# Fatou’s lemma and more properties of lebesgue’s integral

Now that we are equipped with the monotone convergence theorem we can derive some interesting results from it. The first I want to present is Fatou’s lemma – a weaker version of the theorem that deals with non-monotonic sequences!

## What is Fatou’s lemma?

So we’ve seen that the monotone convergence theorem only works with… monotonic sequences of functions. However, most of the time it will not be the case!

We need to find a way to handle the other cases! First of all, we’ll start with a sequence of measurable functions

f_n:X\to[0,\infty]

The first problem that comes up is – who said that this sequence even has a limit? That’s not usually true at all! Indeed, it may has no limit, but it does always have a limit interior:

g(x)=\underline{\lim}f_n(x)

Since by definition:

\underline{\lim}f_n(x)=\sup_k \inf_{n\geq k}f_n(x)

and infimums and supremums are always defined in $\mathbb{R}$ (they can be infinity though).

Moreover, we’ve already proved that the limit interior is a measurable function in this post. So we are definitly on the right track.

In addition, notice that the sequence of functions:

g_k(x)=\inf_{n\geq k}f_n(x)

Is monotonically increasing so we can use the monotone convergence theorem to conclude that the limit of this sequence exists, defined and measurable. Therefore:

\int_X\lim_{k\to\infty}g_k(x)d\mu=\lim_{k\to\infty}\int_Xg_k(x)d\mu=\underline{\lim}\int_Xg_k(x)d\mu

However, by it’s definition we know that:

g_k(x)=\inf_{n\geq k}f_n(x)\leq f_k(x)

Thus:

\underline{\lim}\int_Xg_k(x)d\mu\leq\underline{\lim}\int_Xf_k(x)d\mu

That’s great and all, but it still not useful at all, we don’t even know what $g_k(x)$ are! However, since their increasing, it’s limit equlas to it’s supremum, hence:

\lim_{k\to\infty}g_k(x)=\sup_k \inf_{n\geq k}f_n(x)

But $\sup_k \inf_{n\geq k}f_n(x)$ is exactly $g(x)$! Thus:

\int_X\lim_{k\to\infty}g_kd\mu=\int_X(\sup_k \inf_{n\geq k}f_n)d\mu=\int_Xgd\mu

In total, we have:

\int_Xgd\mu=\int_X\underline{\lim}g_k(x)d\mu\leq\underline{\lim}\int_Xf_k(x)d\mu\\ \Downarrow \\
\int_Xgd\mu \leq\underline{\lim}\int_Xf_k(x)d\mu

Or we can write is as:

\int_X\underline{\lim}f_kd\mu\leq\underline{\lim}\int_Xf_k(x)

So we found out that we can interchange limit interior with the integral but we have to “pay a price” for it and change the equality, into an inequality!

And that’s exactly what Fatou’s lemma states! How great is that? The super strict monotone convergence theorem actually helped us to prove a great corollary that covers the most general case!

A good question will be to ask “Wait, so you’ve proved an inequality, who says that this is not an equality? maybe we can prove the other direction?”

### A counter example

So by the header you can understand that the inequality is indeed not an equaity, I shall present now a really cool counter example, where the left side of the inequality is 0, and the right side is infinity! I think that would satisfy all those who doubt me.

We’ll define for every $n\in \mathbb{N}$:

f_n(x)=1_{[n,\infty)}=\begin{cases}
1 & x\geq n\\
0 & x < n
\end{cases}

It’s not hard to verify that $f_n$ are measurables and it’s not hard to see that the sequnce $\{f_n\}_{n=1}^\infty$ pointwise convergence to zero.

