From 686e638db3f29e28db6a6ebf12b3001ab213c5af Mon Sep 17 00:00:00 2001 From: Matthew Lancellotti Date: Fri, 5 Jan 2018 15:23:25 -0500 Subject: [PATCH] fixed small errors, typos, and siplified/organized proof the induced uF is Radon --- Real Analysis/real-analysis-notes.tex | 118 ++++++++++++-------------- 1 file changed, 56 insertions(+), 62 deletions(-) diff --git a/Real Analysis/real-analysis-notes.tex b/Real Analysis/real-analysis-notes.tex index b28f436..62d7b1f 100644 --- a/Real Analysis/real-analysis-notes.tex +++ b/Real Analysis/real-analysis-notes.tex @@ -1522,7 +1522,7 @@ \subsection{Regularity} Similarly, $\mu$ is \de{inner regular on $E$} if and only if \[ \mu(E) = \sup \{\mu(K) \st K \text{ is compact}, K \subset E\}. \] - Finally, $\mu$ is called \de{regular} is and only if it is inner and + Finally, $\mu$ is \de{regular} if and only if it is inner and outer regular on every Borel set. \end{defn} \begin{defn} @@ -1555,7 +1555,7 @@ \subsection{Regularity} Let $\epsilon > 0$. Then there exists a sequence of $A_n \in \A$ such that $\sum_{n=1}^\infty \mu(A_n) \leq \mu(E) + \epsilon$ and \(E \subset \Union A_n\). So, - decompoising $A_n$ into elements of $\Ep$ and renumbering, we + decomposing $A_n$ into elements of $\Ep$ and renumbering, we can assume without loss of generality that $A_n \in \Ep$, since if $A_n \not \in \Ep$, we can just take $A_n = A_n \intersect (-m,m)$ using the assumption $E \subset (-m,m)$. So, now, take $A_n = (\alpha_n, @@ -1578,86 +1578,80 @@ \subsection{Regularity} and $g(2k+1) = -k$. Then, we have \[ \mu(E) = \sum_{n \geq 1} \mu(E \intersect (g(n),g(n)+1)). \] - So, for all $n$, let $a_{n,i} < b_{n,i}$. Then, \[ - \sum_{i=1}^\infty \mu((a_{n,i},b_{n,i})) \leq \sum_{n \geq 1} + Each $E \intersect (g(n),g(n)+1)$ is bounded, so by the previous case that we just proved, there exist $a_{n,i} < b_{n,i}$ such that \[ + \sum_{i=1}^\infty \mu((a_{n,i},b_{n,i})) \leq \mu(E \intersect (g(n),g(n)+1)) + \frac{\epsilon}{2^n} \] - by the previous bounded base. Finally, this gives us $E \subset + for each $n$. Finally, this gives us $E \subset \Union_{(n,i) \in \N^2} (a_{n,i},b_{n,i})$ and so \[ \sum_{(n,i) \in \N^2} \mu((a_{n,i},b_{n,i})) \leq \sum_{n \geq - 1} \mu(E - \intersect (g(n),g(n)+1)) + \epsilon + 1} \mu(E \intersect (g(n),g(n)+1)) + \epsilon = \mu(E) + \epsilon \] - thereby completing the proof. + thereby proving the lemma. \end{itemize} \end{proof} \subsection*{(2/28/2017) Lecture 9} With this lemma in hand, we now seek to prove the theorem (\ref{radon-thm}). \begin{proof}[Proof of Theorem] - We have to show $\mu_F$ is finite on compact sets, outer + We have to show $\mu = \mu_F$ is finite on bounded sets, outer regular for all sets in $\B_\R$, and inner regular for all sets in $\B_\R$. \begin{itemize} - \item ($\mu_F$ is finite on compact sets). - $\mu_F$ is finite on compact sets by definition. + \item ($\mu$ is finite on bounded sets). + If $E$ is bounded, then there exists an open bounded interval $(a,b) \supset E$. Then $\mu(E) \leq \mu((a,b)) = F(b) - F(a) < \oo$. \item (Outer regularity). We have the $\leq$ direction from - monotonicity and thus equality if $\mu(E) = \infty$. Now, assume - that $\mu(E) < \infty$. Our lemma tells us that \[ - \mu(E) = \inf \left\{ \sum \mu((a_i,b_i)) \st a_i < b_i - \text{ finite}, E \subset \Union_{i=1}^\infty (a_i,b_i) \right\} + monotonicity, and we want the $\geq$ direction. Our lemma tells us that \[ + \mu(E) = \inf \left\{ \sum \mu((a_i,b_i)) \st E \subset \Union_{i=1}^\infty (a_i,b_i) \right\}. \] - Next, take $\epsilon > 0$. Then, there exists $a_i,b_i$ such - that $\sum \mu((a_i,b_i)) \leq \mu(E) +\epsilon$ by - subadditivity. Now, take $V = \Union_{i=1}^\infty (a_i,b_i)$ - (open) to get $\mu(V) \leq \mu(E) + \epsilon$. But $\epsilon$ - was arbitrary, so $\mu(E) = \inf\{ \cdots \}$ as desired. + Now every $\union_{i=1}^\infty (a_i,b_i)$ is open. Therefore + \[ + \inf \left\{ \sum \mu(\Theta) \st \Theta \:\text{open} \right\} + \leq \inf \left\{ \sum \mu((a_i,b_i)) \st E \subset \Union_{i=1}^\infty (a_i,b_i) \right\} + \] + where the RHS is $\mu(E)$ as desired. \item (Inner regularity). The $\geq$ direction is trivial, so we - only need to show $\leq$. We