# How to prove that $\int_a^b = \int_a^c + \int_c^b$ for Riemann-Stieltjes integrals

Let $f$ be Riemann-Stieltjes integrable with respect to $g$ on $[a,b]$. By Riemann-Stieltjes integrable I mean that $\forall\varepsilon>0\exists\delta$ such that when the norm of any partition $P$ of $[a,b]$ is smaller than $\delta$ then $\mid S(P,T,f,g)-\int_{a}^{b}f(x)dg(x)\mid<\varepsilon$ where $T$ is an arbitrary choice of tags. I'm trying to show that for $\forall c\in[a,b]$ $\int_{a}^{b}f(x)dg(x) = \int_{a}^{c}f(x)dg(x) + \int_{c}^{b}f(x)dg(x)$. This seems to be a bit tricky to show as opposed to some other linear properties. The problem comes down to the fact that the partition of choice is arbitrary as long as its norm is sufficiently small. So if $c$ is not an endpoint of a subinterval of the partition then the Riemann-Stieltjes sum cannot be split into a sum over $[a,c]$ and $[c,b]$ and that way be treated easily. Assuming I found a way to do this, then I would still have to prove that the two integrals on the right hand side exist. To my understanding the proof of this is quite a bit easier if we require $g$ to be monotone or perhaps a bound variation which is an entirely new thing for me. However, I don't want to consider special cases. Perhaps there are cases where the integrals exists without g being monotone or a bound variation, I would like to include those as well in the proof. I'm thinking that I perhaps have to find an equivalent definition that is easier to work with and prove it that way but I don't really know how to go about doing so. Some help would be greatly appreciated.

• Use the supremum-infinum approach. And join partitions together. – user370967 Sep 3 '18 at 10:55
• I guess the main thing to worry about is the case when $f,g$ both have discontinuities at the point $c$. Can your tags be at an endpoint of the subinterval they correspond to? – GEdgar Sep 3 '18 at 12:48
• Yes, endpoints of subintervals can be tags as well. Does this cause a problem? – David Sep 3 '18 at 13:32
• This is a difficult theorem. First one proves that if $\int_{a} ^{b} f(x) \, dg(x)$ exists then $\int_{c} ^{d} f(x) \, dg(x)$ exists where $[c, d]$ is any arbitrary subinterval of $[a, b]$ and then show that the result in question. – Paramanand Singh Sep 4 '18 at 7:40

Expanding my comment into an answer.

You need to use Cauchy condition for integrability:

Cauchy's Condition: $f$ is integrable with respect to $g$ on $[a, b]$ if and only if for every $\epsilon >0$ there is a $\delta>0$ such that $|S(P_1,T_1,f,g)-S(P_2,T_2,f,g)|<\epsilon$ for all partitions $P_1,P_2$ of $[a, b]$ with norm less than $\delta$ and $T_1,T_2$ being arbitrary tags for $P_1,P_2$ respectively.

Using this we can prove the integrability of $f$ wrt $g$ on $[c, d] \subseteq [a, b]$ if $f$ is integrable wrt $g$ on $[a, b]$. Let's begin with an arbitrary $\epsilon >0$ and choose $\delta>0$ which meets the Cauchy condition for $f, g$ on $[a, b]$. If $P_1,P_2$ are arbitrary partitions of $[c, d]$ with norm less than $\delta$ and $T_1,T_2$ corresponding choice of tags then it is possible to extend these to partitions $P'_1,P'_2$ of $[a, b]$ and choose tags $T'_1,T'_2$ accordingly such that $$S(P_1,T_1,f,g)-S(P_2,T_2,f,g)=S(P'_1,T'_1,f,g)-S(P'_2,T'_2,f,g)$$ and then for Cauchy condition the same $\delta$ works for $f, g$ on $[c, d]$.

Now using Riemann Stieltjes sums you can prove your desired result.

• Isn't the Cauchy-Stieltjes integral equivalent to the Darboux-Stieltjes integral which in general is not equivalent to the Riemann-Stieltjes integral? Thus considering the Cauchy-Stieltjes integral may not be enough, or am I wrong? – David Sep 4 '18 at 10:44
• @David: The integral is my answer is Riemann Stieltjes integral. And the name Cauchy is added because such conditions as in the theorem mentioned are usually called Cauchy condition. – Paramanand Singh Sep 4 '18 at 11:05
• Why is it possible to choose $T_1', T_2'$ such that the given equality holds (for arbitrary $T_1, T_2$)? – Yibo Yang Apr 9 at 10:40
• @YiboYang: the partition points in $P_1',P_2'$ as well as the tag points for $T_1',T_2'$ in intervals $[a, c], [d, b]$ are kept same. In interval $[c, d]$ they match those of $P_1,P_2,T_1,T_2$. So the part of Riemann sum for intervals $[a, c], [d, b]$ cancel each other and only the part for $[c, d]$ remains. – Paramanand Singh Apr 10 at 1:50