\part{Measurable functions}



\chapter{Measurable functions}

This part will set the foundations of applications of measures to functions and vice versa; it will be necessary for when we establish Lebesgue integration later on.

\begin{definition}[Lebesgue measurable function]
A \emph{Lebesgue measurable function} is a function $f : X \subseteq \mathbb{R} \to \mathbb{R} \cup \{ -\infty, \infty\}$ such that $\{ x : f(x)>y\}$ is Lebesgue measurable for any $y \in \mathbb{R}$.
\end{definition}

One can prove equivalence to the following definition.
\begin{proposition}
$f : X \subseteq \mathbb{R} \to \mathbb{R}\cup\{\pm \infty\}$ is a Lebesgue measurable function iff for any set $U$ open in $\mathbb{R}$, $f^{-1}(U)$ is measurable.
\end{proposition}


\begin{definition}[Measurable function]
Let $(X,\Sigma_X), (Y,\Sigma_X)$ be measurable spaces.
A \emph{measurable function} is a function $f : X \to Y$ such that for any $E in \Sigma_Y$, $f^{-1}(E) \in \Sigma_X$.
\end{definition}

This is similar to the topological definition of a continuous function; continuous functions maps open sets of one topology from open sets of another. Measurable functions map measurable sets of one measurable space from measurable sets of another.

One step in Lebesgue integration is partitioning the range into intervals (which are Borel sets of the Euclidean topology) and use the preimage of the function to determine what subset of the domain is mapped to each interval. Lebesgue integration will require these domain subsets to be measurable to calculate our integral, so our Lebesgue integral will naturally only work on measurable functions.

That said, I've hinted that the Lebesgue integral will only partition its range by means of intervals (which are open sets, assuming we're talking about open intervals for now), which means that we can restrict our attention to what are called \emph{Borel functions} for simplicity. Sure, the Lebesgue integral looses a tiny bit of power by doing this, but the functions that suffer from this downsizing are so pathological that we kind of don't care.

\begin{definition}[Borel function]
	Let $(X, \Sigma), (Y, \mathcal{B}(\mathcal{T}_Y))$ be a measurable space and Borel space respectively.
A \emph{Borel function} is a measurable function $f : X \to Y$.
\end{definition}

This means that Borel functions are defined such that the preimage of any Borel set is a measurable set. Recall that all open sets are Borel sets, so with Borel functions, preimages of open sets are measurable sets.

Though the converse doesn't hold (Borel sets aren't necessarily open sets), since $f^{-1}(A) \cup f^{-1}(B) = f^{-1}(A \cup B)$ and $f^{-1}(A) \cap f^{-1}(B) = f^{-1}(A \cap B)$ for countable unions and intersections respectively, and Borel sets are indeed countable unions and intersections of open sets, letting open sets map from measurable sets is enough to create an equivalent definition!

\begin{proposition}
	Let $(X, \Sigma), (Y, \mathcal{B}(\mathcal{T}_Y))$ be a measurable space and Borel space respectively.
$f : X \to Y$ is a Borel function iff for any $U$ open in $(Y,\mathcal{T}_Y)$, $f^{-1}(U)$ is measurable.
\end{proposition}


The Lebesgue integral will consider functions with a codomain of $\mathbb{R}$, so we'll now prove the following proposition that will give us some good intuition for when we study Lebesgue integration.


\begin{proposition}
Let $(X, \Sigma), (\mathbb{R}, \mathcal{B}(\mathcal{T}_{\mathbb{R}}))$ be a measurable space and Borel space on the Euclidean topology of $\mathbb{R}$ respectively.
$f : X \to \mathbb{R}$ is a Borel function iff for any $a \in \mathbb{R}$, $f^{-1}((-\infty,a))$ is measurable.
\end{proposition}


However as it turns out, we don't even require the intervals to be strictly open. Borel spaces contain not only all open sets, but also closed sets and therefore any type of interval! Alternatively, using countable unions and intersections on open sets as a Borel $\sigma$-algebra permits, we can indeed create any type of interval.


\begin{proposition}
Let $(X, \Sigma), (\mathbb{R}, \mathcal{B}(\mathcal{T}_{\mathbb{R}}))$ be a measurable space and Borel space on the Euclidean topology of $\mathbb{R}$ respectively.
$f : X \to \mathbb{R}$ is a Borel function iff for any $a \in \mathbb{R}$, $f^{-1}((-\infty,a))$ is measurable.
\end{proposition}

So in the end, it doesn't even matter if we're using closed, open, or even semi-closed and semi-open intervals; any of these will do!


