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%  Lecture 7
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%  Fallback: kashlak.pdf §2.3.2, §2.4, §2.4.1 (OCR/notation/curriculum)
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\section[Lecture 7 -- Lebesgue-Stieltjes Measure; Fubini-Tonelli]{Lecture 7 \textemdash{} Lebesgue--Stieltjes Measure; Fubini--Tonelli}
\label{sec:lec07}

We are halfway through the integration arc. Today's two themes both
push measures \emph{between spaces}. First: a measurable map
\(\psi\colon\Xset\to\Yset\) carries a measure \(\mu\) on \(\Xset\) to
its \emph{image measure} \(\nu=\mu\circ\psi^{-1}\) on \(\Yset\); when
\(\mu\) is Lebesgue measure on \(\R\) and \(\psi\) is built from a
distribution function \(F\), this produces the Lebesgue--Stieltjes
measure \(dF\). Second: given two \(\sigma\)-finite spaces, one
constructs the product measure on \(\Xset\times\Yset\), and the
Fubini--Tonelli theorem licences swapping the order of integration.

\subsection{Image measures and Lebesgue--Stieltjes}

\begin{definition}{Image measure}{image-measure}
Let \((\Xset,\Xcal,\mu)\) and \((\Yset,\Ycal)\) be
measurable spaces and \(\psi\colon\Xset\to\Yset\) measurable. The
\emph{image} (or \emph{push-forward}) measure of \(\mu\) under \(\psi\)
is the set function \(\nu=\mu\circ\psi^{-1}\) on \(\Ycal\),
\[
\nu(B) \;=\; \mu\bigl(\psi^{-1}(B)\bigr),\qquad B\in\Ycal.
\]
\end{definition}

\begin{remark}
Measurability of \(\psi\) is what makes \(\psi^{-1}(B)\in\Xcal\),
so \(\nu(B)\) is defined. Inverse images preserve countable unions and
complements, so \(\nu\) is automatically a measure on \(\Ycal\).
This is the construction that turns Lebesgue measure on \(\R\) into the
Lebesgue--Stieltjes measure attached to a distribution function.
\end{remark}

\begin{theorem}{Lebesgue--Stieltjes measure}{lebesgue-stieltjes}
Let \(F\colon\R\to\R\) be non-constant, right-continuous, and
non-decreasing. There exists a unique measure \(dF\) on \(\Bcal(\R)\)
such that for all \(a<b\) in \(\R\),
\[
dF\bigl((a,b]\bigr) \;=\; F(b)-F(a).
\]
Writing \(F(\infty)=\lim_{x\to\infty}F(x)\), \(F(-\infty)=\lim_{x\to-\infty}F(x)\),
\(I=(F(-\infty),F(\infty))\), and
\[
g(y) \;=\; \inf\{x\in\R : y\le F(x)\},\qquad y\in I,
\]
the measure \(dF\) is realised as the image of Lebesgue measure under
\(g\):
\[
dF \;=\; \lambda\circ g^{-1}.
\]
\end{theorem}

\begin{lemma}{Properties of the left-inverse $g$}{g-properties}
With \(F\) and \(g\) as in \cref{thm:lebesgue-stieltjes}, the function
\(g\) is left-continuous and non-decreasing on \(I\), and
\[
g(y) \le x \iff y \le F(x)\qquad\text{for } y\in I,\;x\in\R.
\]
Concretely, for fixed \(y\in I\) the set
\(J_y=\{x\in\R:y\le F(x)\}\) equals \([g(y),\infty)\), so \(g\) is
Borel measurable.
\end{lemma}

\begin{remark}
Once \(g\) is Borel measurable and left-continuous,
\(\lambda\circ g^{-1}\) is a measure on \(\Bcal(\R)\) and
\[
dF\bigl((a,b]\bigr)
   = \lambda\bigl(\{y:g(y)>a,\;g(y)\le b\}\bigr)
   = \lambda\bigl((F(a),F(b)]\bigr)
   = F(b)-F(a).
\]
Uniqueness follows from the \(\pi\)-\(\lambda\) argument used for
Lebesgue measure: any other measure \(\mu\) with
\(\mu((a,b])=F(b)-F(a)\) must agree with \(dF\) on the
\(\pi\)-system of half-open intervals, hence on every Borel set.
\end{remark}

