% ================================================================= % Lecture 7 % Source: handwritten notes (Mathpix-converted) + Kashlak STAT 571 % ================================================================= \section[Lecture 7 -- Lebesgue-Stieltjes Measure; Fubini-Tonelli]{Lecture 7 \textemdash{} Lebesgue--Stieltjes Measure; Fubini--Tonelli} \label{sec:lec07} We are halfway through, with one more lecture on integration to come. The plan today is to use measurable functions to \emph{transport} measures between spaces: a measurable map $\psi:\mathbb{X}\to\mathbb{Y}$ together with a measure $\mu$ on $\mathbb{X}$ produces a pushforward $\nu=\mu\circ\psi^{-1}$ on $\mathbb{Y}$. Specialising to $\mathbb{X}=\mathbb{Y}=\R$ and $\mu=\lambda$ recovers the family of \emph{Lebesgue--Stieltjes} measures $dF$ generated by distribution-like functions $F$. We then build the \emph{product measure} $\mu\times\nu$ rigorously via a monotone class argument, and conclude with \emph{Fubini--Tonelli}, the workhorse for swapping the order of integration. \subsection{Image measures and Lebesgue--Stieltjes} Given two measurable spaces $(\mathbb{X},\Xcal)$ and $(\mathbb{Y},\Ycal)$ and a measurable $\psi:\mathbb{X}\to\Ycal$, any measure $\mu$ on $\Xcal$ induces an \emph{image measure} on $\Ycal$. \begin{definition}{Image (pushforward) measure}{image-measure} Let $\psi:\mathbb{X}\to\mathbb{Y}$ be measurable and let $\mu$ be a measure on $(\mathbb{X},\Xcal)$. The \emph{image measure} of $\mu$ under $\psi$ is the set function \[ \nu(B) \;=\; \mu\bigl(\psi^{-1}(B)\bigr), \qquad B\in\Ycal, \] also written $\nu=\mu\circ\psi^{-1}$. It is a measure on $(\mathbb{Y},\Ycal)$. \end{definition} The most useful instance turns Lebesgue measure $\lambda$ on $\R$ into a one-parameter family of Borel measures indexed by a ``CDF-like'' function $F$. \begin{theorem}{Lebesgue--Stieltjes measure}{lebesgue-stieltjes} Let $F:\R\to\R$ be non-constant, right-continuous, and non-decreasing. Then there exists a unique measure $dF$ on $(\R,\Bcal(\R))$ such that \[ dF\bigl((a,b]\bigr) \;=\; F(b)-F(a) \qquad\text{for all } a