\chapter{Constructing the Borel and Lebesgue measure} It's very difficult to define in one breath a measure on the Borel space $\SB(\RR^n)$. It is easier if we define a weaker notion first. There are two such weaker notions that we will define: \begin{itemize} \ii A \textbf{pre-measure}: satisfies the axioms of a measure, but defined on \emph{fewer} sets than a measure: they'll be defined on an ``algebra'' rather than the full-fledged ``$\sigma$-algebra''. \ii An \textbf{outer measure}: defined on $2^\Omega$ but satisfies weaker axioms. \end{itemize} It will turn out that pre-measures yield outer measures, and outer measures yield measures. \section{Pre-measures} \prototype{Let $\Omega = \RR^2$. Then we take $\SA_0$ generated by rectangles, with $\mu_0$ the usual area.} The way to define a pre-measure is to weaken the $\sigma$-algebra to an algebra. \begin{definition} Let $\Omega$ be a set. We define notions of an \vocab{algebra}, which is the same as $\sigma$-algebra except with ``countable'' replaced by finite everywhere. That is: an algebra $\SA_0$ on $\Omega$ is a nonempty subset of $2^\Omega$, which is closed under complement and \emph{finite} union. The smallest algebra containing a subset $\SF \subseteq 2^\Omega$ is the \vocab{algebra generated by $\SF$}. \end{definition} In practice, we will basically always use generation for algebras. \begin{example} When $\Omega = \RR^n$, we can let $\mathcal{L}_0$ be the algebra generated by $[a_1, b_1] \times \dots \times [a_n, b_n]$. A typical element might look like: \begin{center} \begin{asy} size(5cm); filldraw( (0,0)--(9,0)--(9,2)--(6,2)--(6,5)--(0,5)--cycle, opacity(0.1)+lightcyan, heavycyan ); filldraw( (7,3)--(12,3)--(12,6)--(7,6)--cycle, opacity(0.1)+lightcyan, heavycyan ); \end{asy} \end{center} Unsurprisingly, since we have \emph{finitely} many rectangles and their complements involved, in this case we actually \emph{can} unambiguously assign an area, and will do so soon. \end{example} \begin{definition} A \vocab{pre-measure} $\mu_0$ on a algebra $\SA_0$ is a function $\mu_0 \colon \SA_0 \to [0, +\infty]$ which satisfies the axioms \begin{itemize} \ii $\mu_0(\varnothing) = 0$, and \ii \textbf{Countable additivity}: if $A_1$, $A_2$, \dots are disjoint sets in $\SA_0$ and \emph{moreover} the disjoint union $\bigsqcup A_i$ is contained in $\SA_0$ (not guaranteed by algebra axioms!), then \[ \mu_0\left( \bigsqcup_n A_n \right) = \sum_n \mu_0(A_n). \] \end{itemize} \end{definition} \begin{example} [The pre-measure on $\RR^n$] Let $\Omega = \RR^2$. Then, let $\mathcal{L}_0$ be the algebra generated by rectangles $[a_1, a_2] \times [b_1, b_2]$. We then let \[ \mu_0\left( [a_1, a_2] \times [b_1, b_2] \right) = (a_2-a_1)(b_2-b_1) \] the area of the rectangle. As elements of $\mathcal{L}_0$ are simply \emph{finite} unions of rectangles and their complements (picture drawn earlier), it's not difficult to extend this to a pre-measure $\lambda_0$ which behaves as you expect --- although we won't do this. \end{example} Since we are sweeping something under the rug that turns out to be conceptually important, I'll go ahead and blue-box it. \begin{proposition} [Geometry sanity check that we won't prove] \label{prop:lebesgue_rectangle} For $\Omega = \RR^n$ and $\mathcal{L}_0$ the algebra generated by rectangular prisms, one can define a pre-measure $\lambda_0$ on $\mathcal{L}_0$. \end{proposition} From this point forwards, we will basically do almost no geometry\footnote{White lie. Technically, we will use one more fact: that open sets of $\RR^n$ can be covered by countably infinitely many rectangles, as in \Cref{exer:cubes_vs_open}. This step doesn't involve any area assignments, though.