Suppose E [ | X | ] &lt; ∞ and that A n are disjoint sets with ∪ n = 1 ∞ A n = A. Show that ∑ n = 0 ∞ E [ X 1 A n ] = E [ X 1 A ]Here is what I try with Monotone Convergence Theorem: ∑ n = 0 ∞ E [ X 1 A n ] = lim m → ∞ ∑ n = 0 m E [ X 1 A n ] = lim m → ∞ E [ X ∑ n = 0 m 1 A n ] = lim m → ∞ E [ X 1 F n ] = E [ X lim m → ∞ 1 F n ] = E [ X 1 A ] where F n = ∪ i = 1 n A i , so ∪ n = 1 m A n = ∪ n = 1 m F n , and 1 F n is increasing.I am not quite sure if I get it right. My main concern is the first step: ∑ n = 0 ∞ E [ X 1 A n ] = lim m → ∞ ∑ n = 0 m E [ X 1 A n ]. Do we always have ∑ n = 0 ∞ = lim m → ∞ ∑ n = 0 m ?I may also have made other mistakes. Please feel free to point out. Any other hints are also welcome.

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Suppose   E  [      |    X      |    ]  &amp;lt;  ∞ and that       A    n   are disjoint sets with       ∪          n      =      1        ∞        A    n    =  A. Show that       ∑          n      =      0              ∞        E  [  X            1                      A        n              ]  =  E  [  X            1        A    ]Here is what I try with Monotone Convergence Theorem:                              ∑                      n            =            0                                ∞                          E        [        X                              1                                              A              n                                      ]                            =                  lim                      m            →            ∞                                    ∑                      n            =            0                                m                          E        [        X                              1                                              A              n                                      ]                                          =                  lim                      m            →            ∞                          E        [        X                  ∑                      n            =            0                                m                                                1                                              A              n                                      ]                                          =                  lim                      m            →            ∞                          E        [        X                              1                                              F              n                                      ]                                          =        E        [        X                  lim                      m            →            ∞                                                1                                              F              n                                      ]                                          =        E        [        X                              1                                A                          ]            where       F    n    =      ∪          i      =      1        n        A    i  , so       ∪          n      =      1        m        A    n    =      ∪          n      =      1        m        F    n  , and             1                      F        n             is increasing.I am not quite sure if I get it right. My main concern is the first step:       ∑          n      =      0              ∞        E  [  X            1                      A        n              ]  =      lim          m      →      ∞            ∑          n      =      0              m        E  [  X            1                      A        n              ]. Do we always have       ∑          n      =      0        ∞    =      lim          m      →      ∞            ∑          n      =      0        m  ?I may also have made other mistakes. Please feel free to point out. Any other hints are also welcome.

nuvolor8 · Accepted Answer

As explained in comments and Snoop&#039;s answer, the easiest way is to use DCT.Back to your question, the first step (your main concern) is OK, but where it fails is here:      lim          m      →      ∞        E  [  X            1                      F        n              ]  =  E  [  X      lim          m      →      ∞                  1                      F        n              ]The problem is that X might be both positive and negative, thus   {  X            1                      F        n              } is not monotone.To use MCT, you need to work around it, and decompose X in its positive and negative parts (  X  =      X          +        −      X          −      , with       X          +        ,      X          −        ≥  0), and apply MCT on       X          +       and       X          −       separately.

seupeljewj · Accepted Answer

We use DCT, not MCT:                              ∑                      n            ∈                          N                                      E        [                              1                                              A              n                                      X        ]                            =                  lim                      n            →            ∞                          E                              [                                                (                                    ∑                      k            ≤            n                                                1                                              A              k                                                            )                          X                              ]                          =                                          =                  lim                      n            →            ∞                          E        [                              1                                              ∪                              k                ≤                n                                                    A              k                                      X        ]        =                                                                              =                                                      DCT                                                    E        [                              1                                A                          X        ]            This is because             1                      ∪                  k          ≤          n                            A        k                  |    X      |    ≤      |    X      |    ∈            L        1    ,    ∀  n and             1                      ∪                  k          ≤          n                            A        k              →            1        A   pointwise.

Suppose E [ <mrow class="MJX-TeXAtom-ORD"> | </mrow> X <mrow class="MJX-TeXA

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