$f$ is finite a.e if it is locally integrable?












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I know that when a function $f$ is integrable then it is finite almost every where. I was wondering to know if it is true that whenever $f$ is locally integrable then $f$ is finite almost everywhere. It seems to be true because the book Real Analysis by Stein ,page 107, uses this fact without any proof.






real-analysis lebesgue-integral lebesgue-measure almost-everywhere







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    1












    $begingroup$


    I know that when a function $f$ is integrable then it is finite almost every where. I was wondering to know if it is true that whenever $f$ is locally integrable then $f$ is finite almost everywhere. It seems to be true because the book Real Analysis by Stein ,page 107, uses this fact without any proof.






    real-analysis lebesgue-integral lebesgue-measure almost-everywhere







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I know that when a function $f$ is integrable then it is finite almost every where. I was wondering to know if it is true that whenever $f$ is locally integrable then $f$ is finite almost everywhere. It seems to be true because the book Real Analysis by Stein ,page 107, uses this fact without any proof.






      real-analysis lebesgue-integral lebesgue-measure almost-everywhere







      share|cite|improve this question









      $endgroup$




      I know that when a function $f$ is integrable then it is finite almost every where. I was wondering to know if it is true that whenever $f$ is locally integrable then $f$ is finite almost everywhere. It seems to be true because the book Real Analysis by Stein ,page 107, uses this fact without any proof.






      real-analysis lebesgue-integral lebesgue-measure almost-everywhere




      real-analysis lebesgue-integral lebesgue-measure almost-everywhere






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      asked Dec 31 '18 at 5:51









      S_AlexS_Alex

      1159




      1159






















          2 Answers
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          $begingroup$

          Assume that $f$ is not finite a.e., so there is a subset $B$ of the domain with positive measure on which $f$ is infinity. Then assuming inner regularity of the measure, we can find a compact subset $K$ of $B$ of positive measure, and $int_K f mathrm dmu$ would be infinite, contradicting local integrability.






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            You have not specified the domain. If you are talking about a locally intergable function on $mathbb R^{n}$ w.r.t. Lebesgue measure then $f$ is integrable on ${x:|x| leq k}$ for any positive integer $k$. This implies that $f$ is finite almost everywhere on this ball. Since $mathbb R^{n}$ is the union of these balls and since a countable union of sets of measure $0$ has measure $0$ it follows that $f$ is finite almost everywhere.






            share|cite|improve this answer









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            • $begingroup$
              So you mean for some domains the result does not hold?
              $endgroup$
              – S_Alex
              Dec 31 '18 at 6:13










            • $begingroup$
              I don't have Stein's book, so I don't know the context. If you have a regular Borel measure then every locally integrable fucntion is finite almost everywhere. If you have any Borel measure on a sigma compact space then the result is true again.
              $endgroup$
              – Kavi Rama Murthy
              Dec 31 '18 at 6:24











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            2 Answers
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            2 Answers
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            active

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            2












            $begingroup$

            Assume that $f$ is not finite a.e., so there is a subset $B$ of the domain with positive measure on which $f$ is infinity. Then assuming inner regularity of the measure, we can find a compact subset $K$ of $B$ of positive measure, and $int_K f mathrm dmu$ would be infinite, contradicting local integrability.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Assume that $f$ is not finite a.e., so there is a subset $B$ of the domain with positive measure on which $f$ is infinity. Then assuming inner regularity of the measure, we can find a compact subset $K$ of $B$ of positive measure, and $int_K f mathrm dmu$ would be infinite, contradicting local integrability.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Assume that $f$ is not finite a.e., so there is a subset $B$ of the domain with positive measure on which $f$ is infinity. Then assuming inner regularity of the measure, we can find a compact subset $K$ of $B$ of positive measure, and $int_K f mathrm dmu$ would be infinite, contradicting local integrability.






                share|cite|improve this answer









                $endgroup$



                Assume that $f$ is not finite a.e., so there is a subset $B$ of the domain with positive measure on which $f$ is infinity. Then assuming inner regularity of the measure, we can find a compact subset $K$ of $B$ of positive measure, and $int_K f mathrm dmu$ would be infinite, contradicting local integrability.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 31 '18 at 6:03









                Kenny LauKenny Lau

                19.9k2159




                19.9k2159























                    1












                    $begingroup$

                    You have not specified the domain. If you are talking about a locally intergable function on $mathbb R^{n}$ w.r.t. Lebesgue measure then $f$ is integrable on ${x:|x| leq k}$ for any positive integer $k$. This implies that $f$ is finite almost everywhere on this ball. Since $mathbb R^{n}$ is the union of these balls and since a countable union of sets of measure $0$ has measure $0$ it follows that $f$ is finite almost everywhere.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      So you mean for some domains the result does not hold?
                      $endgroup$
                      – S_Alex
                      Dec 31 '18 at 6:13










