why is the Lebesgue-Stieltjes integral well-defined?












3












$begingroup$


A function $g: [a,b] rightarrow mathbb{R}$ a said to be of bounded variation on the interval $[a,b]$ if
$$ sup_{P: a=x_0 < x_1 ldots < x_i < ldots < x_{n_P}=b} sum_{i=1}^{n_P} |g(x_{i}) -g(x_{i-1}))| < infty, $$
where the supremum if taken over all partitions $P$ of the interval $[a,b]$.



One can show that a (right-continous) function $g: [a,b] rightarrow mathbb{R}$ is of bounded variation if and only if there exist two monotone non-decreasing (right-continuous) functions $g^+$ and $g^-$ such that $g= g^+ -g^-$. (However, this decomposition is not unique, as one can for example add any monotone non-decreasing (right-continuous) function to both $g^+$ and $g^-$). This is called a Jordan decomposition of $g$.



Moreover, given any $g: [a,b] rightarrow mathbb{R}$ which is monotone, non-decreasing and right-continuous function, there exists a unique measure $dg$ on $[a,b]$ such that
$$ dg((c,d])=g(d)-g(c) $$
for all $(c,d] in [a,b]$ and $dg({a})=0$. This measure is called the Lebesgue-Stieltjes measure of $g$.



Given a bounded function $f: [a,b] rightarrow mathbb{R}$ and a right-continuous function of bounded variation $g : [a,b] rightarrow mathbb{R}$, one can define
$$ int_{[a,b]} f,dg := int_{[a,b]} f,dg^+ - int_{[a,b]} f,dg^-, $$
where $g= g^+ -g^-$ is a Jordan decomposition as above. This is called the Lebesgue-Stieltjes integral of $f$ w.r.t. $g$.



My question is, why is $int_{[a,b]} f,dg$ well-defined, i.e. independent of the chosen Jordan decomposition?










share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    A function $g: [a,b] rightarrow mathbb{R}$ a said to be of bounded variation on the interval $[a,b]$ if
    $$ sup_{P: a=x_0 < x_1 ldots < x_i < ldots < x_{n_P}=b} sum_{i=1}^{n_P} |g(x_{i}) -g(x_{i-1}))| < infty, $$
    where the supremum if taken over all partitions $P$ of the interval $[a,b]$.



    One can show that a (right-continous) function $g: [a,b] rightarrow mathbb{R}$ is of bounded variation if and only if there exist two monotone non-decreasing (right-continuous) functions $g^+$ and $g^-$ such that $g= g^+ -g^-$. (However, this decomposition is not unique, as one can for example add any monotone non-decreasing (right-continuous) function to both $g^+$ and $g^-$). This is called a Jordan decomposition of $g$.



    Moreover, given any $g: [a,b] rightarrow mathbb{R}$ which is monotone, non-decreasing and right-continuous function, there exists a unique measure $dg$ on $[a,b]$ such that
    $$ dg((c,d])=g(d)-g(c) $$
    for all $(c,d] in [a,b]$ and $dg({a})=0$. This measure is called the Lebesgue-Stieltjes measure of $g$.



    Given a bounded function $f: [a,b] rightarrow mathbb{R}$ and a right-continuous function of bounded variation $g : [a,b] rightarrow mathbb{R}$, one can define
    $$ int_{[a,b]} f,dg := int_{[a,b]} f,dg^+ - int_{[a,b]} f,dg^-, $$
    where $g= g^+ -g^-$ is a Jordan decomposition as above. This is called the Lebesgue-Stieltjes integral of $f$ w.r.t. $g$.



    My question is, why is $int_{[a,b]} f,dg$ well-defined, i.e. independent of the chosen Jordan decomposition?










    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      A function $g: [a,b] rightarrow mathbb{R}$ a said to be of bounded variation on the interval $[a,b]$ if
      $$ sup_{P: a=x_0 < x_1 ldots < x_i < ldots < x_{n_P}=b} sum_{i=1}^{n_P} |g(x_{i}) -g(x_{i-1}))| < infty, $$
      where the supremum if taken over all partitions $P$ of the interval $[a,b]$.



