Continuity of Lebesgue Stieltjes integral

Question

I am trying to prove that Lebesgue-Stieltjes integral defines a cadlag function (i.e. right continuous with left limits) when its integrator is a cadlag function.

Assume that $A(s)$, $s\in \mathbb{R}_+$ is a right-continuous function, has left limits and of finite variation. Also assume that function $H(s)$, $s\in \mathbb{R}_+$ is measurable function (i.e. Borel measurable on $\mathbb{R}_+$).

It is well-known that $A$ defines a signed measure and thus we can define Lebesgue-Stieltjes integral $Y(t):=\int_0^t H(s)dA(s)$ for any $t$ as long as $\int_0^t |H(s)|d|A|(s)<\infty$.

To prove that $Y(t)$ is right continuous, I choose a decreasing sequence $t_n\downarrow t$ and have $Y(t_n)=\int_{\mathbb{R}_+}1_{(0,t_n]}(s)H(s)dA(s)$. We can see that for every fixed $s$ the integrand $1_{(0,t_n]}(s)H(s)$ tends to $1_{(0,t]}(s)H(s)$. As $\int_0^u |H(s)|d|A|(s)<\infty$ for all $u$, we can use dominated convergence and show that indeed $\lim_{n\to \infty}Y(t_n)=Y(t)$. This proves that $Y(t)$ is right continuous.

Using exactly the same argumens (i.e. by choosing $t_n\uparrow t$) we can show that $Y(t)$ is left continuous. However, my intuition says that when $A(s)$ has a jump discontinuity at point $t$, then $Y(t)$ should also have a jump of size $H(t)\Delta A(t)$. I can't find a flaw in my proof above.

Answer 1

Your mistake is that $1_{(0,t_n]}$ with $t_n \uparrow t$ tends only to $1_{(0,t)}$ and not to $1_{(0,t]}$. If $A$ is continuous in $t$, this doesn't matter. However, if $A$ has a jump in $t$, then $$Y(t) = \int 1_{(0,t)}(s) H(s) \, d A(s) + H(t) (A(t^+)-A(t^-)).$$ and then only have $Y(t_n) \rightarrow \int 1_{(0,t)}(s) H(s) \, d A(s) \ne Y(t)$.

The dominated convergence theorem is also true for the Lebesgue-Stieltjes integral. You can deduce it by noting that $$\left| \int (f-g) \, d \mu \right| \le \int |f-g| \, d \|\mu\|,$$ where $\|\mu\|$ denotes the total variation of a measure.