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I'm currently doing an exercise on the following theorem

If $F$ and $G$ have no common points of discontinuity in $(a,b]$, then $$\int_{(a,b]}G(x)dF(x) = F(b)G(b)-F(a)G(a) - \int_{(a,b]}F(x)dG(x)$$

The exercise is to prove that if there are such shared points of discontinuity, this equality does not hold. I'm kind of stuck at this exercise as I have difficulty grasping the intuition behind $dF(x)$ (or $dG(x)$ for that matter). Can anyone help me with this intuition and exercise? Thanks in advance!!

Mittens
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1 Answers1

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Presumably $F$ and $G$ are both right-continuous with-left limits functions of local finite variation in an interval that contains $[a,b]$. Lebesgue integration by parts states that

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 bounded interval $(a,b]\subset I$, \begin{align} \int_{(a,b]}F(t)\mu_G(dt)=F(b)G(b)-F(a)G(a)-\int_{(a,b]}G(t-)\mu_F(dt)\tag{0}\label{zero} \end{align} where $G(t-)=\lim_{s\nearrow t}G(s)$.

It is common to use $dF(x)$ and $dG(x)$ instead of $\mu_F(dx)$ and $\mu_G(dx)$ respectively. Using measure notation is clearer to me.

Now, Notice that for any singleton $\{x\}$, $\Delta G(x):=\mu_G(\{x\})=G(x)-G(x-)$, $\Delta F(x)=\mu_F(\{x\})=F(x)-F(x-)$. Then,

\begin{align} \int_{(a,b]}G(t-)\mu_F(dt)&=\int_{(a,b]}(G(t)-\Delta G(t))\mu_F(dt)\\ &= \int_{(a,b]}G(t)\mu_F(dt)- \int_{(a,b]}\Delta G(t)\mu_F(dt)\\ &=\int_{(a,b]}G(t)\mu_F(dt)-\sum_{a<t\leq b}\Delta G(t)\Delta F(t) \end{align}

Notice that as $F$ and $G$ are of finite local variation, each function $F$ and $G$ has at most a countable set of discontinuities in $(a,b]$ and that $\sum_{a<t\leq b}|\Delta G(t)\Delta F(t)|<\infty$. Consequently

\begin{align} \int_{(a,b]}F(t)\mu_G(dt)&=F(b)G(b)-F(a)G(a)-\int_{(a,b]}G(t)\mu_F(dt)+\sum_{a<t\leq b}\Delta G(t)\Delta F(t)\tag{1}\label{one} \end{align}

If $G$ and $F$ do not have common jumps (i.e. point of discontinuity) the. the last term in the right-hand side of \eqref{one} is $0$.

Mittens
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