====== Lecture 9 ====== We will cover Tao's 7.5 and 8.1 today. Here we will use Tao's definition of measurable set, and Lebesgue integration, which a priori is not the same as Pugh's. ===== Tao 7.5: Measurable function ===== Let $\Omega \In \R^m$ be measurable, and $f: \Omega \to \R^m$ be a function. If for all open sets $V \In \R^m$, we have $f^{-1}(V)$ being measurable, then $f$ is called a measurable function. If $f: \Omega \to \R^m$ is continuous, then $f$ is measurable. Indeed, if $f^{-1}(V)$ is open in $\Omega$, then $f^{-1}(V)$ is an intersection of open subset $U \In \R^m$ and $\Omega$ (recall the definition of topology on $\Omega$), an intersection of two measurable sets. Instead of checking on all open sets $V \In \R^m$, we can just check for all open boxes in $\R^m$. Since any open can be written as a countable union of open boxes. A measurable function $f$, post compose with a continuous function $g$ is still measurable. Since $$(g \circ f)^{-1}(open) = f^{-1} (g^{-1}(open)) = f^{-1}(open) = measurable $$ Lemma: $f: \Omega \to \R$ is measurable if and only if for all $a \in \R$, $f^{-1}( (a, \infty))$ is measurable. \\ Proof: every open set in $\R$ is a countable union of open interval, hence suffice to show that all open intervals $(a,b)$ has pre-image being measurable. We can easily show that $f^{-1}((a, b])$ is measurable for all $a