                    • $begingroup$
                      I don't have Stein's book, so I don't know the context. If you have a regular Borel measure then every locally integrable fucntion is finite almost everywhere. If you have any Borel measure on a sigma compact space then the result is true again.
                      $endgroup$
                      – Kavi Rama Murthy
                      Dec 31 '18 at 6:24
















                    1












                    $begingroup$

                    You have not specified the domain. If you are talking about a locally intergable function on $mathbb R^{n}$ w.r.t. Lebesgue measure then $f$ is integrable on ${x:|x| leq k}$ for any positive integer $k$. This implies that $f$ is finite almost everywhere on this ball. Since $mathbb R^{n}$ is the union of these balls and since a countable union of sets of measure $0$ has measure $0$ it follows that $f$ is finite almost everywhere.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      So you mean for some domains the result does not hold?
                      $endgroup$
                      – S_Alex
                      Dec 31 '18 at 6:13










                    • $begingroup$
                      I don't have Stein's book, so I don't know the context. If you have a regular Borel measure then every locally integrable fucntion is finite almost everywhere. If you have any Borel measure on a sigma compact space then the result is true again.
                      $endgroup$
                      – Kavi Rama Murthy
                      Dec 31 '18 at 6:24














                    1












                    1








                    1





                    $begingroup$

                    You have not specified the domain. If you are talking about a locally intergable function on $mathbb R^{n}$ w.r.t. Lebesgue measure then $f$ is integrable on ${x:|x| leq k}$ for any positive integer $k$. This implies that $f$ is finite almost everywhere on this ball. Since $mathbb R^{n}$ is the union of these balls and since a countable union of sets of measure $0$ has measure $0$ it follows that $f$ is finite almost everywhere.






                    share|cite|improve this answer









                    $endgroup$



                    You have not specified the domain. If you are talking about a locally intergable function on $mathbb R^{n}$ w.r.t. Lebesgue measure then $f$ is integrable on ${x:|x| leq k}$ for any positive integer $k$. This implies that $f$ is finite almost everywhere on this ball. Since $mathbb R^{n}$ is the union of these balls and since a countable union of sets of measure $0$ has measure $0$ it follows that $f$ is finite almost everywhere.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 31 '18 at 6:01









                    Kavi Rama MurthyKavi Rama Murthy

                    53.9k32055




                    53.9k32055












                    • $begingroup$
                      So you mean for some domains the result does not hold?
                      $endgroup$
                      – S_Alex
                      Dec 31 '18 at 6:13










                    • $begingroup$
                      I don't have Stein's book, so I don't know the context. If you have a regular Borel measure then every locally integrable fucntion is finite almost everywhere. If you have any Borel measure on a sigma compact space then the result is true again.
                      $endgroup$
                      – Kavi Rama Murthy
                      Dec 31 '18 at 6:24


















                    • $begingroup$
                      So you mean for some domains the result does not hold?
                      $endgroup$
                      – S_Alex
                      Dec 31 '18 at 6:13










                    • $begingroup$
                      I don't have Stein's book, so I don't know the context. If you have a regular Borel measure then every locally integrable fucntion is finite almost everywhere. If you have any Borel measure on a sigma compact space then the result is true again.
                      $endgroup$
                      – Kavi Rama Murthy
                      Dec 31 '18 at 6:24
















                    $begingroup$
                    So you mean for some domains the result does not hold?
                    $endgroup$
                    – S_Alex
                    Dec 31 '18 at 6:13




                    $begingroup$
                    So you mean for some domains the result does not hold?
                    $endgroup$
                    – S_Alex
                    Dec 31 '18 at 6:13












                    $begingroup$
                    I don't have Stein's book, so I don't know the context. If you have a regular Borel measure then every locally integrable fucntion is finite almost everywhere. If you have any Borel measure on a sigma compact space then the result is true again.
                    $endgroup$
                    – Kavi Rama Murthy
                    Dec 31 '18 at 6:24




                    $begingroup$
                    I don't have Stein's book, so I don't know the context. If you have a regular Borel measure then every locally integrable fucntion is finite almost everywhere. If you have any Borel measure on a sigma compact space then the result is true again.
                    $endgroup$
                    – Kavi Rama Murthy
                    Dec 31 '18 at 6:24


















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