Set theory - Chapter 10: Measurable Cardinals


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

We have {0,2/2}[0,1]\emptyset\subsetneq\{0,\sqrt{2}/2\}\subseteq[0,1] and 2/2-0\sqrt{2}/2-0\notin\mathbb{Q} so MM\neq\emptyset and we can pick yMy\in M. Let x[0,1]x\in[0,1]. If x-yx-y\notin\mathbb{Q} then by maximality of MM, xM=M0x\in M=M_{0}. Otherwise, |x-y|1|x-y|\leq 1 and so x=y+qx=y+q for some q[-1,1]q\in[-1,1]\cap\mathbb{Q} that is xMqx\in M_{q}. Moreover, M[0,1]M\subseteq[0,1] and so if q[-1,1]q\in[-1,1]\cap\mathbb{Q}, we have Mq[-1,2]M_{q}\subseteq[-1,2] and finally


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


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


Let ν\nu be the measure defined in the proof of Lemma 10.9 (i). Suppose ZZ is an atom of ν\nu and let Y=f-1(Z)Y=f_{{-1}}(Z). We will prove by induction on α<ω1\alpha<\omega_{1} that there is a unique element XαX_{\alpha} at level α\alpha such that μ(YXα)0\mu(Y\cap X_{\alpha})\neq 0 and actually μ(YXα)=ν(Z)\mu(Y\cap X_{\alpha})=\nu(Z).

For α=0\alpha=0, we have 0ν(Z)=μ(f-1(Z))=μ(Y)=μ(YS)0\neq\nu(Z)=\mu(f_{{-1}}(Z))=\mu(Y)=\mu(Y\cap S). Suppose that the result is true at level α\alpha. The level α+1\alpha+1 is made of elements TT which are immediate successors of some XXαX\neq X_{\alpha} at level α\alpha and of two immediate successors UU and XαUX_{\alpha}\setminus U of XαX_{\alpha}. We have μ(YT)μ(YX)=0\mu(Y\cap T)\leq\mu(Y\cap X)=0 and 0μ(Y)=μ(YU)+μ(Y(XαU))0\neq\mu(Y)=\mu(Y\cap U)+\mu(Y\cap(X_{\alpha}\setminus U)). In particular we have μ(YU)0\mu(Y\cap U)\neq 0 or μ(Y(XαU))0\mu(Y\cap(X_{\alpha}\setminus U))\neq 0. Then μ(YU)μ(f-1(Z)f-1(f(U)))=μ(f-1(Zf(U)))=ν(Zf(U))\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, μ(Y(XαU))ν(Zf(XαU))\mu(Y\cap(X_{\alpha}\setminus U))\leq\nu(Z\cap f(X_{{\alpha}}\setminus U)). But ν(Z)=ν(Zf(Xα))=ν(Zf(U(XαU)))=ν(Zf(U))+ν(Zf(XαU))\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 ZZ is an atom, ν(Zf(U))=0\nu(Z\cap f(U))=0 or ν(Zf(XαU))=0\nu(Z\cap f(X_{{\alpha}}\setminus U))=0. Finally, we have {μ(YU),μ(Y(XαU))}={0,μ(Z)}\{\mu(Y\cap U),\mu(Y\cap(X_{\alpha}\setminus U))\}=\{0,\mu(Z)\} and the hypothesis is true at level α+1\alpha+1.

If α\alpha is limit and the result is true for all ξ<α<ω1\xi<\alpha<\omega_{1} Each level is at most countable and μ((SXξ)Y)=μ(o(X)=ξ;XXξXY)=o(X)=ξ;XXξμ(XY)=0\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 μ((Sξ<αXξ)Y)=μ(ξ<α(SXξ)Y)=ξ<αμ((SXξ)Y)=0\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,

μ((ξ<αXξ)Y)=μ(Y)-μ((Sξ<αXξ)Y)=μ(Y)=ν(Z)0\mu\left(\left(\bigcap_{{\xi<\alpha}}X_{\xi}\right)\cap Y\right)=\mu(Y)-\mu% \left({\left(S\setminus\bigcap_{{\xi<\alpha}}X_{\xi}\right)}\cap Y\right)=\mu(% Y)=\nu(Z)\neq 0