      One can show that a (right-continous) function $g: [a,b] rightarrow mathbb{R}$ is of bounded variation if and only if there exist two monotone non-decreasing (right-continuous) functions $g^+$ and $g^-$ such that $g= g^+ -g^-$. (However, this decomposition is not unique, as one can for example add any monotone non-decreasing (right-continuous) function to both $g^+$ and $g^-$). This is called a Jordan decomposition of $g$.



      Moreover, given any $g: [a,b] rightarrow mathbb{R}$ which is monotone, non-decreasing and right-continuous function, there exists a unique measure $dg$ on $[a,b]$ such that
      $$ dg((c,d])=g(d)-g(c) $$
      for all $(c,d] in [a,b]$ and $dg({a})=0$. This measure is called the Lebesgue-Stieltjes measure of $g$.



      Given a bounded function $f: [a,b] rightarrow mathbb{R}$ and a right-continuous function of bounded variation $g : [a,b] rightarrow mathbb{R}$, one can define
      $$ int_{[a,b]} f,dg := int_{[a,b]} f,dg^+ - int_{[a,b]} f,dg^-, $$
      where $g= g^+ -g^-$ is a Jordan decomposition as above. This is called the Lebesgue-Stieltjes integral of $f$ w.r.t. $g$.



      My question is, why is $int_{[a,b]} f,dg$ well-defined, i.e. independent of the chosen Jordan decomposition?










      share|cite|improve this question











      $endgroup$




      A function $g: [a,b] rightarrow mathbb{R}$ a said to be of bounded variation on the interval $[a,b]$ if
      $$ sup_{P: a=x_0 < x_1 ldots < x_i < ldots < x_{n_P}=b} sum_{i=1}^{n_P} |g(x_{i}) -g(x_{i-1}))| < infty, $$
      where the supremum if taken over all partitions $P$ of the interval $[a,b]$.



      One can show that a (right-continous) function $g: [a,b] rightarrow mathbb{R}$ is of bounded variation if and only if there exist two monotone non-decreasing (right-continuous) functions $g^+$ and $g^-$ such that $g= g^+ -g^-$. (However, this decomposition is not unique, as one can for example add any monotone non-decreasing (right-continuous) function to both $g^+$ and $g^-$). This is called a Jordan decomposition of $g$.



      Moreover, given any $g: [a,b] rightarrow mathbb{R}$ which is monotone, non-decreasing and right-continuous function, there exists a unique measure $dg$ on $[a,b]$ such that
      $$ dg((c,d])=g(d)-g(c) $$
      for all $(c,d] in [a,b]$ and $dg({a})=0$. This measure is called the Lebesgue-Stieltjes measure of $g$.



      Given a bounded function $f: [a,b] rightarrow mathbb{R}$ and a right-continuous function of bounded variation $g : [a,b] rightarrow mathbb{R}$, one can define
      $$ int_{[a,b]} f,dg := int_{[a,b]} f,dg^+ - int_{[a,b]} f,dg^-, $$
      where $g= g^+ -g^-$ is a Jordan decomposition as above. This is called the Lebesgue-Stieltjes integral of $f$ w.r.t. $g$.



      My question is, why is $int_{[a,b]} f,dg$ well-defined, i.e. independent of the chosen Jordan decomposition?







