\part{Lebesgue integration}


\chapter{Lebesgue integral}

After much anticipation, we now develop the Lebesgue integral. We have much to look forward to; our new integral will have many desirable features, notably the following.
- Ability to integrate larger class of functions
- Ability to integrate over larger class of domains
- Backwards compatibility with Riemann integral
- Easier proof framework for proving theorems about swapping limits and integrals
- Ability to form a \emph{topological vector space} from integrable functions (this is an idea of functional analysis)

We will develop the Lebesgue integrable for progressively more general functions.


\section{Lebesgue integral for ISFs}

 We will do this by means of simple functions, since as we are about to prove, we can use them to 


\begin{definition}[Lebesgue integral of a simple function]
If $f$ is a a simple function $f : X \to \mathbb{R}\cup\{\pm \infty\}$ of the form $f=\sum^{n}_{k=1} c_k \chi_{F_{k}}$ where the $E_k$ are pairwise disjoint and $E \subseteq X$ is a measurable set, the \emph{Lebesgue integral of $f$ on $E$} is the following sum.
	\[ \int_{E} f d\mu = \sum^{n}_{k=1} c_k \mu(F_{k} \cap E) \]
\end{definition}

\begin{definition}[Integrable simple function (ISF)]
An \emph{ISF on $E$} is a simple function $f : X \to \mathbb{R}$ of the form $f=\sum^{n}_{k=1} c_k \chi_{F_{k}}$ where the $E_k$ are pairwise disjoint such that its Lebesgue integral on $E$ is finite.
\end{definition}

We'll require a few propositions about ISFs to aid our future proofs.
\begin{proposition}
Let $f,g$ be ISFs on $E$ and let $c\in \mathbb{R}$ be a constant.
\[f+g\]
\[fg\]
\[cf\[
\[ | \int f d\mu | =  \int |f| d\mu \]
\[ \int |f+g| d\mu \leq  \int |f| d\mu + \int |g| d\mu \]
\begin{proposition}

\section{Lebesgue integral for nonnegative Borel functions}

This is a good start, but simple functions are rather... simple; we want our Lebesgue integral to be applicable to a wider class of functions. The idea is to see what functions that sequences of ISFs converge to; this class is much better.


We'll use the following proposition to link our definition of the Lebesgue integral for ISFs to that of nonnegative Borel functions defined on a measurable set.

\begin{proposition}
Let $(X,\Sigma)$ be a measurable space and $E \in \Sigma$
Given a nonnegative Borel function on measurable set $f : E \to \mathbb{R}$, there exists a monotone increasing sequence of ISFs $\{f_n\}$ with $f_n : E \to \mathbb{R}$ that converges pointwise.
\end{proposition}

\begin{proposition}
Let $(X,\Sigma)$ be a measurable space and $E \in \Sigma$
Given a bounded nonnegative Borel function on measurable set $f : E \to \mathbb{R}$, there exists a monotone increasing sequence of ISFs $\{f_n\}$ with $f_n : E \to \mathbb{R}$ that converges uniformly to $f$.
\end{proposition}

There is no better way to prove these than to actually construct such an ISF.

\[\psi_n = n \chi_{[n,\infty)} + \sum_{k=0}^{n2^n -1} \frac{k}{2^n} \chi_{[\frac{k}{2^n},\frac{k+1}{2^n})}\]
\[f_n = \psi_n \circ f = n \chi_{f^{-1}([n,\infty))} + \sum_{k=0}^{n2^n -1} \frac{k}{2^n} \chi_{f^{-1}([\frac{k}{2^n},\frac{k+1}{2^n}))} \]

Notice that when $f$ is bounded, eventually $f^{-1}([n,\infty))$ becomes the null set, and this is the only part of the function in the way of uniform convergence.




Since nonnegative Borel functions not only ensure the exisstence of convergent sequences of ISFs, but ensure the existence of a monotone increasing one, we are left with this very convenient definition for a Lebesgue integral for nonnegative Borel functions.
Ideally, we would like to define the Lebesgue integral for nonnegative Borel functions as such.

\[\int_{E} f d\mu = \lim_{n \to \infty} n \mu(E \cap f^{-1}([n,\infty))) + \sum_{k=0}^{n2^n -1} \frac{k}{2^n} \mu(E \cap f^{-1}([\frac{k}{2^n},\frac{k+1}{2^n})) \]

This is reminiscent of a Riemann integral, although rather than any interval partition of the domain, we partition the domain based on mappings to to an interval partition of the image, using our more sophisticated theory of measures to measure whatever sets arise from this process.

