# Almost everywhere – the new everywhere

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 $(x,S,\mu)$ is a measurable space. We say that a property $P$ is true almost everywhere if and only if the set:

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

Has zero measure, $\mu(E) = 0$.

For example, if $X=\mathbb{R}, S=\mathcal{L}(\mathbb{R}),\mu = m$ (recall that $\mathcal{L}(\mathbb{R})$ is the lebesgue measure on the real line) – and the property is “$x$ is irrrational” then it is false when $x$ 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 $f,g$ are non-negative mesurable functions. Moreover, suppose that $f(x)=g(x)$ 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 $f(x)=0$ almost everywhere, thus, we can just denote $g(x)=0$ 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 $\int_Xfd\mu=0$, then $f=0$ almost everywhere. Let’s prove it:

I’l denote the collection of all the $x\in X$ such that $f(x)>0$ as $E$. My goal is to prove that $\mu(E)=0$. In order to do so, I’ll define:

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

Note that: $E=\bigcup_{n=1}^\infty E_n$. Thus:

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

But for every $n\in\mathbb{N}$:

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 $\mu(E) = 0$. Let’s do it then:

Suppose not, then $\mu(E)>0$. Since $f(x)=\infty$ in $E$, then for every $M>0$ and $x\in E$:

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

## 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 $f_n$ be a non-negative measurable function for every $n\in\mathbb{N}$, moreover, suppose that almost everywhere:

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

Then, the a non-negative function $f$ exists such that $f(x)=\lim_{n\to\infty} f_n(x)$ 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 – $\mu(E)=0$, 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 $f_n$ be a non-negative measurable function for every $n\in\mathbb{N}$. Moreover suppose that there exists a non-negative measurable function $f$ 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!