\begin{figure}[h]
\centering
\begin{tikzpicture}[>=Stealth, scale=1.0]
  % axes
  \draw[->, gray] (-0.3,0) -- (5.0,0) node[right] {\small \(x\)};
  \draw[->, gray] (0,-0.3) -- (0,3.4) node[above] {\small \(F(x)\)};
  % a non-decreasing right-continuous F with a jump
  \draw[thick, deepnavy]
    (0.2,0.4) .. controls (1.2,0.6) and (1.6,1.2) .. (2.0,1.4);
  \draw[thick, deepnavy] (2.0,1.4) -- (2.0,2.2);
  \draw[thick, deepnavy]
    (2.0,2.2) .. controls (2.6,2.4) and (3.4,2.7) .. (4.5,2.9);
  % filled / open dots at jump (right-continuous: filled at top)
  \fill[deepnavy] (2.0,2.2) circle (1.5pt);
  \draw[deepnavy, fill=white] (2.0,1.4) circle (1.5pt);
  % horizontal level y meeting the jump
  \draw[dashed, exampleblue] (-0.05,1.8) node[left] {\small \(y\)} -- (2.0,1.8);
  \draw[dashed, exampleblue] (2.0,1.8) -- (2.0,0) node[below] {\small \(g(y)\)};
\end{tikzpicture}
\caption{A right-continuous non-decreasing \(F\) with a jump.
\(g(y)=\inf\{x:y\le F(x)\}\) reads the picture sideways: a level
\(y\) inside the jump still resolves to a single \(x\)-value.}
\label{fig:cdf-leftinverse}
\end{figure}

\begin{definition}{Radon measure}{radon}
A measure \(\mu\) on \((\Omega,\Bcal)\), with \(\Bcal\) the Borel
\(\sigma\)-field, is a \emph{Radon measure} if \(\mu(K)<\infty\) for
every compact \(K\in\Bcal\).
\end{definition}

\begin{remark}
\(dF\) is a Radon measure on \(\R\), and conversely every non-zero
Radon measure on \(\Bcal(\R)\) can be written as \(dF=\lambda\circ
g^{-1}\) for some non-decreasing right-continuous \(F\): take
\[
F(x) \;=\;
\begin{cases}
\;\;\mu\bigl((0,x]\bigr) & \text{if } x\ge 0,\\
-\mu\bigl((x,0]\bigr)    & \text{if } x<0.
\end{cases}
\]
Then \(F(b)-F(a)=\mu((a,b])\) for \(a<b\), so \(\mu=dF\) by uniqueness.
Most measures one meets in practice are Radon.
\end{remark}

\subsection{Product \texorpdfstring{$\sigma$}{sigma}-fields and the product measure}

Switch settings: two measurable spaces, with the goal of building a
measure on the Cartesian product.

\begin{definition}{Rectangles and product $\sigma$-field}{prod-sigma}
Given measurable spaces \((\Xset,\Xcal)\) and
\((\Yset,\Ycal)\), a \emph{rectangle} is a set
\(A\times B\) with \(A\in\Xcal\), \(B\in\Ycal\). Write
\(\Rcal\) for the collection of all rectangles. The \emph{product
\(\sigma\)-field} is
\[
\Xcal\times\Ycal \;=\; \sigma(\Rcal).
\]
\end{definition}

\begin{theorem}{Existence and uniqueness of the product measure}{product-measure}
Let \((\Xset,\Xcal,\mu)\) and \((\Yset,\Ycal,\nu)\) be
\(\sigma\)-finite measure spaces, and let \(\pi\) be the set function
on rectangles defined by
\[
\pi(A\times B) \;=\; \mu(A)\,\nu(B),\qquad A\in\Xcal,\;B\in\Ycal.
\]
Then \(\pi\) extends uniquely to a measure on
\((\Xset\times\Yset,\,\Xcal\times\Ycal)\), and for every
\(E\in\Xcal\times\Ycal\),
\[
\pi(E) \;=\; \iint \indic_E(x,y)\,d\mu(x)\,d\nu(y)
       \;=\; \iint \indic_E(x,y)\,d\nu(y)\,d\mu(x).
\]
\end{theorem}

\begin{remark}
\(\sigma\)-finiteness cannot be dropped: without it, the extension
above need not be unique. The strategy of proof is (i)~prove the result
for finite measures, then (ii)~stitch together \(\sigma\)-finite
exhaustions \(\{A_i\}\subseteq\Xcal\), \(\{B_j\}\subseteq\Ycal\)
with \(\mu(A_i),\nu(B_j)<\infty\). The technical engine for step (i) is
not Dynkin's \(\pi\)-\(\lambda\) theorem but its sibling, the
\emph{monotone class theorem}.
\end{remark}

\subsection{The monotone class theorem}

The monotone class theorem is the natural ``\(\pi\)-\(\lambda\)''
tool when one starts from a field rather than a \(\pi\)-system.

