Let f be Lebesgue integrable on [0,b]. Then for the purposes of integration, we can assume that f is non-negative.(Sorry my χ's are coming out very small, not sure how to fix this latex issue).Then (1) ∫ R ∫ R f ( t ) t χ ( x , b ] ( t ) χ [ 0 , b ] ( x ) d t d x = ∫ R ( ∫ R f ( t ) t χ [ 0 , b ] ( x ) d x ) χ ( x , b ] ( t ) d t (2*) = ∫ 0 b ( ∫ 0 t f ( t ) t d x ) d t Where (1) follows from Fubini's thereom.But I don't understand the move to (2*) -- specifically why we get ∫ 0 t on the inside, rather than ∫ 0 b coming from the χ [ 0 , b ] , but then I'm also not making sense of how χ ( x , b ] affects the interval of integration on the outside.I know I'm missing something elementary here.

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Let   f be Lebesgue integrable on [0,b]. Then for the purposes of integration, we can assume that   f is non-negative.(Sorry my   χ&#039;s are coming out very small, not sure how to fix this latex issue).Then                    (1)                              ∫                                    R                                                ∫                                    R                                                            f            (            t            )                    t                                      χ                          (              x              ,              b              ]                                (          t          )                      χ                          [              0              ,              b              ]                                (          x          )          d          t                     d          x                                    =                  ∫                                    R                                                (                      ∫                                          R                                                                        f              (              t              )                        t                                              χ                              [                0                ,                b                ]                                      (            x            )            d            x                    )                                      χ                          (              x              ,              b              ]                                (          t          )          d          t                                    (2*)                                  =                  ∫          0          b                          (                      ∫            0            t                                              f              (              t              )                        t                    d          x          )                d        t            Where (1) follows from Fubini&#039;s thereom.But I don&#039;t understand the move to (2*) -- specifically why we get       ∫    0    t   on the inside, rather than       ∫    0    b   coming from the       χ          [      0      ,      b      ]      , but then I&#039;m also not making sense of how             χ              (        x        ,        b        ]             affects the interval of integration on the outside.I know I&#039;m missing something elementary here.

szilincsifs · Accepted Answer

The problem here is not in deriving (2) from (1). The problem is that (1) is wrong. They made a mistake when they wrote it out. x is a dummy variable of the inside integration, yet is referenced outside that integration, where it is undefined. The RH side of (1) does not make sense.What they are actually depending on is this                    χ                    (        x        ,        b        ]              (    t    )                            χ                          [          0          ,          b          ]                    (      x      )      =                                    χ                                [            0            ,            b            ]                          (        t        )                                            χ                                      [              0              ,              t              )                                (          x          )                    Which follows because both expressions are 1 if and only if   0  ≤  x  &amp;lt;  t  ≤  b. Otherwise they are both 0. What they should have written is                              ∫                                    R                                                ∫                                    R                                                            f            (            t            )                    t                                                    χ                                            (            x            ,            b            ]                          (        t        )                                            χ                                            [            0            ,            b            ]                          (        x        )                 d        t                 d        x                            =                  ∫                                    R                                                ∫                                    R                                                            f            (            t            )                    t                                                    χ                                            [            0            ,            b            ]                          (        t        )                                            χ                                            [            0            ,            t            )                          (        x        )                 d        t                 d        x                                          =                  ∫                                    R                                                (                      ∫                                          R                                                                        f              (              t              )                        t                                                              χ                                                    [              0              ,              t              )                                (          x          )                     d          x          )                                                    χ                                            [            0            ,            b            ]                          (        t        )                 d        t                                          =                  ∫          0          b                          (                      ∫            0            t                                              f              (              t              )                        t                               d          x          )                         d        t

Let f be Lebesgue integrable on [0,b]. Then for the purposes of integration, we can assume tha

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