Set theory - Chapter 10: Measurable Cardinals

Let $M$ be a maximal subset of $[0,1]$ satisfying $\forall x,y\in M,x\neq y\implies x-y\notin\mathbb{Q}$. For all $q\in\mathbb{Q}$, define $M_{q}=\{x+q:x\in M\}$. If $M_{q}\cap M_{r}\neq\emptyset$ for $q,r\in\mathbb{Q}$ then there are $x,y\in M$ such that $x+q=y+q^{{\prime}}$. So $x-y=q^{{\prime}}-q\in\mathbb{Q}$ and by definition of $M$ $x=y$. Hence $q=q^{{\prime}}$ and the $M_{q}$ are pairwise disjoint.

We have $\emptyset\subsetneq\{0,\sqrt{2}/2\}\subseteq[0,1]$ and $\sqrt{2}/2-0\notin\mathbb{Q}$ so $M\neq\emptyset$ and we can pick $y\in M$. Let $x\in[0,1]$. If $x-y\notin\mathbb{Q}$ then by maximality of $M$, $x\in M=M_{0}$. Otherwise, $|x-y|\leq 1$ and so $x=y+q$ for some $q\in[-1,1]\cap\mathbb{Q}$ that is $x\in M_{q}$. Moreover, $M\subseteq[0,1]$ and so if $q\in[-1,1]\cap\mathbb{Q}$, we have $M_{q}\subseteq[-1,2]$ and finally

Suppose $M$ is Lebesgue measurable. $\mu$ is translation invariant and so all the $M_{q}$ are Lebesgue measurable with measure $\mu(M_{q})=\mu(M)$. Let $\{q_{n}:n\in\mathbb{N}\}$ an enumeration of $[-1,1]\cap\mathbb{Q}$. $M$ is $\sigma$-complete and the $M_{{q_{n}}}$ are pairwise disjoint so

If $\mu(M)=0$ then the inner sum is zero and if $\mu(M)>0$ the inner sum is infinite. A contradiction.

Let $\nu$ be the measure defined in the proof of Lemma 10.9 (i). Suppose $Z$ is an atom of $\nu$ and let $Y=f_{{-1}}(Z)$. We will prove by induction on $\alpha<\omega_{1}$ that there is a unique element $X_{\alpha}$ at level $\alpha$ such that $\mu(Y\cap X_{\alpha})\neq 0$ and actually $\mu(Y\cap X_{\alpha})=\nu(Z)$.

For $\alpha=0$, we have $0\neq\nu(Z)=\mu(f_{{-1}}(Z))=\mu(Y)=\mu(Y\cap S)$. Suppose that the result is true at level $\alpha$. The level $\alpha+1$ is made of elements $T$ which are immediate successors of some $X\neq X_{\alpha}$ at level $\alpha$ and of two immediate successors $U$ and $X_{\alpha}\setminus U$ of $X_{\alpha}$. We have $\mu(Y\cap T)\leq\mu(Y\cap X)=0$ and $0\neq\mu(Y)=\mu(Y\cap U)+\mu(Y\cap(X_{\alpha}\setminus U))$. In particular we have $\mu(Y\cap U)\neq 0$ or $\mu(Y\cap(X_{\alpha}\setminus U))\neq 0$. Then $\mu(Y\cap U)\leq\mu(f_{{-1}}(Z)\cap f_{{-1}}(f(U)))=\mu(f_{{-1}}(Z\cap f(U)))=\nu(Z\cap f(U))$. Similarly, $\mu(Y\cap(X_{\alpha}\setminus U))\leq\nu(Z\cap f(X_{{\alpha}}\setminus U))$. But $\nu(Z)=\nu(Z\cap f(X_{\alpha}))=\nu(Z\cap f(U\cup(X_{{\alpha}}\setminus U)))=\nu(Z\cap f(U))+\nu(Z\cap f(X_{{\alpha}}\setminus U))$ and because $Z$ is an atom, $\nu(Z\cap f(U))=0$ or $\nu(Z\cap f(X_{{\alpha}}\setminus U))=0$. Finally, we have $\{\mu(Y\cap U),\mu(Y\cap(X_{\alpha}\setminus U))\}=\{0,\mu(Z)\}$ and the hypothesis is true at level $\alpha+1$.

