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Let $\alpha$ be Stieltjes weight function (monotone increasing), and $\alpha = \alpha_1+\alpha_2$, where $\alpha_1,\alpha_2$ are also monotone increasing. suppose $f$ is Stieltjes-integrable on $[a,b]$

$$\int_a^bf d\alpha = \int_a^b f d\alpha_1 + \int_a^b f d\alpha_2$$

Under what conditions is this statement true? How can I prove it easily using basic knowledge of upper and lower Stieltjes integral? If it is not true all the time counter examples?

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It should be sufficient that measure $\alpha(M)= \alpha_1(M) + \alpha_2(M)$. –  tom Feb 7 '13 at 22:53
    
I didn't learn measure yet, just in the basic analysis course –  mez Feb 7 '13 at 22:56
    
Try to show that $\alpha(M) = \text{inf}\{ \sum_i \alpha(I_i): M\subset \bigcup_i I_i\} = \text{inf}\{ \sum_i \alpha_1(I_i): M\subset \bigcup_i I_i\} + \text{inf}\{ \sum_i \alpha_2(I_i): M\subset \bigcup_i I_i\} = \alpha_1(M) +\alpha_2(M) $. Where $I$ are intervals of type $ (c,d] $. Plus functions $\alpha,\alpha_1, \alpha_2$ should be right continuous. $\alpha((c,d]) = \alpha(d) - \alpha(c) $ –  tom Feb 7 '13 at 23:01

1 Answer 1

up vote 3 down vote accepted

$\newcommand\uint{\overline{\int_a^b}}\newcommand\lint{\underline{\int_a^b}}$

Notations

Suppose $\alpha$ is monotone increasing on $[a,b]$, $P=\{\,x_0,x_1,\dotsc,x_n\,\}$ is a partition of $[a,b]$, where $a=x_0\le x_1\le\dotsb\le x_n=b$, $f$ is a bounded function on $[a,b]$, we'll use notations here:

\begin{align} &\Delta\alpha_k=\alpha(x_k)-\alpha(x_{k-1})\\\\ &M_k=\sup_{x_{k-1}\le x\le x_k}f(x) &m_k=\inf_{x_{k-1}\le x\le x_k}f(x)\\ &U(P,f,\alpha)=\sum_{k=1}^nM_k\Delta\alpha_k &L(P,f,\alpha)=\sum_{k=1}^nm_k\Delta\alpha_k\\ &\uint fd\alpha=\inf_P U(P,f,\alpha) &\lint fd\alpha=\sup_P U(P,f,\alpha) \end{align}

Stronger results

\begin{align} \uint fd(\alpha_1+\alpha_2)&=\uint fd\alpha_1+\uint fd\alpha_2\\ \lint fd(\alpha_1+\alpha_2)&=\lint fd\alpha_1+\lint fd\alpha_2 \end{align}

Proof

We'll only prove the first one. For each partition $P$, it's easier to check that $$U(P,f,\alpha_1+\alpha_2)=U(P,f,\alpha_1)+U(P,f,\alpha_2)$$ Take $\inf_P$ on both sides, we have $$\uint fd(\alpha_1+\alpha_2)\ge\uint fd\alpha_1+\uint fd\alpha_2$$

Next, for each $u_1>\uint fd\alpha_1$, $u_2>\uint fd\alpha_2$, there's some partitons $P_1$ and $P_2$ such that $U(P_1,f,\alpha_1)<u_1$ and $U(P_2,f,\alpha_2)<u_2$.

Consider $P=P_1\cup P_2$, we have $U(P,f,\alpha_1)\le U(P_1,f,\alpha_1)<u_1$ and $U(P,f,\alpha_2)<u_2$, then $$U(P,f,\alpha_1+\alpha_2)=U(P,f,\alpha_1)+U(P,f,\alpha_2)<u_1+u_2$$ therefore $$\uint fd(\alpha_1+\alpha_2)<u_1+u_2$$

Hence $$\uint fd(\alpha_1+\alpha_2)\le\uint fd\alpha_1+\uint fd\alpha_2$$

Proved.

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Very good proof. thanks. –  mez Feb 8 '13 at 10:00
    
Your proof partially finishes my question. I have further question, what if now we only assume that $f$ is stieltjes integrable with respect to $\alpha = \alpha_1+\alpha_2$, and only that $\alpha_1$,$\alpha_2$ are monotone increasing bounded functions, can we deduce $$\int_a^b f d\alpha = \int_a^b f d\alpha_1 +\int_a^b f d\alpha_2$$? We are not sure whether $$\overline{\int}_a^b f d\alpha_1 = \underline{\int}_a^b f d\alpha_1$$, is it true to begin with? –  mez Feb 8 '13 at 10:43
    
@mezhang According to the definition, $\uint fd\alpha=\lint fd\alpha$ if and only if $f\in\mathcal R(\alpha)$, i.e. integrable (according to $\alpha$). –  Frank Science Feb 9 '13 at 11:14
    
This I know, but how does $\overline{\int}_a^b f d(\alpha_1+\alpha_2) = \underline{\int}_a^b f d(\alpha_1+\alpha_2)$ imply $\overline{\int}_a^b f d\alpha_1 = \underline{\int}_a^b f d\alpha_1$? –  mez Feb 9 '13 at 13:01
    
@mezhang Since $\uint fd\alpha_1+\uint fd\alpha_2=\int_a^bfd(\alpha_1+\alpha_2)=\lint fd\alpha_1+\lint fd\alpha_2$, and $\uint fd\alpha_k\ge\lint fd\alpha_k$, so $\uint fd\alpha_k=\lint fd\alpha_k$ for $k=1,2$. –  Frank Science Feb 10 '13 at 5:58

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