After defining the lebesgue integral for non-negative measurable functions, proving the monotone convergence theorem, fatou’s lemma and lots of properties of the integral – we are finally ready to move on to the last step of the definition. We will define the integral for ‘**integrable**‘ functions. We don’t know what this means yet but this is really not complicated – all the ‘heavy stuff’ is already past us.

But, before I’ll go into that, I want to present a new term that I will use from now on – “**Almost everywhere**“. So let’s begin:

## What does that even mean?

Suppose that is a measurable space. We say that a property is true **almost everywhere** if and only if the set:

E=\{x\in X:P(x)=\text{false}\}

Has **zero** measure, .

For example, if (recall that is the lebesgue measure on the real line) – and the property is “ is irrrational” then it is false when is rational, therefore:

m(E)=m(\{x\in\mathbb{R}|x\in\mathbb{Q}\}) =0

So this is a property that is true almost everywhere.

Let’s see exactly what can we do with this property:

#### Some nice properties

**almost the same functions**

Suppose that are non-negative mesurable functions. Moreover, suppose that **almost everywhere**. So as it turns out – their integral is the **same**:

\int_Xfd\mu=\int_Xg d\mu

Why? Let’s denote:

E=\{x\in X:f(x)=g(x)\}

Since the function are the same almost everywhere, the set:

E^c=\{x\in X:f(x)\neq g(x)\}

as zero measure. Thus:

\int_Xfd\mu=\int_{E\cup E^C}fd\mu=\int_Efd\mu+\int_{\underbrace{E^C}_{0 \text{ measure}}}fd\mu=\int_Efd\mu

\overset{f=g\text{ on } E}{=}\int_Egd\mu=\int_Egd\mu+\int_{E^C}gd\mu= \int_Xgd\mu

And that’s pretty cool (think about how this property is useful when you measure **probability**).

**almost zero**

From this result we can conclude another nice fact, suppose that almost everywhere, thus, we can just denote and get that:

\int_Xfd\mu=\int_Xgd\mu=\int_X0d\mu=0

And as it turns out, the opposite it true as well! That is, if , then almost everywhere. Let’s prove it:

I’l denote the collection of all the such that as . My goal is to prove that . In order to do so, I’ll define:

E_n=\{x\in X:f(x)\geq\frac{1}{n}\}

Note that: . Thus:

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

But for every :

0=\int_X fd\mu\overset{X\supseteq E_n}{\geq}\int_{E_n}fd\mu\overset{f\geq\frac{1}{n}}{\geq}\int_{E_n}\frac{1}{n}d\mu=\frac{1}{n}\mu(E_n)\geq0 \\ \Downarrow \\ \mu(E_n)=0

Therefore:

0\leq \mu(E)\leq\sum_{n=1}^\infty\mu(E_n)=\sum_{n=1}^\infty0=0 \\ \Downarrow \\ \mu(E)=0

As we wanted.

**almost finite**

The last property I want to present is that:

\int_Xfd\mu<\infty\Rightarrow f(x)<\infty \text{ almost everywhere}

That’s pretty nice, both this property and the previous property show us information about the **function** just from it’s integral! Think about it, that’s not trivial at all!

The idea of the proof is the same as before, we denote:

E=\{x\in X:f(x)=\infty\}

And we will prove that . Let’s do it then:

Suppose not, then . Since in , then for every and :

f(x)\geq M\cdot I_{E}(x)

Thus:

\infty>\int_Xfd\mu\geq\int_{E}fd\mu\geq\int_EM\cdot I_Ed\mu=M\mu(E)>0 \\ \Downarrow \\ 0<\mu(E)\leq \frac{1}{M}\int_Xfd\mu\underset{M\to\infty}{\to}0

So we have:

0<\mu(E)\leq 0

And that’s clearly a contradiction!

## Generalized theorems

The term of almost everywhere allows us to generalize the major theorems we’ve proved such as the monotone convergence theorem and fatou’s lemma. Let’s see exaclty how:

#### Monotone convergence theorem

Let be a non-negative measurable function for every , moreover, suppose that **almost everywhere**:

0\leq f_1(x)\leq f_2(x)\leq\dots

Then, the a non-negative function exists such that **almost everywhere** and:

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

The proof is pretty straightforward:

Denote:

E=\{x\in X:f_n(x)\text{ is not increasing} \lor f(x)\neq\lim_{n\to\infty} f_n(x)\}

By the assumption – , therefore:

\int_Xfd\mu=\int_{E^C}fd\mu+\overbrace{\int_Efd\mu}^0=\int_{E^C}fd\mu\overset{\text{mon. con. thm.}}{=} \lim_{n\to\infty}\int_{E^{C}}f_nd\mu

\lim_{n\to\infty}[\int_{E^{C}}f_nd\mu+0]= \lim_{n\to\infty}[\int_{E^C}f_nd\mu+ \int_Ef_nd\mu] =\lim_{n\to\infty}\int_{X}f_nd\mu

As desired.

#### Fatou’s lemma

Let be a non-negative measurable function for every . Moreover suppose that there exists a non-negative measurable function such that:

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

**Almost everywhere**. Thus:

\int_Xfd\mu\leq\underline{\lim}\int_Xfd\mu

Again, the proof is pretty straightforward and similar to the previous one.Try it yourself – there’s really nothing complicated here…

## Summary

So we’ve met a pretty useful term that allowed us to make our theorems a bit stronger. In fact, beacuse of what we’ve shown here, as I said before, this term is really important when calculating **probabilities** – think about it, we can define a **probabilty measure** that measures the probabilty (What else can it do…). Think about why would we want to ignore event (that happens only on a zero measure set).

Ok, so in the next post, we’ll finally complete our journey and we’ll have a brand new integral to work with!