} whatsoever in defining the measure $\SB(\RR^n)$, and only use set theory to extend our measure. So, \Cref{prop:lebesgue_rectangle} is the only sentry which checks to make sure that our ``initial definition'' is sane. To put the point another way, suppose an \textbf{insane scientist}\footnote{Because ``mad scientists'' are overrated.} tried to define a notion of area in which every rectangle had area $1$. Intuitively, this shouldn't be possible: every rectangle can be dissected into two halves and we ought to have $1+1 \ne 1$. However, the only thing that would stop them is that they couldn't extend their pre-measure on the algebra $\mathcal{L}_0$. If they somehow got past that barrier and got a pre-measure, nothing in the rest of the section would prevent them from getting an entire \emph{bona fide} measure with this property. Thus, in our construction of the Lebesgue measure, most of the geometric work is captured in the (omitted) proof of \Cref{prop:lebesgue_rectangle}. \section{Outer measures} \prototype{Keep taking $\Omega = \RR^2$; see the picture to follow.} The other way to weaken a measure is to relax the countable additivity, and this yields the following: \begin{definition} An \vocab{outer measure} $\mu^\ast$ on a set $\Omega$ is a function $\mu^\ast \colon 2^\Omega \to [0, +\infty]$ satisfying the following axioms: \begin{itemize} \ii $\mu^\ast(\varnothing) = 0$; \ii if $E \subseteq F$ and $E,F \in 2^{\Omega}$ then $\mu^\ast(E) \le \mu^\ast(F)$; \ii for any subsets $E_1$, $E_2$, \dots of $\Omega$ we have \[ \mu^\ast \left( \bigcup_n E_n \right) \le \sum_n \mu^\ast(E_n). \] \end{itemize} (I don't really like the word ``outer measure'', since I think it is a bit of a misnomer: I would rather call it ``fake measure'', since it's not a measure either.) \end{definition} The reason for the name ``outer measure'' is that you almost always obtain outer measures by approximating them from ``outside'' sets. Officially, the result is often stated as follows (as \Cref{pr:construct_outer_measure}). \begin{quote} For a set $\Omega$, let $\mathcal{E}$ be \emph{any} subset of $2^{\Omega}$ and let $\rho \colon \mathcal{E} \to [0,+\infty]$ be \emph{any} function. Then \[ \mu^\ast(E) = \inf \left\{ \sum_{n=1}^\infty \rho(E_n) \mid E_n \in \mathcal{E}, \; E \subseteq \bigcup_{n=1}^\infty E_n \right\} \] is an outer measure. \end{quote} However, I think the above theorem is basically always wrong to use in practice, because it is \emph{way too general}. As I warned with the insane scientist, we really do want some sort of sanity conditions on $\rho$: otherwise, if we apply the above result as stated, there is no guarantee that $\mu^\ast$ will be compatible with $\rho$ in any way. %So, the recipe so far goes like: %define an algebra $\SA_0$ generated by some sets with %``known'' measure (like rectangles) and checking it extends well, %then get an outer measure from it. %In the next section, we will then see every outer measure gives a true measure. So, I think it is really better to apply the theorem to pre-measures $\mu_0$ for which one \emph{does} have some sort of guarantee that the resulting $\mu^\ast$ is compatible with $\mu_0$. In practice, this is always how we will want to construct our outer measures. \begin{theorem} [Constructing outer measures from pre-measures] \label{thm:construct_outer} Let $\mu_0$ be a