1 $bR^n$ 中集合 $E$ 称为可测的 (measurable), 如果 $$eelabel{3.2:Caratheodory} m^*T=m^*(Tcap E)+m^*(Tcap E^c),quad forall Tsubset bR^n. eee$$
(1) 所有可测集构成的集族记为 $scrM$.
(2) 这里的 $T$ 称为试验集 (test set).
(3) eqref{3.2:Caratheodory} 称为 Caratheodory 条件.
(4) 当 $Ein scrM$ 时, 记 $mE$ 为 $E$ 的测度: $mE=m^*E$.
2 $$ex Embox{ 可测}lra m^*I=m^*(Icap E)+m^*(Icap E^c),quad forall Isubset bR^n. eex$$
证明: $ a$ 显然.
$la$ 由外测度的次可数可加性知 $leq$ 成立, 往证 $geq$: $$eex ea m^*T+ve&>sum_{i=1}^infty |I_i|quadsex{Tsubset cup_{i=1}^infty I_i}\ &=sum_{i=1}^infty m^*I_i\ &=sum_{i=1}^infty sez{m^*(I_icap E) +m^*(I_icap E^c)}\ &geq m^*sez{cup_{i=1}^infty (I_icap E)} +m^*sez{cup_{i=1}^infty (I_icap E^c)}\ &geq m^*(Tcap E)+m^*(Tcap E^c)\ &quadsex{Tcap Esubset cup_{i=1}^infty (I_icap E),quad Tcap E^csubset cup_{i=1}^infty (I_icap E^c}. eea eeex$$
3 $$eelabel{3.2:measure_property_sepration} Embox{ 可测}lra m^*(Acup B)=m^*A+m^*B,quadforall Asubset E, Bsubset E^c. eee$$
证明: $ a$ 取试验集 $T=Acup B$ 即可.
$la$ $$ex m^*T=m^*sez{(Tcap E)cup (Tcap E^c)} =m^*(Tcap E)+m^*(Tcap E^c). eex$$
4 可测集的性质:
(1) $E$ 可测 $ a E^c$ 可测.
证明: $$ex m^*T=m^*(Tcap E)+m^*(Tcap E^c) =m^*(Tcap (E^c)^c)+m^*(Tcap E^c). eex$$
(2) $E_1,E_2$ 可测 $ a E_1cup E_2, E_1cap E_2$ 可测.
证明: 由 $$ex E_1cap E_2=sex{E_1^ccup E_2^c}^c eex$$
知仅须证明 $E_1cup E_2$ 可测: $$eex ea m^*T &=m^*(Tcap E_1)+m^*(Tcap E_1^c)\ &=m^*(Tcap E_1) +m^*(Tcap E_1^ccap E_2)+m^*(Tcap E_1^ccap E_2^c)\ &=m^*(Tcap (E_1cup E_2)) +m^*(Tcap(E_1cup E_2)^c)\ &quadsex{mbox{ 由 }eqref{3.2:measure_property_sepration} mbox{ 及 }E_1cup (E_1^ccap E_2)=E_1cup E_2}. eea eeex$$
(3) $sed{E_i}_{i=1}^n$ 可测 $dps{cup_{i=1}^n E_i, cap_{i=1}^n E_i}$ 可测.
证明: 利用性质 (2) 及数学归纳法.
(4) $sed{E_i}_{i=1}^infty$ 可测 $dps{ a cup_{i=1}^infty E_i}$ 可测; 且若 $E_i$ 两两不交, 则 $$eelabel{3.2:measure_property_countably_additivity} msex{cup_{i=1}^infty E_i}=sum_{i=1}^infty mE_i. eee$$
证明: 由 $$ex cup_{i=1}^infty E_i =E_1cup [E_2s E_1]cup [E_3s(E_1cup E_2)]cup cdots eex$$
知仅须验证当 $E_i$ 两两不交时, $dps{cup_{i=1}^infty E_i}$ 可测, 且 eqref{3.2:measure_property_countably_additivity} 成立: $$eex ea m^*T&=m^*sez{Tcap sex{cup_{i=1}^j E_i}} +m^*sez{Tcap sex{cup_{i=1}^j E_i}^c}\ &geq m^*sez{cup_{i=1}^j (Tcap E_i)} +m^*sez{Tcapsex{cup_{i=1}^infty E_i}^c}\ &=sum_{i=1}^j m^*(Tcap E_i) +m^*sez{Tcapsex{cup_{i=1}^infty E_i}^c}\ &quadsex{E_imbox{ 两两不交, 利用 }eqref{3.2:measure_property_sepration}mbox{ 及数学归纳法}}; eea eeex$$ $$eex ea m^*T&geq sum_{i=1}^infty m^*(Tcap E_i) +m^*sez{Tcapsex{cup_{i=1}^infty E_i}^c}\ &geq m^*sez{Tcap sex{cup_{i=1}^infty E_i}} +m^*sez{Tcapsex{cup_{i=1}^infty E_i}^c}. eea eeex$$
(5) $sed{E_i}_{i=1}^infty$ 可测 $dps{ a cap_{i=1}^infty E_i}$ 可测.
(6) $sed{E_i}$ 单增可测 $dps{ a msex{lim_{i oinfty}E_i}=lim_{i oinfty}mE_i}$.
证明: $$eex ea msex{lim_{i oinfty}E_i} &=msex{cup_{i=1}^infty E_i}\ &=msex{cup_{i=1}^infty F_i}quadsex{F_1=E_1,F_2=E_2s E_1,F_3=E_3s E_2,cdots}\ &=sum_{i=1}^infty m F_i\ &=lim_{j oinfty} sum_{i=1}^j m F_i\ &=lim_{j oinfty} sez{m E_1+sum_{i=2}^j (mE_i-mE_{i-1})}\ &=lim_{j oinfty} mE_j. eea eeex$$
(7) $sed{E_i}$ 单减可测, $mE_1<infty$ $dps{ a msex{lim_{i oinfty} E_i}=lim_{i oinfty}mE_i}$.
证明: $$eex ea msex{lim_{i oinfty}E_i} &=msex{cap_{i=1}^infty E_i}\ &=msez{E_1s sex{E_1s cap_{i=1}^infty E_i}}\ &=msez{E_1s cup_{i=1}^infty (E_1s E_i)}\ &=m E_1-msez{cup_{i=1}^infty (E_1s E_i)}\ &quadsex{mbox{由 }mE_1<inftymbox{ 及 }cup_{i=1}^infty (Es E_i)mbox{ 可测}}\&= mE_1-lim_{i oinfty}m (E_1s E_i)\ &=mE_1-lim_{i oinfty}(mE_1-mE_i)\ &=lim_{i oinfty}mE_i. eea eeex$$
5 作业: Page 75 T 6, T 7.