If $\alpha$ is limit and the result is true for all $\xi<\alpha<\omega_{1}$ Each level is at most countable and $\mu((S\setminus X_{\xi})\cap Y)=\mu\left(\bigcup_{{o(X)=\xi;X\neq X_{\xi}}}X\cap Y\right)=\sum_{{o(X)=\xi;X\neq X_{\xi}}}\mu(X\cap Y)=0$ by induction hypothesis. Hence $\mu\left({\left(S\setminus\bigcap_{{\xi<\alpha}}X_{\xi}\right)}\cap Y\right)=\mu\left(\bigcup_{{\xi<\alpha}}(S\setminus X_{\xi})\cap Y\right)=\sum_{{\xi<\alpha}}\mu((S\setminus X_{\xi})\cap Y)=0$. Finally,

In particular, $X_{\alpha}=\bigcap_{{\xi<\alpha}}X_{\xi}$ is in the tree $T$ and $o(X_{\alpha})=\alpha$. If $T_{\alpha}=\bigcap_{{\xi<\alpha}}T_{\xi}$ is another element at level $\alpha$ such that $\forall\xi,o(T_{\xi})=\xi$ then there is $\xi<\alpha$ such that $X_{\xi}\nsubseteq T_{\alpha}$ or $T_{\xi}\nsubseteq X_{\alpha}$. Hence $T_{\xi}\neq X_{\xi}$ and by induction hypothesis $\mu(T_{\alpha}\cap Y)\leq\mu(T_{\xi}\cap Y)=0$. Hence the hypothesis is true at level $\alpha$.

We have obtained a linarly ordered set $\{X_{\alpha}:\alpha<\omega_{1}\}$ with exactly one element at each level: that’s a branch in $T$ of length $\omega_{1}$, a contradiction.

Let $\mu$ be an atomless measure on $S$ and $Z_{0}\subseteq Z$. We define by induction on $\alpha<\omega_{1}$ some sets $S_{0}\supseteq S_{1}\supseteq S_{2}\supseteq...\supseteq S_{\alpha}\supseteq...$ such that $\mu(S_{\alpha})\geq\frac{1}{2}\mu(Z_{0})$ and if $\mu(S_{\alpha})>\frac{1}{2}\mu(Z_{0})$ then $\mu(S_{{\alpha+1}})<\mu(S_{\alpha})$. We let $S_{0}=Z_{0}$.

Suppose that $S_{\alpha}$ is defined. If $S_{\alpha}=\frac{1}{2}\mu(Z_{0})$, we just take $S_{{\alpha+1}}=S_{\alpha}$. Otherwise, $\mu(S_{\alpha})>\frac{1}{2}\mu(Z_{0})\geq 0$ we shall construct by induction on $n\geq 1$ some sets $S_{\alpha}^{n}\subseteq S_{\alpha}$ such that

For $n$ large enough, we get $\frac{1}{2}\mu(Z_{0})<{\left(1-2^{{-n}}\right)}\mu(S_{\alpha})\leq\mu(S_{\alpha}^{n})<\mu(S_{\alpha})$ and we define $S_{{\alpha+1}}=S_{\alpha}^{n}$.

Thus let’s construct $S_{\alpha}^{n}$. $\mu(S_{\alpha})>0$ and $\mu$ is atomless, so $S_{\alpha}$ can be split into two disjoint set of positive measure. If $S_{\alpha}^{1}$ is the one of measure at least $\frac{1}{2}\mu(S_{\alpha})$, then $(1-2^{{-1}})\mu(S_{\alpha})\leq\mu(S_{\alpha}^{1})<\mu(S_{\alpha})$. Similarly, if the induction hypothesis is true then $S_{\alpha}\setminus S_{\alpha}^{n}$ is of positive measure and so can be split into two disjoint set of positive measure. Choose $X$ the one of measure at least $\frac{1}{2}\mu(S_{\alpha}\setminus S_{\alpha}^{n})$ and let $S_{\alpha}^{{n+1}}=S_{\alpha}^{n}\cup X$. Then $\mu(S_{\alpha})>\mu(S_{\alpha}^{{n+1}})=\mu(S_{\alpha}^{n})+\mu(X)\geq\mu(S_{\alpha}^{n})+\frac{1}{2}\left(\mu(S_{\alpha})-\mu(S_{\alpha}^{n})\right)=\frac{1}{2}\left(\mu(S_{\alpha}^{n})+\mu(S_{\alpha})\right)\geq\frac{1}{2}\left(1-2^{{-n}}+1\right)\mu(S_{\alpha})=\left(1-2^{{-(n+1)}}\right)\mu(S_{\alpha})$.