\begin{definition}{Monotone class}{monotone-class}
A collection \(\Mcal\) of subsets of \(\Omega\) is a \emph{monotone
class} if it is closed under monotone limits:
\begin{enumerate}
\item for \(\{A_i\}_{i=1}^{\infty}\subseteq\Mcal\) with \(A_i\uparrow
A=\bigcup_{i=1}^{\infty}A_i\), one has \(A\in\Mcal\);
\item for \(\{A_i\}_{i=1}^{\infty}\subseteq\Mcal\) with \(A_i\downarrow
A=\bigcap_{i=1}^{\infty}A_i\), one has \(A\in\Mcal\).
\end{enumerate}
\end{definition}

Recall that a \emph{field} (algebra) on \(\Omega\) is a non-empty
collection containing \(\Omega\), closed under complements and finite
unions. A field that is also a monotone class is automatically a
\(\sigma\)-field.

\begin{theorem}{Monotone class theorem}{monotone-class-thm}
Let \(\Acal\) be a field on \(\Omega\) and let \(\Mcal\) be a monotone
class with \(\Acal\subseteq\Mcal\). Then
\[
\sigma(\Acal) \;\subseteq\; \Mcal.
\]
\end{theorem}

\begin{remark}
Equivalently, the smallest monotone class containing a field
\(\Acal\) coincides with \(\sigma(\Acal)\). One uses this in the
product-measure proof as follows: the rectangles \(\Rcal\) are a
semi-ring, finite disjoint unions of rectangles form a field
\(\Acal\), and \(\pi\) is countably additive on \(\Acal\). The
monotone class theorem is what lifts ``equality on \(\Acal\)'' to
``equality on \(\sigma(\Acal)=\Xcal\times\Ycal\)'' for any
two candidate extensions.
\end{remark}

\begin{lemma}{Order of integration for indicators}{indic-swap}
Let \((\Xset,\Xcal,\mu)\) and \((\Yset,\Ycal,\nu)\) be
finite measure spaces. Then for every
\(E\in\Xcal\times\Ycal\),
\[
\iint \indic_E(x,y)\,d\mu(x)\,d\nu(y)
   \;=\; \iint \indic_E(x,y)\,d\nu(y)\,d\mu(x).
\]
\end{lemma}

The collection of \(E\) for which the displayed equality holds
contains every rectangle \(A\times B\) (since both sides equal
\(\mu(A)\nu(B)\)), is closed under finite disjoint unions, and is a
monotone class by Monotone Convergence. By
\cref{thm:monotone-class-thm} it equals \(\Xcal\times\Ycal\).

\subsection{The Fubini--Tonelli theorem}

The set-function identity in \cref{thm:product-measure} extends to
integrals of \emph{measurable functions}, not just indicators.

\begin{theorem}{Fubini--Tonelli}{fubini-tonelli}
Let \((\Xset,\Xcal,\mu)\) and \((\Yset,\Ycal,\nu)\) be
\(\sigma\)-finite measure spaces, and let
\(f\colon\Xset\times\Yset\to\R\) be \(\Xcal\times\Ycal\)-measurable.
Suppose either
\[
f\ge 0\qquad\text{(Tonelli)},\qquad\text{or}\qquad
\iint |f|\,d(\mu\times\nu)<\infty\quad\text{(Fubini)}.
\]
Then
\[
\int f\,d(\mu\times\nu)
  \;=\; \iint f(x,y)\,d\mu(x)\,d\nu(y)
  \;=\; \iint f(x,y)\,d\nu(y)\,d\mu(x).
\]
Moreover \(\int f(x,y)\,d\mu(x)\) is \(\Ycal\)-measurable in
\(y\), and \(\int f(x,y)\,d\nu(y)\) is \(\Xcal\)-measurable in
\(x\).
\end{theorem}

\begin{remark}
The proof slots together the three convergence theorems of Lecture~6
with the indicator case: the identity is immediate for indicators
(\cref{lem:indic-swap}), extends by linearity to simple functions,
extends by Monotone Convergence to non-negative measurable \(f\)
(Tonelli), and finally extends to integrable \(f=f^+-f^-\) by
splitting positive and negative parts (Fubini). Both halves are needed
in practice: Tonelli to \emph{check} integrability by computing one of
the iterated integrals of \(|f|\); Fubini to then \emph{swap} the
order in the actual integral.
\end{remark}

\begin{remark}
By induction the theorem extends to any finite product
\(\Xcal_1\times\cdots\times\Xcal_n\) of \(\sigma\)-finite spaces. The
infinite product story is genuinely different: a countably infinite
product of \(\sigma\)-finite spaces need not be \(\sigma\)-finite, but
a countable product of \emph{probability} spaces is again a probability
space. We will return to this when constructing stochastic processes.
\end{remark}