      measure-theory bounded-variation stieltjes-integral






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 7 at 21:15









      Bernard

      119k740113




      119k740113










      asked Jan 7 at 20:58









      MortenMorten

      605




      605






















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          $begingroup$

          Let $g:[a,b]rightarrowmathbb{R}$ be right-continuous and of bonded
          variation. Suppose that we have decompositions $g=g_{1}-g_{2}=g_{3}-g_{4}$
          for some right-continuous, increasing (non-strict sense) functions
          $g_{1},g_{2},g_{3},g_{4}$ defined on $[a,b]$. For each $i=1,2,3,4$,
          let $mu_{i}:mathcal{B}((a,b])rightarrowmathbb{R}$ be a Borel
          measure induced by $g_{i}$ such that for any $aleq c<dleq b$,
          $mu_{i}left((c,d]right)=g_{i}(d)-g_{i}(c)$. We go to show that
          $int fdmu_{1}-int fdmu_{2}=int fdmu_{3}-int fdmu_{4}$ for
          any bounded Borel function $f:(a,b]rightarrowmathbb{R}$. Define
          Borel measures $nu=mu_{1}+mu_{4}$ and $nu'=mu_{2}+mu_{3}$.
          Note that $nu$ and $nu'$ are finite measures. Let $mathcal{P}={(c,d]mid aleq c<dleq b}cup{emptyset}$
          and $mathcal{L}={Ainmathcal{B}left((a,b]right)midnu(A)=nu'(A)}$.
          Clearly $mathcal{P}$ is a $pi$-class (in the sense that $Acap Binmathcal{P}$
          whenever $A,Binmathcal{P}$) and $mathcal{L}$ is a $lambda$-class
          (in the sense that: $emptysetinmathcal{L}$; $A^{c}inmathcal{L}$
          whenever $Ainmathcal{L}$; $cup_{n=1}^{infty}A_{n}inmathcal{L}$
          whenever $A_{1},A_{2}ldotsinmathcal{L}$ are pairwisely disjoint.).
          It is routine to show that $mathcal{P}subseteqmathcal{L}$. For,
          let $(c,d]inmathcal{P}$, then
          begin{eqnarray*}
          mu_{1}(c,d]-mu_{2}(c,d] & = & left[g_{1}(d)-g_{1}(c)right]-left[g_{2}(d)-g_{2}(c)right]\
          & = & left[g_{1}(d)-g_{2}(d)right]-left[g_{1}(c)-g_{2}(c)right]\
          & = & g(d)-g(c)\
          & = & left[g_{3}(d)-g_{4}(d)right]-left[g_{3}(c)-g_{4}(c)right]\
          & = & left[g_{3}(d)-g_{3}(c)right]-left[g_{4}(d)-g_{4}(c)right]\
          & = & mu_{3}(c,d]-mu_{4}(c,d].
          end{eqnarray*}

          Re-arranging terms, we have $nu(c,d]=nu'(c,d]$. This shows that
          $mathcal{P}subseteqmathcal{L}$. By Dynkin's $pi-lambda$ theorem,
          we have $sigma(mathcal{P})subseteqmathcal{L}$. However, it is
          well known that $sigma(mathcal{P})=mathcal{B}((a,b])$, so we have
          $sigma(mathcal{P})=mathcal{L=mathcal{B}}(a,b])$.



          For a Borel function of the form $f=1_{A}$, where $Ainmathcal{B}((a,b])$,
          we have
          begin{eqnarray*}
          int fdmu_{1}+int fdmu_{4} & = & mu_{1}(A)+mu_{4}(A)\
          & = & nu(A)\
          & = & nu'(A)\
          & = & int fdmu_{2}+int fdmu_{3}.
          end{eqnarray*}

          By linearlity, the above holds for all simple functions $f$. If $f$
          is a non-negative, bounded Borel function, we may choose a sequence
          of simple functions $(f_{n})$ such that $0leq f_{1}leq f_{2}leqldotsleq f$
          and $f_{n}rightarrow f$ pointwisely. By monotone convergence theorem,
          the above inequlity holds for all non-negative, bounded Borel functions.
          Finally, if $f$ is a bounded Borel function, consider $f=f^{+}-f^{-}$
          and we can prove that the above also holds for bounded Borel function
          $f$. Finally, re-arrange terms, we have $int fdmu_{1}-int fdmu_{2}=int fdmu_{3}-int fdmu_{4}$
          or in the Stieltjes notation: $int fdg_{1}-int fdg_{2}=int fdg_{3}-int fdg_{4}$.