In particular, Xα=ξ<αXξX_{\alpha}=\bigcap_{{\xi<\alpha}}X_{\xi} is in the tree TT and o(Xα)=αo(X_{\alpha})=\alpha. If Tα=ξ<αTξT_{\alpha}=\bigcap_{{\xi<\alpha}}T_{\xi} is another element at level α\alpha such that ξ,o(Tξ)=ξ\forall\xi,o(T_{\xi})=\xi then there is ξ<α\xi<\alpha such that XξTαX_{\xi}\nsubseteq T_{\alpha} or TξXαT_{\xi}\nsubseteq X_{\alpha}. Hence TξXξT_{\xi}\neq X_{\xi} and by induction hypothesis μ(TαY)μ(TξY)=0\mu(T_{\alpha}\cap Y)\leq\mu(T_{\xi}\cap Y)=0. Hence the hypothesis is true at level α\alpha.

We have obtained a linarly ordered set {Xα:α<ω1}\{X_{\alpha}:\alpha<\omega_{1}\} with exactly one element at each level: that’s a branch in TT of length ω1\omega_{1}, a contradiction.


Let μ\mu be an atomless measure on SS and Z0ZZ_{0}\subseteq Z. We define by induction on α<ω1\alpha<\omega_{1} some sets S0S1S2SαS_{0}\supseteq S_{1}\supseteq S_{2}\supseteq...\supseteq S_{\alpha}\supseteq... such that μ(Sα)12μ(Z0)\mu(S_{\alpha})\geq\frac{1}{2}\mu(Z_{0}) and if μ(Sα)>12μ(Z0)\mu(S_{\alpha})>\frac{1}{2}\mu(Z_{0}) then μ(Sα+1)<μ(Sα)\mu(S_{{\alpha+1}})<\mu(S_{\alpha}). We let S0=Z0S_{0}=Z_{0}.

Suppose that SαS_{\alpha} is defined. If Sα=12μ(Z0)S_{\alpha}=\frac{1}{2}\mu(Z_{0}), we just take Sα+1=SαS_{{\alpha+1}}=S_{\alpha}. Otherwise, μ(Sα)>12μ(Z0)0\mu(S_{\alpha})>\frac{1}{2}\mu(Z_{0})\geq 0 we shall construct by induction on n1n\geq 1 some sets SαnSαS_{\alpha}^{n}\subseteq S_{\alpha} such that


For nn large enough, we get 12μ(Z0)<(1-2-n)μ(Sα)μ(Sαn)<μ(Sα)\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 Sα+1=SαnS_{{\alpha+1}}=S_{\alpha}^{n}.

