functional-analysisintegrationmeasure-theoryprobability theoryreal-analysis
In probability theory, when having $ E(f(X))=\int_{-\infty}^\infty f(x)\, dg(x) $, an expectation of a measurable function $f$ of a random variable $X$ with respect to its cumulative distribution function $g$,
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I don't think my understanding is completely correct and should have posted as a comment, but it has length restrictions.
It seems to me that when $g$ is BV and right continuous, and $f$ is Borel measurable, $f$ does not have to be bounded. There are unbounded Lebesgue integrable functions, so the same should be true for Lebesgue-Stieltjes integral, which is just Lebesgue integral w.r.t. the signed measure $\mu_g$ on $\mathcal{B}(\mathbb{R})$ induced by $g$.
It should be OK for function $g$ with bounded variation. We can write $g$ as the difference of two non-decreasing functions.
I don't think Riemann-Stieltjes integral requires the integrator to be non-decreasing. It might be BV or possibly an even broader class of functions. I guess the author of Wikipedia entry mentions only nondecreasing functions because s/he has CDF of a random variable in mind and wants to discuss its application in probability theory. I also have the impression that these two integrals agree whenever the Riemann-Stieltjes integral exists. (Just like the relation between the Lebesgue integral and the Riemann integral.)
If these two notions agree, there's no danger of using the same notation. Otherwise, I've seen authors using prefix to distinguish different types of integrals, e.g., $(R)\int_a^b...$ for Riemann(-Stieltjes) integrals.
As described below, this question is very closely related to the question Meaning of non-existence of expectation?.
There is a difference between existence of integral and integrability in Lebesgue (but not Riemann) integration. A measurable function $f$ (taking values in $[-\infty,\infty]$, on some measure space with measure $\mu$) is said to be Lebesgue integrable if $\int {|f|d\mu } < \infty $. As usual, let $f^+$ and $f^-$ be the (nonnegative) functions defined by $f^ + (x) = \max \{ f(x),0\}$ and $f^ - (x) = -\min \{ f(x),0\}$, so that $|f|=f^+ + f^-$. Then, $f$ is integrable if and only if $\int {f^ + d\mu } < \infty$ and $\int {f^ - d\mu } < \infty$; in this case, $$ \int {fd\mu } : = \int {f^ + d\mu } - \int {f^ - d\mu } $$ (in particular, the integral of an integrable function is finite). However, $f$ need not be integrable in order that the integral $\int {fd\mu }$ be defined. Indeed, the definition of $\int {fd\mu }$ for nonnegative measurable $f$ allows it to take the value $\infty$.
With the convention that $\infty - c = \infty$ and $c - \infty = -\infty$, for $c$ finite, the above definition of $\int {fd\mu }$ holds whenever $\int {f^ + d\mu } \leq \infty$ and $\int {f^ - d\mu } < \infty$ or $\int {f^ + d\mu } < \infty$ and $\int {f^ - d\mu } \leq \infty$. The integral $\int {fd\mu }$ does not exist if and only if $\int {f^ + d\mu } = \infty$ and $\int {f^ - d\mu } = \infty$ (since $\infty - \infty$ is not defined).
Examples: $\int_{(0,1]} {x^{ - 1} dx} = \infty$ (where $dx$ stands for Lebesgue measure), so the integral exists but $x^{-1}$ is not integrable on $(0,1]$. On the other hand, the integral $\int_{(0,1]} {\frac{{\sin (1/x)}}{x}dx}$ does not exist, since $\int_{(0,1]} {\big[\frac{{\sin (1/x)}}{x}\big]^ + }dx = \infty $ and $\int_{(0,1]} {\big[\frac{{\sin (1/x)}}{x}\big]^ - }dx = \infty $ (in particular, $\frac{{\sin (1/x)}}{x}$ is not integrable on $(0,1]$).
Relation to expectations of random variables.
