# Integrable functions

We’ve finally reached the last part of the construction of the lebesgue integral! So far we have successfully defined the integral on non-negative measurable functions. However, we now want to generalize the defintion to function that can be negative.

How would you do it? If you were told by your professor to find a defintion for any measurable function, do you think you would be able to find one?

Let’s see where a naive way of thinking leads us:

“So I know how to integrate non-negative function… However, what if the function $f$ is negative? well then I can represent it as $f=-g$ where $g$ is non-negative… But I know how to do it! So why not just define the the integral of $f$ to be minus the integral of $g$?”

Well, that’s exactly what we are going to do! For each function we’ll define two function:

f^+(x)=\begin{cases}
f(x) & f(x)\geq0\\
0 & f(x)<0
\end{cases} \ \ , \ \ f^-(x)=\begin{cases}
-f(x) & f(x)<0\\
0 & f(x)\geq0
\end{cases}

Those are both non-negative functions and:

f(x)=f^+(x)-f^-(x) ,\\
|f(x)|=|f^{+}(x)|+|f^-(x)|

Ok, we are finally ready for the definition:

### What is an integrable function

Suppose that $f:X\to \mathbb{R}\cup\{\pm\infty\}$ is a measurable function. We say that $f$ is integrable if:

\int_Xf^-d\mu,\int_Xf^+d\mu<\infty

(recall that those functions are positive). If it is the case, then we define the integral as:

\int_Xfd\mu=\int_Xf^+d\mu-\int_Xf^-d\mu

This is that simple!

#### Criteria for integrable functions

After we’ve finally met the definition, I want to prove a really nice statement, and after that, present a list of all the properties this integral has. Ok, let’s see the statement:

Suppose that $f:X\to \mathbb{R}\cup\{\pm\infty\}$ is measureable. Then:

f\text{ is integrable }\iff |f|\text{ is integrable}

And if so:

|\int _Xfd\mu|\leq\int_X|f|d\mu

Let’s prove it:

#### $\Leftarrow$$\Leftarrow$

Suppose that $f$ is integrable, that is:

\int_Xf^-d\mu,\int_Xf^+d\mu<\infty

Since both $f^-,f+$ are non-negative, we know that $|f(x)|=|f^{+}(x)|+|f^-(x)|$. Therefore:

\int_X|f|d\mu=\int_X(f^++f^-)d\mu=\int_Xf^+d\mu+\int_Xf^-d\mu<\infty

Therefore, $|f|$ is integrable.

#### $\Rightarrow$$\Rightarrow$

Suppose that $|f|$ is integrable. Now:

0< f^+(x),f^-(x)\leq  f^+(x)+f^-(x)=|f(x)|

Therefore:

\int_Xf^+d\mu\ ,\ \int_Xf^-d\mu\leq\int_X|f|d\mu<\infty

Thus, $f$ is measurable, as we wanted.

#### Inequality

Note that:

|\int_Xfd\mu|=|\int_Xf^+d\mu-\int_Xf^-d\mu|\overset{\text{triangle inequality}}{\leq} |\int_Xf^+d\mu|+|\int_Xf^-d\mu|

But both $\int_Xf^+d\mu,\int_Xf^-d\mu$ are non-negative, so we can ingnore the absolute value and get:

=\int_Xf^+d\mu+\int_Xf^-d\mu=\int_X|f|d\mu

As desired!

## Properties of the integral

Ok, get ready for a long list – I am going to prove the first two properties, however the rest are really similar and everything is based on what we proved on non-negative functions so I’ll skip those. Trust me, seeing all those proofs is exhausting and those are really not complicated, so if you really want to see a proof of them – try to do it yourself, It’s not hard at all!

So after this ‘preperation’ that basically explains why I am lazy, it’s time for the list:

Suppose that $f,g:X\to\mathbb{R}\cup\{\pm\infty\}$ are integrable functions and $c\in\mathbb{R}$ is some constant. Then;