pre-measure on an algebra $\SA_0$ on a set $\Omega$. \begin{enumerate}[(a)] \ii The map $\mu^\ast \colon 2^\Omega \to [0,+\infty]$ defined by \[ \mu^\ast(E) = \inf \left\{ \sum_{n=1}^\infty \mu_0(A_n) \mid A_n \in \SA_0, \; E \subseteq \bigcup_{n=1}^\infty A_n \right\} \] is an outer measure. \ii Moreover, this measure agrees with $\mu_0$ on sets in $\SA_0$. \end{enumerate} \end{theorem} Intuitively, what is going on is that $\mu^\ast(A)$ is the infimum of coverings of $A$ by countable unions of elements in $\SA_0$. Part (b) is the first half of the compatibility condition I promised; the other half appears later as \Cref{prop:cm_compatible}. \begin{proof} [Proof of \Cref{thm:construct_outer}] As alluded to already, part (a) is a special case of \Cref{pr:construct_outer_measure} (and proving it in this generality is actually easier, because you won't be distracted by unnecessary properties). We now check (b), that $\mu^\ast(A) = \mu_0(A)$ for $A \in \SA_0$. One bound is quick: \begin{ques} Show that $\mu^\ast(A) \le \mu_0(A)$. \end{ques} For the reverse, suppose that $A \subseteq \bigcup_n A_n$. Then, define the sets \begin{align*} B_1 &= A \cap A_1 \\ B_2 &= (A \cap A_2) \setminus B_1 \\ B_3 &= (A \cap A_3) \setminus B_2 \\ &\vdotswithin= \end{align*} and so on. Then the $B_n$ are disjoint elements of $\SA_0$ with $B_n \subset A_n$, and we have rigged the definition so that $\bigsqcup_n B_n = A$. Thus by definition of pre-measure, \[ \mu_0(A) = \sum_n \mu_0(B_n) \le \sum_n \mu_0(A_n) \] as desired. \end{proof} \begin{example} Let $\Omega = \RR^2$ and $\lambda_0$ the pre-measure from before. Then $\lambda^\ast(A)$ is, intuitively, the infimum of coverings of the set $A$ by rectangles. Here is a picture you might use to imagine the situation with $A$ being the unit disk. \missingfigure{circles covered by rectangles} \end{example} \section{Carath\'{e}odory extension for outer measures} We will now take any outer measure and turn it into a proper measure. To do this, we first need to specify the $\sigma$-algebra on which we will define the measure. \begin{definition} Let $\mu^\ast$ be an outer measure. We say a set $A$ is \vocab{Carath\'{e}odory measurable with respect to $\mu^\ast$}, or just \vocab{$\mu^\ast$-measurable}, if the following condition holds: for any set $E \in 2^{\Omega}$, \[ \mu^\ast(E) = \mu^\ast(E \cap A) + \mu^\ast(E \setminus A). \] \end{definition} This definition is hard to motivate, but turns out to be the right one. One way to motivate is this: it turns out that in $\RR^n$, it will be equivalent to a reasonable geometric condition (which I will state in \Cref{prop:lebesgue_geo}), but since that geometric definition requires information about $\RR^n$ itself, this is the ``right'' generalization for general measure spaces. Since our goal was to extend our $\SA_0$, we had better make sure this definition lets us measure the initial sets that we started with! \begin{proposition} [Carath\'{e}odory measurability is compatible with the initial $\SA_0$] \label{prop:cm_compatible} Suppose $\mu^\ast$ was obtained from a pre-measure $\mu_0$ on an algebra $\SA_0$, as in \Cref{thm:construct_outer}. Then every set in $\SA_0$ is $\mu^\ast$-measurable. \end{proposition} This is the second half of the compatibility condition that we get if we make sure our initial $\mu_0$ at least satisfies the pre-measure axioms. (The first half was (b) of \Cref{thm:construct_outer}.) \begin{proof} Let $A \in \SA_0$ and $E \in 2^{\Omega}$; we wish to prove $\mu^\ast(E) = \mu^\ast(E \cap A) + \mu^\ast(E \setminus A)$. The definition of outer measure already requires $\mu^\ast(E) \le \mu^\ast(E \cap A) + \mu^\ast(E \setminus A)$ and so it's enough to prove the reverse inequality. By definition of infimum, for any $\eps > 0$, there is a covering $E \subset \bigcup_n A_n$ with $\mu^\ast(E) + \eps \ge \sum_n \mu_0(A_n)$. But \[ \sum_n \mu_0(A_n) = \sum_n \left( \mu_0(A_n \cap A) + \mu_0(A_n \setminus A) \right) \ge \mu^\ast(E \cap A) + \mu^\ast(E \setminus A) \] with the first equality being the definition of pre-measure on $\SA_0$, the second just being by definition of $\mu^\ast$ (since $A_n \cap A$ certainly covers $E \cap A$, for example). Thus $\mu^\ast(E) + \eps \ge \mu^\ast(E \cap A) + \mu^\ast(E \setminus A)$. Since the inequality holds for any $\eps > 0$, we're done. \end{proof} To add extra icing onto the cake, here is one more niceness condition which our constructed measure will happen to satisfy. \begin{definition} A \vocab{null set} of a measure space $(\Omega, \SA, \mu)$ is a set $A \in \SA$ with $\mu(A) = 0$. A measure space $(\Omega, \SA, \mu)$ is \vocab{complete} if whenever $A$ is a null set, then all subsets of $A$ are in $\SA$ as well (and hence null sets). \end{definition} This is a nice property to have, for obvious reasons. Visually, if I have a bunch of dust which I \emph{already} assigned weight zero, and I blow away some of the dust, then the remainder should still have an assigned weight --- zero. The extension theorem will give us $\sigma$-algebras with this property. \begin{theorem} [Carath\'{e}odory extension theorem for outer measures] \label{thm:cara_outer} If $\mu^\ast$ is an outer measure, and $\SA\cme$ is the set of $\mu^\ast$-measurable sets with respect to $\mu^\ast$, then $\SA\cme$ is a $\sigma$-algebra on $\Omega$, and the restriction $\mu\cme$ of $\mu^\ast$ to $\SA\cme$ gives a \emph{complete} measure space. \end{theorem} (Phonetic remark: you can think of the superscript ${}\cme$ as standing for either ``Carath\'{e}odory measurable'' or ``complete''. Both are helpful for remembering what this represents. This notation is not standard but the pun was too good to resist.) Thus, if we compose \Cref{thm:construct_outer} with \Cref{thm:cara_outer}, we find that every pre-measure $\mu_0$ on an algebra $\SA_0$ naturally gives a $\sigma$-algebra $\SA\cme$ with a complete measure $\mu\cme$, and our two compatibility results (namely (b) of \Cref{thm:construct_outer}, together with \Cref{prop:cm_compatible}) means that $\SA\cme \supset \SA_0$ and $\mu\cme$ agrees with $\mu$. Here is a table showing the process, where going down each row of the table corresponds to restriction process. \begin{center} \begin{tabular}[h]{llcl} & & Construct order & Notes \\ \hline $2^\Omega$ & $\mu^\ast$ & Step 2 & $\mu^\ast$ is outer measure obtained from $\mu_0$ \\[1em] $\SA\cme$ & $\mu\cme$ & Step 3 & $\SA\cme$ defined as $\mu^\ast$-measurable sets, \\ &&& $(\SA\cme, \mu\cme)$ is complete. \\[1em] $\SA_0$ & $\mu_0$ & Step 1 & $\mu_0$ is a pre-measure \end{tabular} \end{center} \section{Defining the Lebesgue measure} This lets us finally define the Lebesgue measure on $\RR^n$. We wrap everything together at once now. \begin{definition} We create a measure on $\RR^n$ by the following procedure. \begin{itemize} \ii Start with the algebra $\mathcal{L}_0$ generated by rectangular prisms, and define a \emph{pre-measure} $\lambda_0$ on this $\mathcal{L}_0$ (this was glossed over in the example). \ii By \Cref{thm:construct_outer}, this gives the \vocab{Lebesgue outer measure} $\lambda^\ast$ on $2^{\RR^n}$, which is compatible on all the rectangular prisms. \ii By Carath\'{e}odory (\Cref{thm:cara_outer}), this restricts to a complete measure $\lambda$ on the $\sigma$-algebra $\mathcal{L}(\RR^n)$ of $\lambda^\ast$-measurable sets (which as promised contains all rectangular prisms).