Notice that for every $n$:

\int_\mathbb{R}f_nd m=m([n,\infty))=\infty
\\ \Downarrow \\
\underline{\lim}\int_\mathbb{R}f_nd m=\infty

However:

\int_\mathbb{R}\underline{\lim}f_ndm=\int_\mathbb{R}0dm=0

So, we now have:

\int_\mathbb{R}\underline{\lim}f_ndm=0\neq\infty=\underline{\lim}\int_\mathbb{R}f_nd m

Anyways, Fatou’s lemma still hold here since:

\int_\mathbb{R}\underline{\lim}f_ndm<\underline{\lim}\int_\mathbb{R}f_nd m

## Results of the monotone convergence theorem

So after seeing Fatou’s lemma, which is a really nice one. It’s time to fill holes that I promised I’ll fill. First, let’s see how the monotone convergence theorem allows us to ‘prevent’ interaction with the supremum in the definition of the integral:

## The integral as limit

Suppose that $f:X\to [0,\infty]$ is measurable. So, there exists an increasing sequnce of simple functions:

0\leq\varphi_1(x)\leq\varphi_2(x)\leq\varphi_3(x)\leq\dots

Such that for every $x\in X$:

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

And:

\int_Xfd\mu=\lim_{n\to\infty}\int_X\varphi_nd\mu

The proof here is not long but it’s a little bit complicated. However, this is one that gives a lot of information, so it’s worth suffering a bit for it.

#### The proof

We are going to construct those simple functions. Let $n$ be a natural number.

For this specific $n$ we will define $2^{2n}$ sets.

Each one of the sets is going to represent a pre-image of an interval on the y-axis.

For every $1\leq k\leq 2^{2n}$ we define:

E_{n_k}=\{x\in X:f(x)\in[\frac{k-1}{2^n},\frac{k}{2^n})\}=f^{-1}([\frac{k-1}{2^n},\frac{k}{2^n}))

Those are measurable sets since $f$ is measurable. For example, if $n=1$ the sets are:

E_{1_1}=\{x\in X:f(x)\in[0,\frac{1}{2})\} \\
E_{1_2}=\{x\in X:f(x)\in[\frac{1}{2},1)\} \\
E_{1_3}=\{x\in X:f(x)\in[1,\frac{3}{2})\} \\
E_{1_4}=\{x\in X:f(x)\in[\frac{3}{2},2)\}

Notice that the measure of each interval of the form $[\frac{k-1}{2^n},\frac{k}{2^n})$ is $\frac{1}{2^n}$. So, when $n$ is getting bigger, the intervals become smaller (hence will induce better approximation…)

Moreover, The “highest” value that those sets approch is $2^n$ which is increasing at every step, so eventually, points with high value will be in one of those sets.

I also want to metion that ${E_{n_{k}}}$ are pairwise disjoint.

Now we can define:

\varphi_n(x)=\sum_{k=1}^{2^{2n}}\frac{k-1}{2^n}\cdot 1_{E_{n_{k}}}(x)

Let’s figure out what’s going on here…

First, $\varphi_n$ is a simple function. Note that when $k$ is increasing, the ‘weight’ of the indicator $1_{E_{n_{k}}}$ is getting bigger.

That is, if $x\in {E_{n_{k}}}$, then $\varphi_n(x)=\frac{k-1}{2^n}$ and this value is getting bigger when $k$ is getting bigger. In addition, this value is always smaller or equals the value of $f(x)$.

Here is a cool animation of those sums:

Let’s prove two main properties of the sequence $\{\varphi_n\}_{n=1}^\infty$

##### It’s mononically increasing

Pick some arbitrary $x\in X$ such that $f(x)\in E_{n_k}$. Thus, $\varphi_n(x)=\frac{k-1}{2^n}$.

In $n+1$, the set $E_{n_{k}}=f^{-1}([\frac{k-1}{2^n},\frac{k}{2^n}))$ splits into two sets:

• $E_{{(n+1)}_{2k-1}}=f^{-1}([\frac{2k-2}{2^{n+1}},\frac{2k-1}{2^{n+1}}))$
• $E_{{(n+1)}_{2k}}=f^{-1}([\frac{2k-1}{2^{n+1}},\frac{2k}{2^{n+1}}))$

If $x\in E_{{(n+1)}_{2k-1}}$, then:

\varphi_{n+1}(x)=\frac{2k-2}{2^{n+1}}=\frac{k-1}{2^{n+1}}=\varphi_n(x)

If $x\in E_{{(n+1)}_{2k}}$, then:

\varphi_{n+1}(x)=\frac{2k-1}{2^{n+1}}=\frac{k-{1\over2}}{2^{n+1}}>\varphi_n(x)

Thus, $\varphi_{n+1}(x)\geq \varphi_n(x)$. So it is indeed an increasing sequence.