Now suppose $\alpha$ is limit and that $S_{\beta}$ is defined for all $\beta<\alpha$. If $S_{\alpha}=\bigcap_{{\beta<\alpha}}S_{\beta}$ then clearly $S_{\beta}\supseteq S_{\alpha}$ for all $\beta<\alpha$. Let $\beta_{n}$ a countable increasing sequence which is cofinal in $\alpha$ with $\beta_{0}=0$. Clearly, $S_{\alpha}\subseteq\bigcap_{{n<\omega}}S_{{\beta_{n}}}$. Conversely, let $x\in\bigcap_{{n<\omega}}S_{{\beta_{n}}}$. For all $\beta<\alpha$, there is $n<\omega$ such that $\beta<\beta_{n}$ and so $x\in S_{{\beta_{n}}}\subseteq S_{\beta}$. Hence $S_{\alpha}=\bigcap_{{n<\omega}}S_{{\beta_{n}}}$. We have $\mu(S_{\alpha})=\mu(Z_{0})-\mu\left(\bigcup_{{n<\omega}}(S_{0}\setminus S_{{\beta_{n}}})\right)=\mu(Z_{0})-\sum_{{n=0}}^{{+\infty}}\mu(T_{n})$ where $T_{n}=(S_{0}\setminus S_{{\beta_{{n+1}}}})\setminus(S_{0}\setminus S_{{\beta_{n}}})$. For all $N<\omega$, we have $\sum_{{n=0}}^{{N}}\mu(T_{n})=\mu(S_{0}\setminus S_{{\beta_{N}}})=\mu(Z_{0})-\mu(S_{{\beta_{N}}})\leq\frac{1}{2}\mu(Z_{0})$. Hence $\sum_{{n=0}}^{{+\infty}}\mu(T_{n})\leq\frac{1}{2}\mu(Z_{0})$ and so $\mu(S_{\alpha})\geq\frac{1}{2}\mu(Z_{0})$.

Finally, suppose that for all $\alpha<\omega_{1}$, we have $\mu(S_{\alpha})>\frac{1}{2}\mu(Z_{0})$. Then $\mu(S_{{\alpha+1}})<\mu(S_{{\alpha}})$ and we can pick $x_{\alpha}\in\mathbb{Q}\cap(\mu(S_{{\alpha+1}}),\mu(S_{{\alpha}}))$. We get a one-to-one mapping $\alpha\mapsto x_{\alpha}$ from $\omega_{1}$ to $\mathbb{Q}$. A contradiction. Hence there is some $\alpha<\omega_{1}$ such that $\mu(S_{\alpha})=\frac{1}{2}\mu(Z_{0})$

If $\mu$ is a measure on $S$ then the ideal $I_{\mu}=\{X\subseteq S:\mu(X)=0\}$ is $\kappa$-complete if and only if $\mu$ is $\kappa$-complete by Lemma 10.6. If moreover $\mu$ is a two-valued measure then the ultrafilter $U=\{X\subseteq S:\mu(X)=1\}$ is the dual filter of $I_{\mu}$. Hence $U$ is $\kappa$-complete iff $I_{\mu}$ is $\kappa$-complete iff $\mu$ is $\kappa$-additive.

Let $U$ be a nonprincipal $\kappa$-complete ultrafilter on $\kappa$. By theorem 10.20, if $f$ is the least function such that for all $\gamma<\kappa$, $\{\alpha:f(\alpha)>\gamma\}\in U$ then $f_{*}(U)$ is a normal measure. If $f=d$ is the diagonal function, then $f_{*}(U)=d_{*}(U)=U$ so $U$ is a normal measure.

Conversely, suppose that $U$ is a normal measure. Let $\gamma<\kappa$. $\left|\kappa\setminus\{\alpha<\kappa:d(\alpha)>\gamma\}\right|=\left|\{\alpha<\kappa:\alpha\leq\gamma\}\right|=|\gamma|<\kappa$. Hence, since $U$ is $\kappa$-complete and $U$ is an ultrafilter, $\{\alpha<\kappa:d(\alpha)>\gamma\}\in U$. Suppose that there is $f<d$ with the property that for all $\gamma<\kappa$, $\{\alpha:f(\alpha)>\gamma\}\in U$. Then for all $\gamma<\kappa$, $X_{\gamma}=\{\alpha<\kappa:\gamma<f(\alpha)<d(\alpha)=\alpha\}\in U$. Because $U$ is normal, $\Delta_{{\gamma<\kappa}}X_{\gamma}\in U$. But if $\alpha\in\Delta_{{\gamma<\kappa}}X_{\gamma}$ then for all $\gamma<\alpha$, $\alpha\in X_{\gamma}$ that is $\gamma<f(\alpha)<\alpha$. We get a contradiction for $\gamma=f(\alpha)$. So $d$ is the least function with the property mentioned above.