Thus let’s construct SαnS_{\alpha}^{n}. μ(Sα)>0\mu(S_{\alpha})>0 and μ\mu is atomless, so SαS_{\alpha} can be split into two disjoint set of positive measure. If Sα1S_{\alpha}^{1} is the one of measure at least 12μ(Sα)\frac{1}{2}\mu(S_{\alpha}), then (1-2-1)μ(Sα)μ(Sα1)<μ(Sα)(1-2^{{-1}})\mu(S_{\alpha})\leq\mu(S_{\alpha}^{1})<\mu(S_{\alpha}). Similarly, if the induction hypothesis is true then SαSαnS_{\alpha}\setminus S_{\alpha}^{n} is of positive measure and so can be split into two disjoint set of positive measure. Choose XX the one of measure at least 12μ(SαSαn)\frac{1}{2}\mu(S_{\alpha}\setminus S_{\alpha}^{n}) and let Sαn+1=SαnXS_{\alpha}^{{n+1}}=S_{\alpha}^{n}\cup X. Then μ(Sα)>μ(Sαn+1)=μ(Sαn)+μ(X)μ(Sαn)+12(μ(Sα)-μ(Sαn))=12(μ(Sαn)+μ(Sα))12(1-2-n+1)μ(Sα)=(1-2-(n+1))μ(Sα)\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 α\alpha is limit and that SβS_{\beta} is defined for all β<α\beta<\alpha. If Sα=β<αSβS_{\alpha}=\bigcap_{{\beta<\alpha}}S_{\beta} then clearly SβSαS_{\beta}\supseteq S_{\alpha} for all β<α\beta<\alpha. Let βn\beta_{n} a countable increasing sequence which is cofinal in α\alpha with β0=0\beta_{0}=0. Clearly, Sαn<ωSβnS_{\alpha}\subseteq\bigcap_{{n<\omega}}S_{{\beta_{n}}}. Conversely, let xn<ωSβnx\in\bigcap_{{n<\omega}}S_{{\beta_{n}}}. For all β<α\beta<\alpha, there is n<ωn<\omega such that β<βn\beta<\beta_{n} and so xSβnSβx\in S_{{\beta_{n}}}\subseteq S_{\beta}. Hence Sα=n<ωSβnS_{\alpha}=\bigcap_{{n<\omega}}S_{{\beta_{n}}}. We have μ(Sα)=μ(Z0)-μ(n<ω(S0Sβn))=μ(Z0)-n=0+μ(Tn)\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 Tn=(S0Sβn+1)(S0Sβn)T_{n}=(S_{0}\setminus S_{{\beta_{{n+1}}}})\setminus(S_{0}\setminus S_{{\beta_{% n}}}). For all N<ωN<\omega, we have n=0Nμ(Tn)=μ(S0SβN)=μ(Z0)-μ(SβN)12μ(Z0)\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 n=0+μ(Tn)12μ(Z0)\sum_{{n=0}}^{{+\infty}}\mu(T_{n})\leq\frac{1}{2}\mu(Z_{0}) and so μ(Sα)12μ(Z0)\mu(S_{\alpha})\geq\frac{1}{2}\mu(Z_{0}).

Finally, suppose that for all α<ω1\alpha<\omega_{1}, we have μ(Sα)>12μ(Z0)\mu(S_{\alpha})>\frac{1}{2}\mu(Z_{0}). Then μ(Sα+1)<μ(Sα)\mu(S_{{\alpha+1}})<\mu(S_{{\alpha}}) and we can pick xα(μ(Sα+1),μ(Sα))x_{\alpha}\in\mathbb{Q}\cap(\mu(S_{{\alpha+1}}),\mu(S_{{\alpha}})). We get a one-to-one mapping αxα\alpha\mapsto x_{\alpha} from ω1\omega_{1} to \mathbb{Q}. A contradiction. Hence there is some α<ω1\alpha<\omega_{1} such that μ(Sα)=12μ(Z0)\mu(S_{\alpha})=\frac{1}{2}\mu(Z_{0})


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


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

Conversely, suppose that UU is a normal measure. Let γ<κ\gamma<\kappa. |κ{α<κ:d(α)>γ}|=|{α<κ:αγ}|=|γ|<κ\left|\kappa\setminus\{\alpha<\kappa:d(\alpha)>\gamma\}\right|=\left|\{\alpha<% \kappa:\alpha\leq\gamma\}\right|=|\gamma|<\kappa. Hence, since UU is κ\kappa-complete and UU is an ultrafilter, {α<κ:d(α)>γ}U\{\alpha<\kappa:d(\alpha)>\gamma\}\in U. Suppose that there is f<df<d with the property that for all γ<κ\gamma<\kappa, {α:f(α)>γ}U\{\alpha:f(\alpha)>\gamma\}\in U. Then for all γ<κ\gamma<\kappa, Xγ={α<κ:γ<f(α)<d(α)=α}UX_{\gamma}=\{\alpha<\kappa:\gamma<f(\alpha)<d(\alpha)=\alpha\}\in U. Because UU is normal, Δγ<κXγU\Delta_{{\gamma<\kappa}}X_{\gamma}\in U. But if αΔγ<κXγ\alpha\in\Delta_{{\gamma<\kappa}}X_{\gamma} then for all γ<α\gamma<\alpha, αXγ\alpha\in X_{\gamma} that is γ<f(α)<α\gamma<f(\alpha)<\alpha. We get a contradiction for γ=f(α)\gamma=f(\alpha). So dd is the least function with the property mentioned above.


