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For Lebesgue integral of a measurable function, its existence and integrability are different, as the latter requires the integral not only to exist but also to be finite.

  1. I wonder if there is similar difference between existence of Riemann integral and Riemann integrability of a function over an interval? Or there is no difference, i.e., both require the Riemann integral to exist and to be finite? My question arose when I tried to understand a comment by Shai Covo following this reply

    Consider X uniform(0,1), with distribution function F and probability distribution μ. Then, $\int {f(x)dF(x)} = \int_0^1 {f(x)dx}$ is an ordinary Riemann integral, whereas $\int {f(x)d\mu (x)} = \int_0^1 {f(x)dx}$ is Lebesgue integral. The former is not defined if f is (for example) unbounded.

  2. Is the answer for Riemann integral same for Riemann-Stieltjes Integral?

  3. Is the answer for Lebesgue integral same for Lebesgue-Stieltjes Integral? I guess yes, because the latter is actually defined in terms of the former?

Thanks and regards!

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In my experience, to say something is X integrable is identical to saying it has a finite integral, of type X. –  Jon Beardsley Apr 11 '11 at 2:38
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My understanding is that when we say an integral (any type) exists, it means the function is integrable (finite integral) or the integral is infinite.

However, Riemann integral is only defined for bounded functions defined on a bounded interval $[a,b]$. (At least that's the case I've seen in all the text books.) So when the integral exists, it must be finite. One way to define Riemann integral is through upper and lower Riemann integrals. The upper Riemann integral is defined as $$U(f)=\inf\left\{\sum_i \left(\sup_{x_{i-1}\le x \le x_i}f(x)\right)\cdot (x_i-x_{i-1}):\text{ all partitions } (x_0,x_1,\cdots,x_n)\right\},$$ and the lower Riemann integral is defined to be $$L(f)=\sup\left\{\sum_i \left(\inf_{x_{i-1}\le x \le x_i}f(x)\right)\cdot (x_i-x_{i-1}):\text{ all partitions } (x_0,x_1,\cdots,x_n)\right\}.$$ The function $f$ is called Riemann integrable if $U(f)=L(f)$, and the common value is called the Riemann integral. Clearly, since we only consider bounded functions $f$: $|f|\le M$, we always have $$-M\le L(f)\le U(f)\le M,$$ and the Riemann integral is always finite if $L(f)=U(f)$.

The case of Riemann-Stieltjes integral should be similar.

If you take the definitions from Wikipedia, there's no infinite Riemann(-Stieltjes) integral because the definition requires the existence of a real number $A$ which is the limit of Riemann sum.

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Thanks! Is the answer for Riemann integral same for Riemann-Stieltjes Integral? Is the answer for Lebesgue integral same for Lebesgue-Stieltjes Integral? I guess yes, because the latter is actually defined in terms of the former? –  Tim Apr 11 '11 at 3:37
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The reason why RI is defined only for bounded functions is because it can be proven that unbounded functions on compact intervals are not RI..... To see this, pick any partition, then at least on one of the intervals of this partition the function is unbounded, which means that we can make the Riemann sum as high as we want by picking the right sample points in this partition. –  N. S. Apr 11 '11 at 3:42
    
(1) Yes. Riemann or Riemann-Stieltjes integrals, existence = finite (=integrable). (2) Lebesgue or Lebesgue-Stieltjes integrals can exist with infinite value. The reason is exactly as you said. –  GWu Apr 11 '11 at 3:43
    
@user9176 This is a good point! That's what I suspected but couldn't prove it off the top of my head, so I didn't mention it. –  GWu Apr 11 '11 at 3:43
    
I added a short argument which explains why does that happen. –  N. S. Apr 11 '11 at 3:46
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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.

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