\footnote{If I wanted to be consistent with the previous theorems, I might prefer to write $\mathcal{L}\cme$ and $\lambda\cme$ for emphasis. It seems no one does this, though, so I won't.} \end{itemize} The resulting complete measure, denoted $\lambda$, is called the \vocab{Lebesgue measure}. The algebra $\mathcal{L}(\RR^n)$ we obtained will be called the \vocab{Lebesgue $\sigma$-algebra}; sets in it are said to be \vocab{Lebesgue measurable}. \end{definition} Here is the same table from before, with the values filled in for the special case $\Omega = \RR^n$, which gives us the Lebesgue algebra. \begin{center} \begin{tabular}[h]{llcl} & & Construct order & Notes \\ \hline $2^{\RR^n}$ & $\lambda^\ast$ & Step 2 & $\lambda^\ast$ is Lebesgue outer measure \\[1em] $\mathcal L(\RR^n)$ & $\lambda$ & Step 3 & Lebesgue $\sigma$-algebra (complete) \\[1em] $\mathcal L_0$ & $\lambda_0$ & Step 1 & Define pre-measure on rectangles \end{tabular} \end{center} Of course, now that we've gotten all the way here, if we actually want to \emph{compute} any measures, we can mostly gleefully forget about how we actually constructed the measure and just use the properties. The hard part was to showing that there \emph{is} a way to assign measures consistently; actually figuring out what that measure's value is \emph{given that it exists} is often much easier. Here is an example. \begin{example} [The Cantor set has measure zero] The standard \vocab{middle-thirds Cantor set} is the subset $[0,1]$ obtained as follows: we first delete the open interval $(1/3, 2/3)$. This leaves two intervals $[0,1/3]$ and $[2/3,1]$ from which we delete the middle thirds again from both, i.e.\ deleting $(1/9,2/9)$ and $(7/9,8/9)$. We repeat this procedure indefinitely and let $C$ denote the result. An illustration is shown below. \begin{center} \includegraphics[width=0.8\textwidth]{media/cantor-thirds.png} \\ \footnotesize Image from \cite{img:cantor} \end{center} It is a classic fact that $C$ is uncountable (it consists of ternary expansions omitting the digit $1$). But it is measurable (it is an intersection of closed sets!) and we contend it has measure zero. Indeed, at the $n$th step, the result has measure $(2/3)^n$ leftover. So $\mu(C) \le (2/3)^n$ for every $n$, forcing $\mu(C) = 0$. \end{example} This is fantastic, but there is one elephant in the room: how are the Lebesgue $\sigma$-algebra and the Borel $\sigma$-algebra related? To answer this question briefly, I will state two results (but another answer is given in the next section). The first is a geometric interpretation of the strange Carath\'{e}odory measurable hypothesis. \begin{proposition} [A geometric interpretation of Lebesgue measurability] \label{prop:lebesgue_geo} A set $A \subseteq \RR^n$ is Lebesgue measurable if and only if for every $\eps > 0$, there is an open set $U \supset A$ such that \[ \lambda^\ast(U \setminus A) < \eps \] where $\lambda^\ast$ is the Lebesgue outer measure. \end{proposition} I want to say that this was Lebesgue's original formulation of ``measurable'', but I'm not sure about that. In any case, we won't need to use this, but it's good to see that our definition of Lebesgue measurable