Sequence of functions who are bounded from below, change of limit and integral












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I have a question about changing limit and integral. I know of the monotone convergence theorem, so if my sequence is greater zero and increasing, I can change integral and limit. My question now is, what if my sequence is only bounded from below, by a possible negative number, and increasing. Can I change integral and limit is this case? I am pretty sure I can't. Does something change if I have a probability space? Thanks in advance!










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    0












    $begingroup$


    I have a question about changing limit and integral. I know of the monotone convergence theorem, so if my sequence is greater zero and increasing, I can change integral and limit. My question now is, what if my sequence is only bounded from below, by a possible negative number, and increasing. Can I change integral and limit is this case? I am pretty sure I can't. Does something change if I have a probability space? Thanks in advance!










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I have a question about changing limit and integral. I know of the monotone convergence theorem, so if my sequence is greater zero and increasing, I can change integral and limit. My question now is, what if my sequence is only bounded from below, by a possible negative number, and increasing. Can I change integral and limit is this case? I am pretty sure I can't. Does something change if I have a probability space? Thanks in advance!










      share|cite|improve this question









      $endgroup$




      I have a question about changing limit and integral. I know of the monotone convergence theorem, so if my sequence is greater zero and increasing, I can change integral and limit. My question now is, what if my sequence is only bounded from below, by a possible negative number, and increasing. Can I change integral and limit is this case? I am pretty sure I can't. Does something change if I have a probability space? Thanks in advance!







      real-analysis probability measure-theory






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      asked Dec 14 '18 at 11:38









      Testname420Testname420

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          $-I_{(n, infty)}$ is a counter example for the general case. In the case of finite measure $f_n$ increasing and $geq -a$ implies $f_n+a $ is non-negative and increasing so $lim int (f_n+a) to int (f+a)$ and we can subtract $a mu (X)$ from both sides to get $lim int f_n to int f$






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          • $begingroup$
            Alright makes a lot of sense thank you
            $endgroup$
            – Testname420
            Dec 14 '18 at 11:53











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          $begingroup$

          $-I_{(n, infty)}$ is a counter example for the general case. In the case of finite measure $f_n$ increasing and $geq -a$ implies $f_n+a $ is non-negative and increasing so $lim int (f_n+a) to int (f+a)$ and we can subtract $a mu (X)$ from both sides to get $lim int f_n to int f$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Alright makes a lot of sense thank you
            $endgroup$
            – Testname420
            Dec 14 '18 at 11:53
















          0












          $begingroup$

          $-I_{(n, infty)}$ is a counter example for the general case. In the case of finite measure $f_n$ increasing and $geq -a$ implies $f_n+a $ is non-negative and increasing so $lim int (f_n+a) to int (f+a)$ and we can subtract $a mu (X)$ from both sides to get $lim int f_n to int f$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Alright makes a lot of sense thank you
            $endgroup$
            – Testname420
            Dec 14 '18 at 11:53














          0












          0








          0





          $begingroup$

          $-I_{(n, infty)}$ is a counter example for the general case. In the case of finite measure $f_n$ increasing and $geq -a$ implies $f_n+a $ is non-negative and increasing so $lim int (f_n+a) to int (f+a)$ and we can subtract $a mu (X)$ from both sides to get $lim int f_n to int f$






          share|cite|improve this answer









          $endgroup$



          $-I_{(n, infty)}$ is a counter example for the general case. In the case of finite measure $f_n$ increasing and $geq -a$ implies $f_n+a $ is non-negative and increasing so $lim int (f_n+a) to int (f+a)$ and we can subtract $a mu (X)$ from both sides to get $lim int f_n to int f$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 14 '18 at 11:43









          Kavi Rama MurthyKavi Rama Murthy

          59.4k42161




          59.4k42161












          • $begingroup$
            Alright makes a lot of sense thank you
            $endgroup$
            – Testname420
            Dec 14 '18 at 11:53


















          • $begingroup$
            Alright makes a lot of sense thank you
            $endgroup$
            – Testname420
            Dec 14 '18 at 11:53
















          $begingroup$
          Alright makes a lot of sense thank you
          $endgroup$
          – Testname420
          Dec 14 '18 at 11:53




          $begingroup$
          Alright makes a lot of sense thank you
          $endgroup$
          – Testname420
          Dec 14 '18 at 11:53


















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