          share|cite|improve this answer











          $endgroup$













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            $begingroup$

            Let $g:[a,b]rightarrowmathbb{R}$ be right-continuous and of bonded
            variation. Suppose that we have decompositions $g=g_{1}-g_{2}=g_{3}-g_{4}$
            for some right-continuous, increasing (non-strict sense) functions
            $g_{1},g_{2},g_{3},g_{4}$ defined on $[a,b]$. For each $i=1,2,3,4$,
            let $mu_{i}:mathcal{B}((a,b])rightarrowmathbb{R}$ be a Borel
            measure induced by $g_{i}$ such that for any $aleq c<dleq b$,
            $mu_{i}left((c,d]right)=g_{i}(d)-g_{i}(c)$. We go to show that
            $int fdmu_{1}-int fdmu_{2}=int fdmu_{3}-int fdmu_{4}$ for
            any bounded Borel function $f:(a,b]rightarrowmathbb{R}$. Define
            Borel measures $nu=mu_{1}+mu_{4}$ and $nu'=mu_{2}+mu_{3}$.
            Note that $nu$ and $nu'$ are finite measures. Let $mathcal{P}={(c,d]mid aleq c<dleq b}cup{emptyset}$
            and $mathcal{L}={Ainmathcal{B}left((a,b]right)midnu(A)=nu'(A)}$.
            Clearly $mathcal{P}$ is a $pi$-class (in the sense that $Acap Binmathcal{P}$
            whenever $A,Binmathcal{P}$) and $mathcal{L}$ is a $lambda$-class
            (in the sense that: $emptysetinmathcal{L}$; $A^{c}inmathcal{L}$
            whenever $Ainmathcal{L}$; $cup_{n=1}^{infty}A_{n}inmathcal{L}$
            whenever $A_{1},A_{2}ldotsinmathcal{L}$ are pairwisely disjoint.).
            It is routine to show that $mathcal{P}subseteqmathcal{L}$. For,
            let $(c,d]inmathcal{P}$, then
            begin{eqnarray*}
            mu_{1}(c,d]-mu_{2}(c,d] & = & left[g_{1}(d)-g_{1}(c)right]-left[g_{2}(d)-g_{2}(c)right]\
            & = & left[g_{1}(d)-g_{2}(d)right]-left[g_{1}(c)-g_{2}(c)right]\
            & = & g(d)-g(c)\
            & = & left[g_{3}(d)-g_{4}(d)right]-left[g_{3}(c)-g_{4}(c)right]\
            & = & left[g_{3}(d)-g_{3}(c)right]-left[g_{4}(d)-g_{4}(c)right]\
            & = & mu_{3}(c,d]-mu_{4}(c,d].
            end{eqnarray*}

            Re-arranging terms, we have $nu(c,d]=nu'(c,d]$. This shows that
            $mathcal{P}subseteqmathcal{L}$. By Dynkin's $pi-lambda$ theorem,
            we have $sigma(mathcal{P})subseteqmathcal{L}$. However, it is
            well known that $sigma(mathcal{P})=mathcal{B}((a,b])$, so we have
            $sigma(mathcal{P})=mathcal{L=mathcal{B}}(a,b])$.



            For a Borel function of the form $f=1_{A}$, where $Ainmathcal{B}((a,b])$,
            we have
            begin{eqnarray*}
            int fdmu_{1}+int fdmu_{4} & = & mu_{1}(A)+mu_{4}(A)\
            & = & nu(A)\
            & = & nu'(A)\
            & = & int fdmu_{2}+int fdmu_{3}.
            end{eqnarray*}

            By linearlity, the above holds for all simple functions $f$. If $f$
            is a non-negative, bounded Borel function, we may choose a sequence
            of simple functions $(f_{n})$ such that $0leq f_{1}leq f_{2}leqldotsleq f$
            and $f_{n}rightarrow f$ pointwisely. By monotone convergence theorem,
            the above inequlity holds for all non-negative, bounded Borel functions.
            Finally, if $f$ is a bounded Borel function, consider $f=f^{+}-f^{-}$
            and we can prove that the above also holds for bounded Borel function
            $f$. Finally, re-arrange terms, we have $int fdmu_{1}-int fdmu_{2}=int fdmu_{3}-int fdmu_{4}$
            or in the Stieltjes notation: $int fdg_{1}-int fdg_{2}=int fdg_{3}-int fdg_{4}$.