Let $D$ be a normal measure on $\kappa$ and let $f:{[\kappa]}^{{<\omega}}\rightarrow\kappa$ such that for all $x\in{[\kappa]}^{{<\omega}}$ $f(x)=0$ or $f(x)<\min x$. We show by induction on $n$ that for any $g:{[\kappa]}^{n}\rightarrow\kappa$ with the previous property, there is there is $H\in D$ such that $g$ is constant on ${[H]}^{n}$. Then given $f$ as above, we consider $H_{n}\in D$ homogeneous for $f_{{|{[\kappa]}^{n}}}$. Then $H=\bigcup_{{n<\omega}}H_{n}\in D$ and $f$ is constant on ${[H]}^{{<\omega}}$.

We start with $n=1$. Let $h:\kappa\rightarrow{[\kappa]}^{1}$ be the canonical injection. Then for all $0<\alpha<\kappa$, $g(h(\alpha))=0<\alpha$ or $g(h(\alpha))<\min h(\alpha)=\min{\{\alpha\}}=\alpha$. Hence $g\circ h$ is regressive on $\kappa\setminus\{0\}\in D$ and because $D$ is normal there is $H\in D$ such that $g\circ h$ is constant on $H$. Hence $g$ is constant on ${[H]}^{1}$.

Suppose that the result is true for some $n<\omega$. Let $\alpha<\kappa$ and define Define $g_{{\alpha}}:{[\kappa]}^{n}\rightarrow\kappa$ by $g_{{\alpha}}(x)=g(\{\alpha\}\cup x)$ if $\alpha<\min x$ and $g_{{\alpha}}(x)=0$ otherwise. We have $g_{{\alpha}}(x)=0$ or $g_{{\alpha}}(x)<\min(\{\alpha\}\cup x)=\alpha<\min x$. By induction hypothesis, there is $X_{\alpha}\in D$ such that $g_{\alpha}$ is constant on ${[D]}^{n}$ with value $\gamma_{\alpha}$. Let $X=\Delta_{{\alpha<\kappa}}X_{\alpha}$. If $\alpha_{0}<\alpha_{1}<...<\alpha_{n}$ are elements of $X$ then $\alpha_{1},...\alpha_{n}\in X_{{\alpha_{0}}}$ and $g(\{\alpha_{0},\alpha_{1},...,\alpha_{n}\})=g_{{\alpha_{0}}}(\alpha_{1},...,\alpha_{n})=\gamma_{{\alpha_{0}}}$. For all $\alpha_{0}\in X\setminus\{0\}\in D$, we have $\gamma_{{\alpha_{0}}}=0$ or $\gamma_{{\alpha_{0}}}<\alpha_{0}$ so there is $H\in D$ and $\gamma<\kappa$ such that for all $\alpha_{0}\in H$, $\gamma_{{\alpha_{0}}}=\gamma$. Finally, $g$ is constant on ${[H]}^{{n+1}}$.

Let $\kappa$ be a measurable cardinal and $U$ a normal measure on $\kappa$. Let $f:\kappa\rightarrow P_{\kappa}(\kappa)$ be the canonical injection and consider the ultrafilter $D=f_{*}(U)=\{X\subseteq P_{\kappa}(\kappa):f_{{-1}}(X)\in U\}$. $U$ is $\kappa$-complete, so $D$ is $\kappa$-complete too. For all $x\in P_{\kappa}(\kappa)$, $f_{{-1}}(\{x\})$ does not belong to $U$ since it is either or $\{x\}$ and so $D$ is nonprincipal.

Let $g:P_{\kappa}(\kappa)\rightarrow\kappa$ and $X\in D$ such that $\forall x\in X,g(x)\in x$. Then $g\circ f:\kappa\rightarrow\kappa$ is regressive on $f_{{-1}}(X)\in U$ since $g(f(y))\in f(y)=y$. Because $U$ is normal, there is $Y\in U$ such that $g\circ f$ is constant on $Y$ and so $g$ is constant on $f(Y)\in D$. Finally, $D$ is normal.