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Suppose that $F,G \in BV[a, b]$, where $-\infty<a<b<\infty$. I want to prove that if there are no points on $[a,b]$ where $F$ and $G$ are both discontinuous, then $$\int_{[a,b]} F d G+\int_{[a,b]} G dF= F(b)G(b)-F(a-)G(a-)$$

I know that $$\int_{[a,b]}\frac{F(x)+F(x-)}{2}dG(x)+\int_{[a,b]} \frac{G(x)+G(x-)}{2}dF(x)=F(b)G(b)-F(a-)G(-b).$$

How I can use the hypothesis that there are no points in $[a,b]$ where $F$ and $G$ are both discontinuous??

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    $\begingroup$ Why don't you explain to the reader what NBV means? $\endgroup$
    – amsmath
    Aug 1, 2021 at 1:00
  • $\begingroup$ Why would one cares about the limit to $-\infty$ when the statement is about the interval $[a, b]$? @OliverDiaz $\endgroup$
    – Arctic Char
    Aug 4, 2021 at 13:30
  • $\begingroup$ @ArcticChar: I don't , since as you noted the result is of a local nature. Now, if both $F$ and $G$ were of total bounded variation over $\mathbb{R}$, then it would probably be desirable to normalize so that $G(-\infty)=\mu_G(\emptyset)=0=\mu_F(\emptyset)=F(-\infty)$, in which case the integration by parts formula takes the form \begin{align}\int_{\mathbb{R}}F(t)\mu_G(dt)&=F(\infty)G(\infty)-\int_{\mathbb{R}}G(t-)\mu_F(dt)\\&=\mu_F(\mathbb{R})\mu_G(\mathbb{R})-\int_{\mathbb{R}}G(t-)\mu_F(dt)\end{align} $\endgroup$
    – Oliver Díaz
    Aug 4, 2021 at 15:19
  • $\begingroup$ @OliverDiaz I have edited the post. I guess one could also change it to $-\infty \le a < b \le +\infty$ and keeps the N. $\endgroup$
    – Arctic Char
    Aug 4, 2021 at 17:11
  • $\begingroup$ @ArcticChar: under the assumtion that $F$ and $G$ are both of finite total variation over $\mathbb{R}$, that is correct. That is why the normalization is convenient; also this way we have $F(+\infty)=\mu_F(\mathbb{R})$ and $G(+\infty)=\mu_G(\mathbb{R})$, which is what one typically does in say, probability theory, when once discusses cumulative distributions. OF course, here $\mu_G$ and $\mu_F$ are in general signed measures (with local or total finite variation) $\endgroup$
    – Oliver Díaz
    Aug 4, 2021 at 17:17

1 Answer 1

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Recall that any rights continuous function of finite variation $F$ on an interval $\alpha,\beta$ generates a unique (possibly signed) measure $\mu_F$ if local finite variation such that $\mu_F((a,b])=F(b)-F(a)$. This is the so called Lebesgue-Stieltjes measure associated to $F$, see for example Klenke, A. Probability theory, Universitext, Springer-Verlag, London 2008, pp 26-27.)

Theorem: Let $F$, $G$ be right--continuous functions of locally finite variation on an interval $I$ (bound or unbounded) Let $\mu_F$ and $\mu_G$ the Stieltjes-Lebesgue measures generated by $F$ and $G$ respectively. For any compact $[a,b]\subset I$ $$ \int_{(a,b]}F(t)\mu_G(dt)=F(b)G(b)-F(a)G(a)-\int_{(a,b]}G(t-)\mu_F(dt) $$ where $G(t-)=\lim_{s\nearrow t}G(s)$.

A proof can be obtained using Fubini's theorem \begin{aligned} F(b)-F(a))(G(b)-G(a))&=\int_{(a,b]\times(a,b]}\mu_F\otimes\mu_G(dt,ds)\\ &=\int_{(a,b]}\Big(\int_{(a,s]}\mu_F(dt)\Big)\mu_G(s) +\int_{(a,b]}\Big(\int_{(s,b]}\mu_F(dt)\Big)\mu_G(ds)\\ &=\int_{(a,b]}\Big(\int_{(a,s]}\mu_F(dt)\Big)\mu_G(s) +\int_{(a,b]}\Big(\int_{(a,t)}\mu_G(ds)\Big)\mu_F(dt)\\ &=\int_{(a,b]} F(s)-F(a)\mu_G(s) +\int_{(a,b]}G(t-)-G(a)\mu_F(dt)\\ \end{aligned} Algegraic simplifications yields the result in the Theorem.

If in addition $F$ and $G$ are continuous, then $F(a)=F(a-)$ and $G(a)=G(a-)$. Hence $\mu_F(\{a\})=\mu_G(\{a\})=0$, and we can substitute $(a,b]$ by $[a,b]$ in the integration.

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