In theory, we do not need to use this exact sequence approaching from below; more generally, the supremum of any Lebesgue integral of an ISF dominated by $f$ will suffice.

\begin{definition}[Lebesgue integral of a nonnegative Borel function]
Let $(X,\Sigma,\mu)$ be a measure space and $E \in \Sigma$
For a nonnegative Borel function on a measurable set $f : E \to [0,\infty)\cup\{\infty\}$, the \emph{Lebesgue integral of $f$ on $E$} is the following limit.
\[ \int_{E} f d\mu = \sup \{ \int_{E} g d\mu : g \text{ is an ISF} \land \forall x \in E [ 0 \leq g(x) \leq f(x) ] \} \]
%Where $f_n$ is sequence of ISFs that converge to $f$ in measure
\end{definition}







\section{Lebesgue integral for Borel functions}

The only major undesirable feature of our current Lebesgue integral is the fact that it only works for nonnegative functions. Fortunately, there is a simple decomposition that can be made for any Borel function that easily extends our definition to its final form.


\begin{definition}[Lebesgue integral of a Borel function]
	For a Borel function $f : X \to \mathbb{R}\cup\{\pm \infty\}$ and measurable set $E \subseteq X$, the \emph{Lebesgue integral of $f$ on $E$} is defined as such.
\[ \int_{E} f d\mu =  \int_{E} f^{+} d\mu - \int_{E} f^{-} d\mu \]
Where $f^{+}(x) = \begin{cases} f(x) & f(x) > 0 \\ 0 & f(x) \leq 0 \end{cases}$ and $f^{-}(x) = \begin{cases} -f(x) & f(x) < 0 \\ 0 & f(x) \geq 0 \end{cases}$.
\end{definition}

We define a Lebesgue integrable function as such.

\begin{definition}[Lebesgue integrable function]
A \emph{Lebesgue integrable function on $E$} is a Borel function $f : X \to \mathbb{R}$ with $E \subseteq X$ a measurable set, such that the Lebesgue integral of $|f|$ is finite. 
\end{definition}

One simple observation is that if the function isn't finite almost everywhere, it's not Lebesgue integrable.




There is an alternative way to construct this Lebesgue integral based on any sequence of ISFs converging to the nonnegative Borel function, though it requires proving that the result of the integral is independent of the way that the sequence converges.

\begin{proposition}
For any sequence of ISFs $f_n$ converging to the nonnegative Borel function $f : X \to \mathbb{R}$ in measure on a measurable set $E \subseteq X$, we hve the following.
\[ \int_{E} f d\mu = \lim_{n\to \infty} \int_{E} f_n d\mu \]
\end{proposition}

One could have defined the Lebesgue integral directly from this, however it requires checking whether this definition is independent of the sequence and only on its limit (like done in the proof). I chose not to officially define it like this for the sake of having an easier path directly to the Lebesgue integral; imagine it is Christmas morning, you're 8, and you want to unwrap your present under the tree but your mom tells you you need to verify that your present is 'well defined' first. What a bummer!





\section{Alternative definition by improper Riemann integral}


\chapter{Properties of Lebesgue integral}

f=g almost everywhere means same integral

\begin{definition}[Complex Lebesgue integral of a Borel function]
For a Borel function $f : X \to \mathbb{C}$ and measurable set $E \subseteq X$, the \emph{Lebesgue integral of $f$ on $E$} is defined as such.
\[ \int_{E} f d\mu =  \int_{E} u d\mu + i \int_{E} v d\mu \]
\end{definition}


\begin{proposition}
Let $f,g$ be Borel functions on $E$ and let $a,b\in \mathbb{R}$ be a constant.
\[\int_{E} (af+bg)d\mu = a \int_{E} f d\mu + b \int_{E} g d\mu \]
\[ \int |f+g| d\mu \leq  \int |f| d\mu + \int |g| d\mu \]
\[fg\]
\[cf\[
\[ | \int f d\mu | \leq^  \int |f| d\mu \]
\[  \int f(x-a) d\mu(x) = \int f d\mu \]
\begin{proposition}


\begin{definition}[Complex Lebesgue integral]
A \emph{Complex Lebesgue integrable function on $E$} is a Borel function $f : X \to \mathbb{C}$ with $E \subseteq X$ a measurable set, such that the Lebesgue integral of $|f|$ is finite. 
\end{definition}

\begin{proposition}
Let $f$ be Riemann integrable on $I$, then $f$ is Lebesgue integrable on $I$.
\end{proposition}



\section{Radon-Nikodiým theorem}


The Lebesgue integral provides us with a really powerful point of view to understand measures. Note that any measure $\mu$ can be represented in the following form.

\[ \mu(E) = \int_{E} d\mu\]

We can actually form new measures on the same measurable space by inserting a 'kernel' (Borel function to integrate over) to this integral!

\[\nu (E) = \int_{E} f d\mu\]

Note that when we have a $\mu$-null set $N$, it will also be a $\nu$-null set if $\nu$ is constructed in this way.
\[\nu (N) = \int_{N} f d\mu=0\]

This means that $\nu << \mu$ for any $\nu$ constructed in this way.

