[Math] Is there difference between existence of integral and integrability

integrationmeasure-theoryreal-analysis

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!

Best Answer

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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