has a down-to-earth geometric interpretation. \begin{ques} Deduce that every open set is Lebesgue measurable. Conclude that the Lebesgue $\sigma$-algebra contains the Borel $\sigma$-algebra. (A different proof is given later on.) \end{ques} However, the containment is proper: there are more Lebesgue measurable sets than Borel ones. Indeed, it can actually be proven using transfinite induction (though we won't) that $\left\lvert \SB(\RR) \right\rvert = \left\lvert \RR \right\rvert$. Using this, one obtains: \begin{exercise} Show the Borel $\sigma$-algebra is not complete. (Hint: consider the Cantor set. You won't be able to write down an example of a non-measurable set, but you can use cardinality arguments.) Thus the Lebesgue $\sigma$-algebra strictly contains the Borel one. % It should contain every subset of the Cantor set, % since Lebesgue is complete. \end{exercise} Nonetheless, there is a great way to describe the Lebesgue $\sigma$-algebra, using the idea of completeness. \begin{definition} Let $(\Omega, \SA, \mu)$ be a measure space. The \vocab{completion} $(\Omega, \ol{\SA}, \ol{\mu})$ is defined as follows: we let \[ \ol{\SA} = \left\{ A \cup N \mid A \in \SA, N \text{ subset of null set} \right\}. \] and $\ol{\mu}(A \cup N) = \mu(A)$. One can check this is well-defined, and in fact $\ol{\mu}$ is the unique extension of $\mu$ from $\SA$ to $\ol{\SA}$. This looks more complicated than it is. Intuitively, all we are doing is ``completing'' the measure by telling $\ol{\mu}$ to regard any subset of a null set as having measure zero, too. \end{definition} Then, the saving grace: \begin{theorem} [Lebesgue is completion of Borel] For $\RR^n$, the Lebesgue measure is the completion of the Borel measure. \end{theorem} \begin{proof} This actually follows from results in the next section, namely \Cref{exer:cubes_vs_open} and part (c) of Carath\'{e}odory for pre-measures (\Cref{thm:cara_premeasure}). \end{proof} \section{A fourth row: Carath\'{e}odory for pre-measures} \prototype{The fourth row for the Lebesgue measure is $\SB(\RR^n)$.} In many cases, $\SA\cme$ is actually bigger than our original goal, and instead we only need to extend $\mu_0$ on $\SA_0$ to $\mu$ on $\SA$, where $\SA$ is the $\sigma$-algebra generated by $\SA_0$. Indeed, our original goal was to get $\SB(\RR^n)$, and in fact: \begin{exercise} Show that $\SB(\RR^n)$ is the $\sigma$-algebra generated by the $\mathcal{L}_0$ we defined earlier. \label{exer:cubes_vs_open} \end{exercise} Fortunately, this restriction is trivial to do. \begin{ques} Show that $\SA\cme \supset \SA$, so we can just restrict $\mu\cme$ to $\SA$. \end{ques} We will in a moment add this as the fourth row in our table. However, if this is the end goal, than a somewhat different Carath\'{e}odory theorem can be stated because often one more niceness condition holds: \begin{definition} A pre-measure or measure $\mu$ on $\Omega$ is \vocab{$\sigma$-finite} if $\Omega$ can be written as a countable union $\Omega = \bigcup_n A_n$ with $\mu(A_n) < \infty$ for each $n$. \end{definition} \begin{ques} Show that the pre-measure $\lambda_0$ we had, as well as the Borel measure $\SB(\RR^n)$, are both $\sigma$-finite. \end{ques} Actually, for us, $\sigma$-finite is basically always going to be true, so you can more or less just take it for granted. \begin{theorem} [Carath\'{e}odory extension theorem for pre-measures] \label{thm:cara_premeasure} Let $\mu_0$ be a pre-measure on an algebra $\SA_0$ of $\Omega$, and let $\SA$ denote the $\sigma$-algebra