##### It approaches to $f$$f$

Since each interval is getting smaller and samller, the approximation is getting better and better. At the $n$-th level, the error is at most $\frac{1}{n}$.

#### The punch line

Those properties allow us to activate the monotone convergence theorem and conclude that:

\int_Xfd\mu=\lim_{n\to\infty}\int_X\varphi_nd\mu

As we wanted!

## Now we can solve our problems

This theorem now allows us to work with limits instead of supremums, and it also showed us what exactly I talked about when I said that we divide the y-axis instead of the x-axis.

Since we now work with limits, life is going to be much easier! Proving the integral is linear becomes an easy task:

##### Linearity

Suppose that $f,g:X\to [0,\infty]$ are two measurable functions. We know that we can find increasing sequences $\{\varphi_n\}_{n=1}^\infty,\{\psi_n\}_{n=1}^\infty$ of non-negative simple functions such that:

f=\lim_{n\to\infty}\varphi_n \\ g=\lim_{n\to\infty}\psi_n

And:

\int_Xfd\mu=\lim_{n\to\infty}\int_X\varphi_nd\mu
\int_Xgd\mu=\lim_{n\to\infty}\int_X\psi_nd\mu

We can use the monotone convergence theorem (MCT) to get:

\int_X(f+g)d\mu\overset{\text{MCT}}{=}\lim_{n\to\infty}\int_X\varphi_n+\psi_nd\mu

But $\int_X\varphi_n+\psi_nd\mu$ is an integral of simple functions, and we know it’s linear. Thus:

=\lim_{n\to\infty}[\int_X\varphi_nd\mu+\int_X\psi_nd\mu]=\int_Xfd\mu+\int_Xd\mu

As we wanted!

##### Interchanging infinte sum and integral

Notice that if $f(x)=\sum_{n=1}^\infty f_n(x)$, then:

f(x)=\lim_{n\to\infty}S_n(x)\ \ \  (S_n(x)=\sum_{k=1}^nf_n(x))

(This is pointwise convergence!) Notice that $S_n(x)$ is an increasing sequnce, thus, we can use monotone convergence theorem and the linearity to get:

\int_Xfd\mu=\lim_{n\to\infty}\int_XS_nd\mu=\lim_{n\to\infty}\int_X\sum_{k=1}^nf_nd\mu
\overset{\text{linearity}}{=}\lim_{n\to\infty}\sum_{k=1}^n\int_Xf_nd\mu=\sum_{n=1}^\infty\int_Xf_nd\mu

And this is really, really strong! We can interchange summation and integration even if the convergence is ‘only’ pointwise.

##### Linearity in the sets

Suppose that $E_n$ are pairwise disjoint. Let’s define:

E=\bigcup_{n=1}^\infty E_n

Thus, for every $x\in X$:

1_E(x)=\sum_{n=1}^\infty1_{E_n}(x)

Now:

\int_Efd\mu=\int_Xf\cdot 1 _Ed\mu=\int_X\sum_{n=1}^\infty f\cdot1_{E_n}d\mu
=\sum_{n=1}^\infty\int_X f\cdot1_{E_n}d\mu=\sum_{n=1}^\infty\int_{E_n} fd\mu

So not only we have linearity – we have countable lineraity which is pretty awesome.

## Summary

That’s it! We’ve finally finished the third part of our process to define an integral. Before I move to the 4th though, I want to inroduce a new improtant term that is going to be really useful. The term “Almost everywhere”.