Let DD be a normal measure on κ\kappa and let f:[κ]<ωκf:{[\kappa]}^{{<\omega}}\rightarrow\kappa such that for all x[κ]<ωx\in{[\kappa]}^{{<\omega}} f(x)=0f(x)=0 or f(x)<minxf(x)<\min x. We show by induction on nn that for any g:[κ]nκg:{[\kappa]}^{n}\rightarrow\kappa with the previous property, there is there is HDH\in D such that gg is constant on [H]n{[H]}^{n}. Then given ff as above, we consider HnDH_{n}\in D homogeneous for f|[κ]nf_{{|{[\kappa]}^{n}}}. Then H=n<ωHnDH=\bigcup_{{n<\omega}}H_{n}\in D and ff is constant on [H]<ω{[H]}^{{<\omega}}.

We start with n=1n=1. Let h:κ[κ]1h:\kappa\rightarrow{[\kappa]}^{1} be the canonical injection. Then for all 0<α<κ0<\alpha<\kappa, g(h(α))=0<αg(h(\alpha))=0<\alpha or g(h(α))<minh(α)=min{α}=αg(h(\alpha))<\min h(\alpha)=\min{\{\alpha\}}=\alpha. Hence ghg\circ h is regressive on κ{0}D\kappa\setminus\{0\}\in D and because DD is normal there is HDH\in D such that ghg\circ h is constant on HH. Hence gg is constant on [H]1{[H]}^{1}.

Suppose that the result is true for some n<ωn<\omega. Let α<κ\alpha<\kappa and define Define gα:[κ]nκg_{{\alpha}}:{[\kappa]}^{n}\rightarrow\kappa by gα(x)=g({α}x)g_{{\alpha}}(x)=g(\{\alpha\}\cup x) if α<minx\alpha<\min x and gα(x)=0g_{{\alpha}}(x)=0 otherwise. We have gα(x)=0g_{{\alpha}}(x)=0 or gα(x)<min({α}x)=α<minxg_{{\alpha}}(x)<\min(\{\alpha\}\cup x)=\alpha<\min x. By induction hypothesis, there is XαDX_{\alpha}\in D such that gαg_{\alpha} is constant on [D]n{[D]}^{n} with value γα\gamma_{\alpha}. Let X=Δα<κXαX=\Delta_{{\alpha<\kappa}}X_{\alpha}. If α0<α1<<αn\alpha_{0}<\alpha_{1}<...<\alpha_{n} are elements of XX then α1,αnXα0\alpha_{1},...\alpha_{n}\in X_{{\alpha_{0}}} and g({α0,α1,,αn})=gα0(α1,,αn)=γα0g(\{\alpha_{0},\alpha_{1},...,\alpha_{n}\})=g_{{\alpha_{0}}}(\alpha_{1},...,% \alpha_{n})=\gamma_{{\alpha_{0}}}. For all α0X{0}D\alpha_{0}\in X\setminus\{0\}\in D, we have γα0=0\gamma_{{\alpha_{0}}}=0 or γα0<α0\gamma_{{\alpha_{0}}}<\alpha_{0} so there is HDH\in D and γ<κ\gamma<\kappa such that for all α0H\alpha_{0}\in H, γα0=γ\gamma_{{\alpha_{0}}}=\gamma. Finally, gg is constant on [H]n+1{[H]}^{{n+1}}.


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

Let g:Pκ(κ)κg:P_{\kappa}(\kappa)\rightarrow\kappa and XDX\in D such that xX,g(x)x\forall x\in X,g(x)\in x. Then gf:κκg\circ f:\kappa\rightarrow\kappa is regressive on f-1(X)Uf_{{-1}}(X)\in U since g(f(y))f(y)=yg(f(y))\in f(y)=y. Because UU is normal, there is YUY\in U such that gfg\circ f is constant on YY and so gg is constant on f(Y)Df(Y)\in D. Finally, DD is normal.

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