The following question stands however; if we have any two measures $\mu,\nu$ with $\nu << \mu$, does this idea work the other way? That is, with these conditions is there always such a measurable function $f$ that can connect them in the sense that $\nu(E)=\int_{E} f d\mu$. 

We've got some good news in this respect.


\begin{theorem}[Radon-Nikodiým theorem]
Let $\nu,\mu$ be $\sigma$-finite measures on $(X,\Sigma)$ where $\nu << \mu$. There exists an almost everywhere nonnegative, measurable function $f : X \to [0,\infty)$ such that the following holds for any measurable set $E$.
\[\nu (E) = \int_{E} f d\mu \]
\end{theorem}

There exists a particularly elegant notation to denote this that is justified by this theorem, expressing a similar intuition to the reverse chain rule.

\begin{definition}[Radon-Nikodým derivative]
Radon-Nikodým derivative is an almost everywhere nonnegative Borel function $\frac{d\nu}{d\mu}$ such that the following holds.
\[\nu (E) = \int_{E} \frac{d\nu}{d\mu} d\mu = \int_{E} d\nu\]
\end{definition}

\chapter{Convergence theorems}

A major limitation of the Riemann integral is that it does not provide sufficient tools to discuss when a limit and integral commute. The Lebesgue integral offers.



\begin{theorem}[Dominated convergence theorem (DCT)]
Let $\{f_n\}$ be a sequence of $\mu$-measurable functions converging pointwise to $f$ satisfying (for all $n$ and $x$) $|f_n| \leq g$ for some Lebesgue integrable $g$, then
\[ \lim_{n \to \infty} \int_E |f_n  -f| d\mu = 0\]
Which implies that
\[ \lim_{n \to \infty} \int_E f_n d\mu = \int_E f d\mu \]
\end{theorem}


Noting that constant functions are Lebesgue integrable on measurable sets with finite measure, a simple corollary of the DCT the BCT, which finds a case where the function $g$ can be replaced with a constant $c$.

\begin{theorem}[Bounded convergence theorem (BCT)]
Let $(X,\Sigma,\mu)$ be a finite measure space, and $\{f_n\}$ be a sequence of $\mu$-measurable functions converging pointwise to $f$ satisfying (for all $n$ and $x$) $|f_n| \leq c$ for some constant $c$, then
\[ \lim_{n \to \infty} \int_E |f_n  -f| d\mu = 0\]
Which implies that
\[ \lim_{n \to \infty} \int_E f_n d\mu = \int_E f d\mu \]
\end{theorem}




\begin{theorem}[Monotone convergence theorem (MCT)]
Let $\{f_n\}$ be a sequence of $\mu$-measurable functions converging pointwise to $f$ satisfying (for all $n$ and $x$) $f_n \leq f_{n+1}$, then
\[ \lim_{n \to \infty} \int_E f_n d\mu =  \sup \int_E f_n d\mu  = \int_E f d\mu \]
\end{theorem}



\begin{lemma}[Fatou's lemma]
Let $\{f_n\}$ be a sequence of nonnegative $\mu$-Lebesgue integrable functions converging pointwise almost everywhere to $f$ on $[a,b]$, then
	\[\int_{[a,b]} f d\mu \leq \lim \inf \int_{[a,b]} f_n d\mu\]
\end{lemma}


The theory of Lebesgue integration is backwards compatible with Riemann integration; any Riemann integrable function is $\lambda$-Lebesgeu integrable.

\begin{proposition}[Lebesgue-Riemann integral relation]
Let $f$ be Riemann integrable on $[a,b]$
\[\int_{[a,b]} f d\lambda = \int^{b}_{a} f(x) dx\]
\end{proposition}

The lower Riemann sum naturally gives rise to an ISF that, when integrated on according to the Lebesgue measure, gives the lower Riemann sum.

\[L(f,\mathcal{P}) = \sum_{k=1}^{n} \sup \{ f(x) : x \in [x_{k-1},x_{k})\} \chi_{[x_{k-1} , x_{k})}\]

\[ \int^{b}_{a} f(x) dx = \lim_{\|\mathcal{P}\| \to 0} \int_{[a,b]} L(f,\mathcal{P}) d\lambda \]

Since Lower Riemann sums converge monotonly, then MTC implies the follwing, proving the theorem.
\[ \lim_{\|\mathcal{P}\| \to 0} \int_{[a,b]} L(f,\mathcal{P}) d\lambda  = \int_{[a,b]} f d\lambda \]









\chapter{Improper Lebesgue integral}