推 hectorhsu:原來是這樣做＠＠ 感恩 12/06 23:06
※ 引述《hectorhsu (The Hector)》之銘言：
: Suppose that
: [1] ν1 is abolutely contionuous with respect to μ1（ν1<<μ1）
: [2] ν2 is abolutely contionuous with respect to μ2（ν2<<μ2）
: 證明 (ν1xν2) << (μ1xμ2)
: --------
: 應該是卡在不知道如何表示 (μ1xμ2)(E) = 0，來做 (ν1xν2)(E)？
Assume (X,M1,μ1), (X,N1,ν1) and (Y,M2,μ2), (Y,N2,ν2)
E \in M1 x N1, (μ1xμ2)(E) = 0
Let Ex = {y \in Y | (x,y) \in E}, x \in X
(Ey = {x \in X | (x,y) \in E}, y \in Y)
(μ1xμ2)(E) = 0 implies
μ1{x|μ2(Ex) > 0} = 0
(μ2{y|μ1(Ey) > 0} = 0)
ν1<<μ1, ν2<<μ2 imply
ν1{x|μ2(Ex) > 0} = 0
(ν2{y|μ1(Ey) > 0} = 0)
which also imply
ν1{x|ν2(Ex) > 0} = 0
(ν2{y|ν1(Ey) > 0} = 0)
since {x|ν2(Ex) > 0} is contined in {x|μ2(Ex) > 0}
(and {y|ν1(Ey) > 0} is contined in {y|μ1(Ey) > 0})
Hence (ν1xν2)(E)= 0
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