Let $(\Omega,\mathcal{F},P)$ be a probability space, that is, a measure space with $P(\Omega)=1$. A random variable is a measurable function $X:\Omega \to \mathbb{R}$. So, $X(\omega)$ ($\omega \in \Omega$) and $P$ play the same role as $f(x)$ and $\mu$ above, respectively. The expectation of $X$ is defined by ${\rm E}(X) = \int_\Omega {XdP} $, which is just a special case of $\int {fd\mu }$ above. Accordingly, $X$ is said to be integrable if ${\rm E}|X|:=\int_\Omega {|X|dP} < \infty$, and has expectation if ${\rm E}(X^+) := \int_\Omega {X^ + dP} \le \infty $ and ${\rm E}(X^-) := \int_\Omega {X^ - dP} < \infty $ or ${\rm E}(X^+) < \infty $ and ${\rm E}(X^-) \leq \infty $ (in either case, ${\rm E}(X)= {\rm E}(X^+) - {\rm E}(X^-)$); $X$ does not admit an expectation if and only if ${\rm E}(X^+) = \infty$ and ${\rm E}(X^-) = \infty$.
It is instructive to consider the above examples in the setting of expectations. Let the probability space $(\Omega,\mathcal{F},P)$ be defined by $\Omega = (0,1]$, $\mathcal{F} = \mathcal{B}((0,1])$, and $P$ Lebesgue measure on $(0,1]$ (thus $dP(\omega)=d\omega$). Note that the random variable $X$ defined by $X(\omega)=\omega$ is a uniform$(0,1]$ random variable. Define $X_1$ by $X_1 = 1/X$, that is $X_1 (\omega) = 1/\omega$. Then, ${\rm E}(X_1) = \int_{(0,1]} {\frac{1}{\omega }d\omega } = \infty$; so, $X_1$ has (infinite) expectation but is not integrable. On the other hand, the random variable $X_2$ defined by $X_2 = \frac{{\sin (1/X)}}{X}$, that is $X_2 (\omega) = \frac{{\sin (1/\omega)}}{\omega}$, does not admit an expectation, since the corresponding integral, $\int_{(0,1]} {\frac{{\sin (1/\omega )}}{\omega }d\omega }$, does not exist (as noted above).
Finally, it is interesting to consider the above examples with connection to the strong law of large numbers (SLLN). Let $X_1^1,X_2^1,\ldots$ be a sequence of i.i.d. random variables distributed as $X_1$, and let $S_n^1 = X_1^1 + \cdots + X_n^1$. Then, almost surely, $n^{-1}S_n^1 \to {\rm E}(X_1) = \infty$ (this follows from the standard SLLN by using the monotone convergence theorem). So, the existence of ${\rm E}(X_1) := \int_{(0,1]} {\frac{1}{\omega }d\omega } = \infty$ agrees with SLLN. As for the second example, let $X_1^2,X_2^2,\ldots$ be a sequence of i.i.d. random variables distributed as $X_2$, and let $S_n^2 = X_1^2 + \cdots + X_n^2$. Since $X_2$ does not admit an expectation, with $X_2 ^+$ and $X_2^-$ behaving the same, one may expect that, almost surely, $\lim \sup n^{ - 1} S_n^2 = \infty $ and $\lim \inf n^{ - 1} S_n^2 = -\infty$; see Theorem 1 in the paper The strong law of large numbers when the mean is undefined, by K. Bruce Erickson. Here, it is important to note that $\frac{{\sin (1/x)}}{x}$ is (improperly) Riemann integrable on $(0,1]$; so SLLN may justify the Lebesgue non-integrability of this function.
Best Answer
If $f$ is continuous, then $\int_a^b {f(x)dF(x)} = \int_a^b {f(x)d\mu (x)} $, for any $-\infty < a < b < \infty$, where $F$ and $\mu$ are the distribution function and probability distribution of $X$, respectively (they are related by $\mu((s,t])=F(t)-F(s)$, for any $-\infty < s < t < \infty$).
A drawback of the Riemann-Stieltjes integral is illustrated in the following simple example. Suppose that $X=0$ almost surely. Then, $\int {F(x)dF(x)} $ is not defined, whereas $\int {F(x)d\mu (x)} = F(0) = 1$. Of course, ${\rm E}[F(X)] = {\rm E}[F(0)]= 1$.
Another (more significant) drawback is indicated in GWu's answer.