\section{Constructing Borel functions}


When we finish constructing our Lebesgue integral, we'd like to know what class of functions it can work its magic on; I've spoiled that one condition it will require is that the function be Borel. However we currently haven't seen concrete examples of Borel functions; we hope that this class of functions is rich so that our Lebesgue integral isn't a downgrade from the Riemann integral.

Besides, with an intuitive notion of what this new class of function looks like, we could form work arounds. We'll start by examining characteristic functions and simple functions (these will be building blocks in the study of Lebesue integration).

Note that the codomain is taken as the Euclidean topology with its Borel $\sigma$-algebra.
\begin{proposition}
Let $(X,\Sigma)$ be a measurable space.
$\chi_{E} : X \to \{0,1\}$ is a Borel function iff $E$ is measurable.
\end{proposition}


\begin{proposition}
Real continuous functions are Borel functions
\end{proposition}


The property of being measurable is closed under many arithmetic operations.
\begin{proposition}
f+g is measurable
f-g is measurable
fg is measurable
cf is measurable
|f| is measurable
min(f,g) is measurable
max(f,g) is measurable
\end{proposition}

continuous functions are Borel functions 


\chapter{Sequences of measurable functions}
pointwise convergent sequence of measurable function converges to measurable function


\[ \sup \{ f_1 , \hdots , f_n \} \]
\[ \inf \{ f_1 , \hdots , f_n \} \]
\[ \sup_{n} ( f_n ) \]
\[ \inf_{n} ( f_n ) \]

\section{Simple functions}

\begin{proposition}
Simple functions are Borel iff all its indicator sets are made from Borel sets.
\end{proposition}

The following is a simple corollary of a more general proposition we have proven. However, it will be a necessary fact down the line.
\begin{corollary}
Convergent sequences of simple functions are Borel.
\end{corollary}




\chapter{Theorems on measurable functions}

\section{Littlewood's 3 principles}

\item For any Lebesgue measurable set $E$,  there exists a cover of intervals $\bigcup^{\infty}_{n=1}I_n$ such that for any $\varepsilon > 0$, $\lambda(E \setminus \bigcup^{\infty}_{n=1}I_n) < \varepsilon$
\item Convergent sequences of Borel functions converge uniformly almost everywhere
\item Borel functions are continuous almost everywhere

\section{Egorov's theorem}

\begin{theorem}[Egorov's theorem]
Measurable sequences of functions that converge pointwise converge uniformly almost everywhere.
\end{theorem}

There exists a (much less exciting) partial converse to Egorov's theorem

Measurable sequences of functions that converge uniformly almost everywhere converge pointwise almost everywhere.


I suppose this is interesting in that it means that almost everywhere pointwise and uniform convergence are the same for sequnces of measurable functions.

\section{Luzin's theorem}


\begin{theorem}[Luzin's theorem (Euclidean topology)]
Let $f : X \to Y$ be an almost everywhere finite Borel function and $E\subseteq X$ be a set, then for any $\varepsilon >0$, $f$ is continuous on some closed set $F$ where $\mu (E \setminus F ) < \varepsilon$.
\end{theorem}

This instantly implies Littlewood's 3rd principle, however the reason I don't say it is continuous almost everywhere here is because that looses insight of the fact that we are dealing with closed sets.


\begin{theorem}[Luzin's theorem]
Let $(X,\Sigma,\mu)$ be a Radon measure space
Let $(Y,\mathcal{T})$ be a second-countable topological space
Let $f : X \to Y$ be an almost everywhere finite Borel function and $E\subseteq X$ be a measurable set, then $f$ continuous almost everywhere on $E$.
\end{theorem}



\chapter{Convergence in measure}
\section{Convergence in measure}
\section{Cauchy in measure}