            share|cite|improve this answer











            $endgroup$


















              1












              $begingroup$

              Let $g:[a,b]rightarrowmathbb{R}$ be right-continuous and of bonded
              variation. Suppose that we have decompositions $g=g_{1}-g_{2}=g_{3}-g_{4}$
              for some right-continuous, increasing (non-strict sense) functions
              $g_{1},g_{2},g_{3},g_{4}$ defined on $[a,b]$. For each $i=1,2,3,4$,
              let $mu_{i}:mathcal{B}((a,b])rightarrowmathbb{R}$ be a Borel
              measure induced by $g_{i}$ such that for any $aleq c<dleq b$,
              $mu_{i}left((c,d]right)=g_{i}(d)-g_{i}(c)$. We go to show that
              $int fdmu_{1}-int fdmu_{2}=int fdmu_{3}-int fdmu_{4}$ for
              any bounded Borel function $f:(a,b]rightarrowmathbb{R}$. Define
              Borel measures $nu=mu_{1}+mu_{4}$ and $nu'=mu_{2}+mu_{3}$.
              Note that $nu$ and $nu'$ are finite measures. Let $mathcal{P}={(c,d]mid aleq c<dleq b}cup{emptyset}$
              and $mathcal{L}={Ainmathcal{B}left((a,b]right)midnu(A)=nu'(A)}$.
              Clearly $mathcal{P}$ is a $pi$-class (in the sense that $Acap Binmathcal{P}$
              whenever $A,Binmathcal{P}$) and $mathcal{L}$ is a $lambda$-class
              (in the sense that: $emptysetinmathcal{L}$; $A^{c}inmathcal{L}$
              whenever $Ainmathcal{L}$; $cup_{n=1}^{infty}A_{n}inmathcal{L}$
              whenever $A_{1},A_{2}ldotsinmathcal{L}$ are pairwisely disjoint.).
              It is routine to show that $mathcal{P}subseteqmathcal{L}$. For,
              let $(c,d]inmathcal{P}$, then
              begin{eqnarray*}
              mu_{1}(c,d]-mu_{2}(c,d] & = & left[g_{1}(d)-g_{1}(c)right]-left[g_{2}(d)-g_{2}(c)right]\
              & = & left[g_{1}(d)-g_{2}(d)right]-left[g_{1}(c)-g_{2}(c)right]\
              & = & g(d)-g(c)\
              & = & left[g_{3}(d)-g_{4}(d)right]-left[g_{3}(c)-g_{4}(c)right]\
              & = & left[g_{3}(d)-g_{3}(c)right]-left[g_{4}(d)-g_{4}(c)right]\
              & = & mu_{3}(c,d]-mu_{4}(c,d].
              end{eqnarray*}

              Re-arranging terms, we have $nu(c,d]=nu'(c,d]$. This shows that
              $mathcal{P}subseteqmathcal{L}$. By Dynkin's $pi-lambda$ theorem,
              we have $sigma(mathcal{P})subseteqmathcal{L}$. However, it is
              well known that $sigma(mathcal{P})=mathcal{B}((a,b])$, so we have
              $sigma(mathcal{P})=mathcal{L=mathcal{B}}(a,b])$.



              For a Borel function of the form $f=1_{A}$, where $Ainmathcal{B}((a,b])$,
              we have
              begin{eqnarray*}
              int fdmu_{1}+int fdmu_{4} & = & mu_{1}(A)+mu_{4}(A)\
              & = & nu(A)\
              & = & nu'(A)\
              & = & int fdmu_{2}+int fdmu_{3}.
              end{eqnarray*}

              By linearlity, the above holds for all simple functions $f$. If $f$
              is a non-negative, bounded Borel function, we may choose a sequence
              of simple functions $(f_{n})$ such that $0leq f_{1}leq f_{2}leqldotsleq f$
              and $f_{n}rightarrow f$ pointwisely. By monotone convergence theorem,
              the above inequlity holds for all non-negative, bounded Borel functions.
              Finally, if $f$ is a bounded Borel function, consider $f=f^{+}-f^{-}$
              and we can prove that the above also holds for bounded Borel function
              $f$. Finally, re-arrange terms, we have $int fdmu_{1}-int fdmu_{2}=int fdmu_{3}-int fdmu_{4}$
              or in the Stieltjes notation: $int fdg_{1}-int fdg_{2}=int fdg_{3}-int fdg_{4}$.






