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I want to use Integration by parts for general Lebesgue-Stieltjes integrals. The following theorem can be found in the literature:

Theorem: If $F$ and $G$ are right-continuous and non-decreasing functions, we have that: $$ \int_{(a,b]}G(x)\text{d}F(x)=F(b)G(b)-F(a)G(a)- \int_{(a,b]}F(x-)\text{d}G(x),$$ where $F(x-)$ is the left limit of $F$ in $x$.

Does the following result hold:

Theorem: If $F$ and $G$ are left-continuous and non-decreasing functions, we have that: $$ \int_{[a,b)}G(x)\text{d}F(x)=F(b)G(b)-F(a)G(a)- \int_{[a,b)}F(x+)\text{d}G(x),$$ where $F(x+)$ is the right limit of $F$ in $x$.

Is it possible to combine these result. So use integration by parts when $F$ is right cont., $G$ is left cont.?

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You can Compare with the following theorem,

Theorem: Suppose $f$ and $g$ are bounded functions with no common discontinuities on the interval $[a,b]$, and the Riemann-Stieltjes integral of $f$ with respect to $g$ exists. Then the Riemann-Stieltjes integral of $g$ with respect to $f$ exists, and

$$\int_{a}^{b} g(x)df(x) = f(b)g(b)-f(a)g(a)-\int_{a}^{b} f(x)dg(x)\,. $$

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But it is not possible get rid of the condition of no common discontinuities? – Daniel Linders Oct 26 '12 at 13:51
    
I guess that common discontinuities can prevent integrability. – Siminore Oct 26 '12 at 14:08
    
For Riemann-Stieltjes common discontinuities are a problem. But I would think that for Lebesgue-Stieltjes integrals, this is condition is not needed. If the common discontinuities are rc, it is no problem. – Daniel Linders Oct 26 '12 at 14:26

Variants of the above Lebesgue--Stieltjes partial integral results are given in Theorem 21.67(iv) (it holds for general BV functions, by Remarks 21.68) of Real and Abstract Analysis: A modern treatment of the theory of functions of ... E. Hewitt, K. Stromberg p. 419

They seem to answer your question, though at the cost of having terms like $(F(x+)+F(x-))/2$ in place of $F(x)$.

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