1. If $h:X\to\mathbb{R}\cup\{\pm\infty\}$ is measurable and $|h(x)|\leq|f(x)|$ almost everywhere, then $h$ is integrable.
2. $f$ is integrable on any measurable subset $E\sub X$ and if $E=A\cup B$ (disjoint union) then: $\int_{E}fd\mu=\int_Afd\mu+\int_Bfd\mu$.
3. $\mu(\{x\in X: |f(x)|=\infty\} = 0$
4. If $E$ is measurable and $\mu(E)=0$ then $\int_Efd\mu=0$
5. $cf$ is integrable and $\int_X cfd\mu=c\int_Xfd\mu$.
6. $f+g$ is integrable and $\int_X(f+g)d\mu+\int_X fd\mu +\int_X gd\mu$
7. If $f_1,\dots,f_n$ are integrable and $c_1,\dots,c_n$ are constants then $\sum_{k=1}^n c_kf_k$ is integrable and $\int_X (\sum_{k=1}^n c_kf_k)d\mu=\sum_{k=1}^n \int_Xc_kf_kd\mu$.
8. If $f(x)\leq g(x)$ almost everywhere then $\int_X\leq \int_X gd\mu$.

Ok, that’s it. Those are more than enough properties that we would like an integral to satisfy. The idea of the proof is similar for all the properites – use the properties you already know about integrals of non-negative function.

For example, here is a proof for the first one: since $|h|,|f|$ are both non-negative and $|h(x)|\leq|f(x)|$ almost everywhere we conclude that:

\int_X|h(x)|d\mu\leq\int_X|f(x)|d\mu<\infty

Thus, $|h|$ is integrable, and we just proved that this is equivalent to $h$ being integrable.

For the second property, we can write $f=f^+-f^-$ where $f^+,f^-$ are non-negative. Thus:

\int_Ef^+d\mu=\int_Af^+d\mu+\int_Bf^+d\mu \\

\int_Ef^-d\mu=\int_Af^-d\mu+\int_Bf^-d\mu

Therefore:

\int_Efd\mu=\int_Ef^+d\mu-\int_Ef^-d\mu=

\int_Af^+d\mu+\int_Bf^+d\mu
-(\int_Af^-d\mu+\int_Bf^-d\mu)
=(\int_Af^+d\mu-\int_Af^-d\mu)+(\int_Bf^+d\mu-\int_Bf^-d\mu) \\=\int_Afd\mu+\int_Bfd\mu

As we wanted.

The rest of them can be easily proved in similar ways…

## The dominated convergence theorem

So we’ve defined the integral for integrable function, and that’s great and all. However, recall that we’ve also had some pretty strong statements that apply only on non-negative functions. Those statement were:

• The monotone convergence theorem
• Fatou’s lemma

Those theorem allowed us to interchange the order of the integration and the is there some sort of an analogue for those theorems? As it turns out, there is, and that’s exactly what the dominated convergence theorem states:

Suppose that $f_n:X\to\mathbb{R}\cup\{\pm\infty\}$ are measurble functions and suppose that the limit:

f(x)=\lim_{n\to\infty}f_n(x)

exists almost everywhere. Moreover, suppose that there exists an integrable non-negative function $g(x)$ such that for every $n\in\mathbb{N}$ and for every $x\in X$:

|f_n(x)|\leq g(x)

Then $f,f_n$ are all integrable and:

\lim_{n\to\infty}\int_Xf_nd\mu=\int_Xfd\mu

This theorem states that: ” if you can find some function that dominates the sequence, then yes, you may interchange the order of limit and integration”

The idea of the proof is to use fatou’s lemma twice, and show that the upper limit of the sequence equals the lower limit of the sequence, how exaclty? we are going to rely on the fact that for every sequence $a_n$:

-\overline{\lim}a_n=\underline{\lim}(-a_n) \ \ \ \ \ \ \ \ (*)

A proof for that is easy and usually seen in calculus class. Moreover, this is a really intuitive statement – think about it!

Ok, let’s start proving:

First of all, since for every $n,x$: $|f_n(x)|\leq g(x)$ and $g$ is integrable, than we conclude that $f_n$ is integrable as well (property 1 in the list – $g=|g|$). Moreover:

|f(x)|=\lim_{n\to\infty}|f_n(x)|\leq\lim_{n\to\infty}g(x)=g(x)

Thus, $f$ is integrable as well. In addition, we know that:

-g(x)\leq f_n(x)\leq g(x) \\ \Downarrow \\
0\leq f_n(x)+g(x) \ \ ,\ \ 0\leq g(x)-f_n(x)

We can now use fatou’s lemma on $f_n(x)+g(x),g(x)-f_n(x)$ which converges to $f(x)+g(x),g(x)-f(x)$ respectivly, and get:

1. \int_{X}(f+g)d\mu\leq\underline{\lim}\int_{X}(f_n+g)d\mu \\

2.  \int_{X}(g-f)d\mu\leq\underline{\lim}\int_{X}(g-f_n)d\mu

We can use the additivity of the integral to get:

1. \int_{X}fd\mu+\int_Xgd\mu=\int_{X}(f+g)d\mu\leq\underline{\lim}\int_{X}(f_n+g)d\mu

=\underline{\lim}\int_{X}f_nd\mu+\underline{\lim}\int_Xgd\mu=\underline{\lim}\int_{X}f_nd\mu+\int_Xgd\mu \\ \Downarrow
\\
\int_{X}fd\mu\leq \underline{\lim}\int_Xgd\mu


And in the second equation:

2. \int_{X}gd\mu-\int_Xfd\mu=\int_{X}(g-f)d\mu\leq\underline{\lim}\int_{X}(g-f_n)d\mu

=\underline{\lim}\int_Xgd\mu+\underline{\lim}(-\int_Xf_nd\mu)=\int_Xgd\mu+\underline{\lim}(-\int_Xf_nd\mu) \\ \Downarrow \\
-\int_Xfd\mu\;\leq\underline{\lim}(-\int_Xf_nd\mu)\overset{(*)}{=}-\overline{\lim}\int_Xf_nd\mu \\ \Downarrow \\
\overline{\lim}\int_Xf_nd\mu\leq\int_Xfd\mu

So we combine both results a to get:

\int_Xfd\mu\leq\underline{\lim}\int_X f_{n}d\mu\leq\overline{\lim}\int_Xf_nd\mu\leq\int_Xfd\mu

But the ‘sides’ of this inequality are the same, so everything there is not an inequality but an equality:

\int_Xfd\mu=\underline{\lim}\int_X f_{n}d\mu=\overline{\lim}\int_Xf_nd\mu

So we’ve proved that the upper and lower limit are the same, thus the limit exists and equals to both of them, therefore:

\int_Xfd\mu=\lim_{n\to\infty}\int_X f_{n}d\mu

As we wanted!

### The bounded convergence theorem

An immideate result from the dominated convergence theorem is the bounded convergence theorem that tell us that as long as the sequence of functions is bounded in a set, you may interchange the order.

However, the dominated convergence theorem is stronger, isn’t it? Well, of course it is! But in most cases, it is easier to bound the function with a constant rather than a function.

The bounded convergence theorem goes like this:

Suppose that $E\sub X$ is measurable and $\mu(E)<\infty$. In addition, for every $n\in \mathbb{N}$ there exists a measurable function $f_n:E\to\mathbb{R}$ and there exists some $M>0$ such that for every $n$ and every $x\in E$: $|f_n(x)|\leq M$. Moreover, for every $x\in E$ the limit $f(x)=\lim_{n\to\infty}f_n(x)$ exists. Thus:

\int_Efd\mu=\lim_{n\to\infty}\int_Ef_nd\mu

The proof is pretty easy if we use the domainated convergence theorem, for every $x\in E$ we’ll define $g(x):=M$. So $g(x)\geq 0$ is a measurable and integrable function since:

\int_Egd\mu=\int_E Md\mu=M\cdot \mu(E)<\infty

Moreover, $|f_n(x)|\leq g(x)$ so we can apply the domainted convergence theorem to get:

\int_Efd\mu=\lim_{n\to\infty}\int_Ef_nd\mu

As we wanted.

## Summary

That’s it! Finally we can say that we’ve defined a better integral, a more sophisticated one that applies for a lot more functions than riemann integral does.

Moreover, we’ve seen three major theoremes that shows us when we can interchange the limit and the integration – which is absoloutly not trivial! The theorems where:

• Lebesgue’s monotone convergence theorem
• Fatou’s lemma
• Lebesgue’s dominant convergence theorem (and the bounded convergence theorem)

I found a really cool illustration about them:

(The image was found in this site) I really like it – There is an innocent dude that all he wants to do is just interchange the order of the limit and the integral… He’s not sure if he can do it, so not 1, not 2… but 3 theorems come to help him, he’s so lucky!

Ok, so it seems like we’re done here, like I can stop posting about measure theory since everything is so perfect, right?

Wrong! even though we’ve defined the most amazing integral ever lived, there is still one major problem – and it’s a really big one… How the hell do I calculate it?

We know how to calculate reimann integral using the fundamental theorem of calculus. But is the theorem applies to leabesgue’s integral as well? Is there some other way to calculate it? Is there even a way to calculate it? Well, we are going to find out, but we’ll work hard for an answer…