              share|cite|improve this answer











              $endgroup$
















                1












                1








                1





                $begingroup$

                Let $g:[a,b]rightarrowmathbb{R}$ be right-continuous and of bonded
                variation. Suppose that we have decompositions $g=g_{1}-g_{2}=g_{3}-g_{4}$
                for some right-continuous, increasing (non-strict sense) functions
                $g_{1},g_{2},g_{3},g_{4}$ defined on $[a,b]$. For each $i=1,2,3,4$,
                let $mu_{i}:mathcal{B}((a,b])rightarrowmathbb{R}$ be a Borel
                measure induced by $g_{i}$ such that for any $aleq c<dleq b$,
                $mu_{i}left((c,d]right)=g_{i}(d)-g_{i}(c)$. We go to show that
                $int fdmu_{1}-int fdmu_{2}=int fdmu_{3}-int fdmu_{4}$ for
                any bounded Borel function $f:(a,b]rightarrowmathbb{R}$. Define
                Borel measures $nu=mu_{1}+mu_{4}$ and $nu'=mu_{2}+mu_{3}$.
                Note that $nu$ and $nu'$ are finite measures. Let $mathcal{P}={(c,d]mid aleq c<dleq b}cup{emptyset}$
                and $mathcal{L}={Ainmathcal{B}left((a,b]right)midnu(A)=nu'(A)}$.
                Clearly $mathcal{P}$ is a $pi$-class (in the sense that $Acap Binmathcal{P}$
                whenever $A,Binmathcal{P}$) and $mathcal{L}$ is a $lambda$-class
                (in the sense that: $emptysetinmathcal{L}$; $A^{c}inmathcal{L}$
                whenever $Ainmathcal{L}$; $cup_{n=1}^{infty}A_{n}inmathcal{L}$
                whenever $A_{1},A_{2}ldotsinmathcal{L}$ are pairwisely disjoint.).
                It is routine to show that $mathcal{P}subseteqmathcal{L}$. For,
                let $(c,d]inmathcal{P}$, then
                begin{eqnarray*}
                mu_{1}(c,d]-mu_{2}(c,d] & = & left[g_{1}(d)-g_{1}(c)right]-left[g_{2}(d)-g_{2}(c)right]\
                & = & left[g_{1}(d)-g_{2}(d)right]-left[g_{1}(c)-g_{2}(c)right]\
                & = & g(d)-g(c)\
                & = & left[g_{3}(d)-g_{4}(d)right]-left[g_{3}(c)-g_{4}(c)right]\
                & = & left[g_{3}(d)-g_{3}(c)right]-left[g_{4}(d)-g_{4}(c)right]\
                & = & mu_{3}(c,d]-mu_{4}(c,d].
                end{eqnarray*}

                Re-arranging terms, we have $nu(c,d]=nu'(c,d]$. This shows that
                $mathcal{P}subseteqmathcal{L}$. By Dynkin's $pi-lambda$ theorem,
                we have $sigma(mathcal{P})subseteqmathcal{L}$. However, it is
                well known that $sigma(mathcal{P})=mathcal{B}((a,b])$, so we have
                $sigma(mathcal{P})=mathcal{L=mathcal{B}}(a,b])$.



                For a Borel function of the form $f=1_{A}$, where $Ainmathcal{B}((a,b])$,
                we have
                begin{eqnarray*}
                int fdmu_{1}+int fdmu_{4} & = & mu_{1}(A)+mu_{4}(A)\
                & = & nu(A)\
                & = & nu'(A)\
                & = & int fdmu_{2}+int fdmu_{3}.
                end{eqnarray*}

                By linearlity, the above holds for all simple functions $f$. If $f$
                is a non-negative, bounded Borel function, we may choose a sequence
                of simple functions $(f_{n})$ such that $0leq f_{1}leq f_{2}leqldotsleq f$
                and $f_{n}rightarrow f$ pointwisely. By monotone convergence theorem,
                the above inequlity holds for all non-negative, bounded Borel functions.
                Finally, if $f$ is a bounded Borel function, consider $f=f^{+}-f^{-}$
                and we can prove that the above also holds for bounded Borel function
                $f$. Finally, re-arrange terms, we have $int fdmu_{1}-int fdmu_{2}=int fdmu_{3}-int fdmu_{4}$
                or in the Stieltjes notation: $int fdg_{1}-int fdg_{2}=int fdg_{3}-int fdg_{4}$.






