’t know how to deal with pointwise convergence. That fact reflects another major advantage of lebegsue integral.

From this theorem we can conclude a lot! But since this theorem is so important – it deserves it’s own post, so I’ll end this one now…