can also reduce to $E$ being - bounded by using continuity from below and intersecting $E - \intersect [-m,m]$ to get $\mu(E) = \lim_{m \to \infty} \mu(E - \intersect [-m,m])$. So, if $\alpha < \mu(E)$, there exists an - $m$ such that $\mu(E \intersect [-m,m]) > \alpha$. Now, if $E$ - is bounded, there exists a - compact $K \subset E \intersect [-m,m] \subset E$ with $\mu(K) > - \alpha$. Thus, for all $\alpha$, we have $\sup \{\mu(K) - \st \text{ compact }K \subset E\} \geq \mu(E)$. Now assume that - $E$ is bounded. Then, there exists an $m \in \N$ such that $E - \subset [-m,m]$. Let $\epsilon > 0, F = [-m,m] \setminus E$, and - $\mu(F) < \infty$. Our previous case tells us that there exists - an open $V \supset F$ with $\mu(V) \leq \mu(F) + \epsilon$. So, - we take $K = [-m,m] \intersect V^c$ which is closed and bounded, - and therefore compact. + only need to show $\leq$. We break into two cases: + \begin{itemize} + \item ($E$ unbounded). We can reduce to $E$ being + bounded by using continuity from below and intersecting $E + \intersect [-m,m]$ to get $\mu(E) = \lim_{m \to \infty} \mu(E + \intersect [-m,m])$. So, if $\alpha < \mu(E)$, there exists an + $m$ such that $\mu(E \intersect [-m,m]) > \alpha$. There exists a + compact $K \subset E \intersect [-m,m] \subset E$ with $\mu(K) > + \alpha$. Thus, for all $\alpha$, we have $\sup \{\mu(K) + \st \text{ compact }K \subset E\} \geq \mu(E)$. - Next, we show that $K - \subset E$. If $x \in K$ and $x \not \in E$, we get $x \in - [-m,m] \setminus E = F \subset V$, yielding that $x \in V^c$ and - so $x \in E$, a contradiction. Similarly, we also have that $V - \supset F$, and so $\mu(V) \leq \mu(F) + \epsilon$ giving us - $\mu(V \setminus F) \leq \epsilon$. + \item ($E$ bounded). There exists an $m \in \N$ such that $E + \subset [-m,m]$. Let $\epsilon > 0, F = [-m,m] \setminus E$, and + $\mu(F) < \infty$. Our previous case tells us that there exists + an open $V \supset F$ with $\mu(V) \leq \mu(F) + \epsilon$ and thus $\mu(V \setminus F) \leq \epsilon$. So, + we take $K = [-m,m] \intersect V^c$ which is closed and bounded, + and therefore compact. By design, $K \subset E$.\\ - Finally, we show $E \setminus K \subset V \setminus F$. We note - that - \begin{align*} - E \setminus F & = E \intersect K^c \\ - & = E \intersect ([-m,m] \intersect V^c)^c \\ - & = E \intersect ([-m,m]^c \union V) \\ - & = (E \intersect [-m,m]^c) \union (E \intersect - V)\\ - & = E \intersect V - \end{align*} - since $E \intersect [-m,m]^c = \emptyset$. Furthermore, - \begin{align*} - V \setminus F & = V \intersect F^c \\ - & = V \intersect ([-m,m] \intersect F^c)^c \\ - & = V \intersect ([-m,m]^c \union F) \\ - & = (V \intersect F) \union (V \intersect [-m,m]^c) - \end{align*} - Thus, we get that $\mu(E \setminus K) \leq \mu(V \setminus F) - \leq \epsilon$ for all $\epsilon$. + Finally, we show $E \setminus K \subset V \setminus F$. We note + that + \begin{align*} + E \setminus F & = E \intersect K^c \\ + & = E \intersect ([-m,m] \intersect V^c)^c \\ + & = E \intersect ([-m,m]^c \union V) \\ + & = (E \intersect [-m,m]^c) \union (E \intersect + V)\\ + & = E \intersect V + \end{align*} + since $E \intersect [-m,m]^c = \emptyset$. Furthermore, + \begin{align*} + V \setminus F & = V \intersect F^c \\ + & = V \intersect ([-m,m] \intersect F^c)^c \\ + & = V \intersect ([-m,m]^c \union F) \\ + & = (V \intersect F) \union (V \intersect [-m,m]^c) + \end{align*} + Thus, we get that $\mu(E \setminus K) \leq \mu(V \setminus F) + \leq \epsilon$ for all $\epsilon$. + \end{itemize} \end{itemize} \end{proof} \begin{defn} - We say $\mu$ is \de{tight} if, for all $\epsilon > 0$, there + We say that a probability measure $\mu$ is \de{tight} if, for all $\epsilon > 0$, there exists a compact $K \subset \R$ such that $\mu(K) \geq 1-\epsilon$. \end{defn} @@ -1674,7 +1668,7 @@ \subsection{Regularity} \begin{defn} (Aside). We say a sequence of Borel probability measures on $\R$, $(\mu_n)_{n \geq 1}$, \de{converges weakly} to a Borel probability - measure such that, for all $f \from \R \to \R$ bounded an + measure $\mu$ if, for all $f \from \R \to \R$ bounded an continuous, \[ \int f d \mu_n \to \int f \d\mu \text{ as } n \to \infty \]