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I am looking for a reference or a direct answer for a proof of this statement:

Assume you have a random variable X. Define F(x) to be $P(X\le x)$, this function is bounded, monotone, non-deacreasing and right-countinuous and $\lim_{x \rightarrow -\infty} =0$. Define $\mu$ to be the Lebesgue-Stieltjes measure of the distribution function $F$. Assume that $-\infty < a \le b <\infty $. Assume that $f$ is a continuous real-valued function. Define $\int_a^b f dF(x)$ to be the Riemann-Stieltjes integral.

We then have

$$\int_a^b f dF(x) = \int_{(a,b]}f d\mu.$$

According to the answers here:

Is expectation Riemann-/Lebesgue–Stieltjes integral?

,this statement seems to be true.

I am able to show this when $f$ is the constant function, but not generally when it is continuous.

PS: I am not sure if I should have written $[a,b]$ instead of $(a,b]$? This is important for the case have $P(X=c)>0$ for a real number c. This also makes me unsure if I should have $a<b$ or if I can have $a \le b$?

user394334
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1 Answers1

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I think this question was asked somewhere, but I couldn't find a link so I write a quick answer here.

By definition

$$ \int^b_a f dF(x) = \lim_{\| \delta_n \| \rightarrow 0} \sum^{n-1}_{i=0} f(c_i)(F(x_{i+1})-F(x_{i})) = \lim_{n\rightarrow 0} \int_{(a,b]}f_nd\mu $$

where $f_n(x) = \sum^{n-1}_{i=0}f(c_i)1_{(x_i,x_{i+1}]}(x)$. Since $f$ is uniformly continuous on $[a,b]$, $f_n$ converge pointwise to $f$ and by DCT, the right-end of the above equation is just $\int_{(a,b]}fd\mu$.

Psychomath
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  • Thank you very much! Do you have a comment to my last statement? If we have that $P(X=c)>0$. Do we then have that $\int_c^c 1\cdot dF(x)=0$, but $\int_{[c,c]}1 \cdot d\mu = P(X=c)>0$? And was it correct to write $(a,b]$ and not $[a,b]$? – user394334 Aug 01 '24 at 01:04
  • Ah, it's always the half open bracket $(a,b]$ since the definition of $f_n$ above has $1_{(a,x_1]}, 1_{(x_1,x_2]},...$ and also remember the measure $\mu$ is defined such that $\mu(a,b] =F(b) - F(a)$ so we need $a<b$ otherwise the interval is empty. – Psychomath Aug 01 '24 at 01:07
  • Yeah so if $a=b$, it still equals to the LS integral since $(a,a] = \emptyset$ – Psychomath Aug 01 '24 at 01:25
  • Thank you very much for your help! – user394334 Aug 01 '24 at 01:27
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    I deleted the comment before your last answer because I didn't want to bother you with further questions, but rather ask a neq quesiton instead. I appreciate all your help! – user394334 Aug 01 '24 at 01:42