generated by $\SA_0$. Let $\SA\cme$, $\mu\cme$ be as in \Cref{thm:cara_outer}. Then: \begin{enumerate}[(a)] \ii The restriction of $\mu\cme$ to $\SA$ gives a measure $\mu$ extending $\mu_0$. \ii If $\mu_0$ was $\sigma$-finite, then $\mu$ is the unique extension of $\mu_0$ to $\SA$. \ii If $\mu_0$ was $\sigma$-finite, then $\mu\cme$ is the completion of $\mu$, hence the unique extension of $\mu_0$ to $\SA\cme$. \end{enumerate} \end{theorem} Here is the updated table, with comments if $\mu_0$ was indeed $\sigma$-finite. \begin{center} \begin{tabular}[h]{llcl} & & Construct order & Notes \\ \hline $2^\Omega$ & $\mu^\ast$ & Step 2 & $\mu^\ast$ is outer measure obtained from $\mu_0$ \\[1em] $\SA\cme$ & $\mu\cme$ & Step 3 & $(\SA\cme, \mu\cme)$ is completion $(\SA, \mu)$, \\ &&& $\SA\cme$ defined as $\mu^\ast$-measurable sets \\[1em] $\SA$ & $\mu$ & Step 4 & $\SA$ defined as $\sigma$-alg.\ generated by $\SA_0$ \\[1em] $\SA_0$ & $\mu_0$ & Step 1 & $\mu_0$ is a pre-measure \end{tabular} \end{center} And here is the table for $\Omega = \RR^n$, with Borel and Lebesgue in it. \begin{center} \begin{tabular}[h]{llcl} & & Construct order & Notes \\ \hline $2^{\RR^n}$ & $\lambda^\ast$ & Step 2 & $\lambda^\ast$ is Lebesgue outer measure \\[1em] $\mathcal L(\RR^n)$ & $\lambda$ & Step 3 & Lebesgue $\sigma$-algebra, completion of Borel one \\[1em] $\SB(\RR^n)$ & $\mu$ & Step 4 & Borel $\sigma$-algebra, generated by $\mathcal{L}_0$ \\[1em] $\mathcal L_0$ & $\lambda_0$ & Step 1 & Define pre-measure on rectangles \end{tabular} \end{center} Going down one row of the table corresponds to restriction, while each of $\mu_0 \to \mu \to \mu\cme$ is a unique extension when $\mu_0$ is $\sigma$-finite. \begin{proof} [Proof of \Cref{thm:cara_premeasure}] For (a): this is just \Cref{thm:construct_outer} and \Cref{thm:cara_outer} put together, combined with the observation that $\SA^\ast \supset \SA_0$ and hence $\SA^\ast \supset \SA$. Parts (b) and (c) are more technical, and omitted. \end{proof} \section{From now on, we assume the Borel measure} \todo{explain why} \section{\problemhead} \begin{dproblem} [Constructing outer measures from arbitrary $\rho$] \label{pr:construct_outer_measure} For a set $\Omega$, let $\mathcal{E}$ be \emph{any} subset of $2^{\Omega}$ and let $\rho \colon \mathcal{E} \to [0,+\infty]$ be \emph{any} function. Prove that \[ \mu^\ast(E) = \inf \left\{ \sum_{n=1}^\infty \rho(E_n) \mid E_n \in \mathcal{E}, \; E \subseteq \bigcup_{n=1}^\infty E_n \right\} \] is an outer measure. \end{dproblem} \begin{problem} [The insane scientist] Let $\Omega = \RR^2$, and let $\mathcal{E}$ be the set of (non-degenerate) rectangles. Let $\rho(E) = 1$ for every rectangle $E \in \mathcal{E}$. Ignoring my advice, the insane scientist uses $\rho$ to construct an outer measure $\mu^\ast$, as in \Cref{pr:construct_outer_measure}. \begin{enumerate}[(a)] \ii Find $\mu^\ast(S)$ for each subset $S$ of $\RR^2$. \ii Which sets are $\mu^\ast$-measurable? \end{enumerate} You should find that no rectangle is $\mu^\ast$-measurable, unsurprisingly foiling the scientist. \begin{hint} Show that \[ \mu^\ast(S) = \begin{cases} 0 & S = \varnothing \\ 1 & S \text{ bounded and nonempty} \\ \infty & S \text{ not bounded}. \end{cases} \] This lets you solve (b) readily; I think the answer is just unbounded sets, $\varnothing$, and one-point sets. \end{hint} \end{problem} \begin{problem} \gim A function $f \colon \RR \to \RR$ is continuous. Must $f$ be measurable with respect to the Lebesgue measure on $\RR$? \end{problem}