                share|cite|improve this answer











                $endgroup$



                Let $g:[a,b]rightarrowmathbb{R}$ be right-continuous and of bonded
                variation. Suppose that we have decompositions $g=g_{1}-g_{2}=g_{3}-g_{4}$
                for some right-continuous, increasing (non-strict sense) functions
                $g_{1},g_{2},g_{3},g_{4}$ defined on $[a,b]$. For each $i=1,2,3,4$,
                let $mu_{i}:mathcal{B}((a,b])rightarrowmathbb{R}$ be a Borel
                measure induced by $g_{i}$ such that for any $aleq c<dleq b$,
                $mu_{i}left((c,d]right)=g_{i}(d)-g_{i}(c)$. We go to show that
                $int fdmu_{1}-int fdmu_{2}=int fdmu_{3}-int fdmu_{4}$ for
                any bounded Borel function $f:(a,b]rightarrowmathbb{R}$. Define
                Borel measures $nu=mu_{1}+mu_{4}$ and $nu'=mu_{2}+mu_{3}$.
                Note that $nu$ and $nu'$ are finite measures. Let $mathcal{P}={(c,d]mid aleq c<dleq b}cup{emptyset}$
                and $mathcal{L}={Ainmathcal{B}left((a,b]right)midnu(A)=nu'(A)}$.
                Clearly $mathcal{P}$ is a $pi$-class (in the sense that $Acap Binmathcal{P}$
                whenever $A,Binmathcal{P}$) and $mathcal{L}$ is a $lambda$-class
                (in the sense that: $emptysetinmathcal{L}$; $A^{c}inmathcal{L}$
                whenever $Ainmathcal{L}$; $cup_{n=1}^{infty}A_{n}inmathcal{L}$
                whenever $A_{1},A_{2}ldotsinmathcal{L}$ are pairwisely disjoint.).
                It is routine to show that $mathcal{P}subseteqmathcal{L}$. For,
                let $(c,d]inmathcal{P}$, then
                begin{eqnarray*}
                mu_{1}(c,d]-mu_{2}(c,d] & = & left[g_{1}(d)-g_{1}(c)right]-left[g_{2}(d)-g_{2}(c)right]\
                & = & left[g_{1}(d)-g_{2}(d)right]-left[g_{1}(c)-g_{2}(c)right]\
                & = & g(d)-g(c)\
                & = & left[g_{3}(d)-g_{4}(d)right]-left[g_{3}(c)-g_{4}(c)right]\
                & = & left[g_{3}(d)-g_{3}(c)right]-left[g_{4}(d)-g_{4}(c)right]\
                & = & mu_{3}(c,d]-mu_{4}(c,d].
                end{eqnarray*}

                Re-arranging terms, we have $nu(c,d]=nu'(c,d]$. This shows that
                $mathcal{P}subseteqmathcal{L}$. By Dynkin's $pi-lambda$ theorem,
                we have $sigma(mathcal{P})subseteqmathcal{L}$. However, it is
                well known that $sigma(mathcal{P})=mathcal{B}((a,b])$, so we have
                $sigma(mathcal{P})=mathcal{L=mathcal{B}}(a,b])$.



                For a Borel function of the form $f=1_{A}$, where $Ainmathcal{B}((a,b])$,
                we have
                begin{eqnarray*}
                int fdmu_{1}+int fdmu_{4} & = & mu_{1}(A)+mu_{4}(A)\
                & = & nu(A)\
                & = & nu'(A)\
                & = & int fdmu_{2}+int fdmu_{3}.
                end{eqnarray*}

                By linearlity, the above holds for all simple functions $f$. If $f$
                is a non-negative, bounded Borel function, we may choose a sequence
                of simple functions $(f_{n})$ such that $0leq f_{1}leq f_{2}leqldotsleq f$
                and $f_{n}rightarrow f$ pointwisely. By monotone convergence theorem,
                the above inequlity holds for all non-negative, bounded Borel functions.
                Finally, if $f$ is a bounded Borel function, consider $f=f^{+}-f^{-}$
                and we can prove that the above also holds for bounded Borel function
                $f$. Finally, re-arrange terms, we have $int fdmu_{1}-int fdmu_{2}=int fdmu_{3}-int fdmu_{4}$
                or in the Stieltjes notation: $int fdg_{1}-int fdg_{2}=int fdg_{3}-int fdg_{4}$.







                share|cite|improve this answer














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                edited Jan 7 at 22:14

























                answered Jan 7 at 22:02









                Danny Pak-Keung ChanDanny Pak-Keung Chan

                2,32138




                2,32138






























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