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Durrett constructs the integral of $f$ with respect to a $\sigma$-finite measure $\mu$ on the space $(\Omega, \mathcal{F}, \mu)$ using the following four steps.


Step 1. Assuming $\mu(A_i) < \infty$, define $$\int \varphi d\mu = \sum\limits_{i = 1}^n a_i \mu(A_i)$$ where $ \varphi = \sum\limits_{i = 1}^n a_i \mathbb{1}_{A_i}$ is a simple function.

Step 2. Suppose $f$ is bounded on $E \subset \Omega$ and vanishes on $E^c$, then defines $$\int\limits f d\mu = \sup\limits_{\varphi\leq f} \int \varphi d\mu$$

Step 3. Assume $f \geq 0$, then define, $$ \int\limits f d\mu = \sup\left\{\int h d\mu: 0 \leq h \leq f, h \text{ bounded }, \mu(\{x: h(x) >0\}) < \infty\right\}$$

Step 4. Assume $f$ is integrable ($\int |f| d\mu < \infty$), then define,

$$ \int f d\mu = \int f^+ d\mu - \int f^- d\mu = \int \max(f(x),0) d\mu - \int\min(-f(x), 0)d\mu$$


Durrett's presentation felt very disjointed and left much to be desired. At first he provides a very nice, computable expression for the integral, but at then end it is so abstract now I have no idea how to take an integral of any function.

Let's backtrack from step 4 to see how it all ties together.

  • Why did Durrett stop the construction of $\int f d\mu$ at integrable functions? He promised to define the meaning of $\int f d\mu$ and not $\int f d\mu$ with the side condition "$|f|$ is integrable". Not all functions are integrable, so how do I take the integral of those?

  • How does step 4 depend on step 3? When he writes, $\int |f| d\mu < \infty$, is the integral $\int |f| d\mu$ defined as $\int |f| d\mu = \sup\left\{\int h d\mu: 0 \leq h \leq |f|, h \text{ bounded }, \mu(\{x: h(x) >0\}) < \infty\right\}$ (from step 3) or $\int\limits |f| d\mu = \sup\limits_{\varphi\leq |f|} \int \varphi d\mu$ (from step 2)?

  • How does step 3 depend on step 2? Recall in step 2, $\int\limits f d\mu = \sup\limits_{\varphi\leq f} \int \varphi d\mu$. However, in step 3, $ \int\limits f d\mu = \sup\left\{\int h d\mu: 0 \leq h \leq f, h \text{ bounded }, \mu(\{x: h(x) >0\}) < \infty\right\}$. The connection is not obvious or clear. For example, $\varphi$ is not required to be bounded, and there is no condition such as $\mu(\{x:\varphi(x) > 0\}) < \infty)$. The definition given in step 2 doesn't logically imply the one given in step 3.

  • In step 2, what did Durrett mean when he wrote $\varphi \leq f \leq\psi$? Functions are not ordered, so $\varphi \leq f \leq\psi$ doesn't make sense. Should I interpret the above as $\forall \omega \in \Omega, \varphi(\omega) \leq f(\omega) \leq \psi(\omega)$?

  • How does it all tie together? Ultimately, I want to use the integral to compute things such as the number of calls being dropped, which is a number, not a math expression. Is it possible to re-expression the definition given in step 4 in terms of the one in step 1?

Thanks in advance!

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    $\begingroup$ The side condition is $|f|$ is integrable, not $f$ is integrable. $\endgroup$ – saulspatz Jul 15 '19 at 2:07
  • $\begingroup$ @saulspatz Thanks corrected $\endgroup$ – Olórin Jul 15 '19 at 2:14
  • $\begingroup$ It still doesn't make sense. You say, "Not all functions are integrable, so how do I take the integral of those?" You can't integrate non-integrable functions, of course. What are you thinking of? $\endgroup$ – saulspatz Jul 15 '19 at 2:26
  • $\begingroup$ @saulspatz $f = sin$ over the real line $\endgroup$ – Olórin Jul 15 '19 at 2:27
  • $\begingroup$ Not integrable. $\endgroup$ – saulspatz Jul 15 '19 at 2:29
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Reading between the lines, I think your confusion stems from the concept of integrability. I believe the simplest way to understand this concept is by starting with a super concrete case: $\mu$ is counting measure on $\mathbb N$ (i.e. $\mu(A)=|A|$), with $\sigma$-algebra consisting of all subsets of $\mathbb N$.

In this case, a measurable function $f$ the same thing as an infinite sequence $(a_n)_{n\in\mathbb N}$. And the Lebesgue integral with respect to $\mu$ is simply the sum of the $a_n$ - but with one caveat: the integral makes no reference to the ordering of $\mathbb N$, whereas the sum of an infinite sequence is defined as the limit of its partial sums, and this manifests in the phenomenon of conditional convergence, where a series can have different sums based on the order you take its values in. (See also this nice historical article on the same topic.)

Thus, due to these reordering issues, it does not make sense to speak of $\int f\ d\mu$ without imposing further conditions on $f$. The simplest way to deal with this problem is to assume that $f\geq 0$: then $\int f\ d\mu$ is well-defined, but could be equal to $\infty$. Taking one step further, one can use absolute convergence instead of conditional convergence, which means to require that $\int |f|\ d\mu<\infty$. (This integral always makes sense since $|f|\geq 0$, although it can equal $\infty$.) There are some results showing that absolute convergence is equivalent to "unconditional convergence" (meaning that all rearrangements of a series converge) in certain situations, but not others.

Finally, the most general condition under which $\int f\ d\mu$ makes sense (that I know of) is the following: $$ \min\left(\int\max(f,0)\ d\mu,\int\max(-f,0)\ d\mu\right)<\infty. $$ In this case, one defines $$ \int f\ d\mu=\int\max(f,0)\ d\mu-\int\max(-f,0)\ d\mu, $$ which makes sense since we are not subtracting $\infty$ from $\infty$. (Note that $\max(f,0)-\max(-f,0)=\max(f,0)+\min(f,0)=f$, so it is not an unreasonable definition.)

In summary, the phrase "$f$ is integrable" has a very specific meaning, namely that $\int |f|\ d\mu<\infty$. There are some non-integrable functions for which $\int f\ d\mu$ can be unambiguously defined. But there are also some functions for which $\int f\ d\mu$ cannot be defined - changing the order in which you traverse the domain of integration changes the result.


Now that I've addressed your first bullet point, let me move on to the rest.

  • Step 4 depends on step 3 to define $\int |f|\ d\mu$ in the condition $\int |f|\ d\mu<\infty$. You cannot use step 2 to define $\int |f|\ d\mu$ in general, since there might not be a set $E$ such that $|f|$ is bounded on $E$ and vanishes off $E$. For example, $f(n)=n$ on the measure space $\mathbb N$ with counting measure $\mu$ has the property that $|f|$ is not of the form that step 2 applies to.

  • Step 3 depends on step 2 since every $h$ appearing in the supremum of step 3 satisfies the condition of a function that step 2 applies to. Moreover, note that all $\phi$ appearing step 2 are, in fact, bounded, since $a_i\in\mathbb R$ and thus $\sup\phi = \max\{a_1,\ldots,a_n\}<\infty$. Note that this construction of the integral applies only to functions $f$ taking values in $\mathbb R$, meaning that $\pm\infty$ is not allowed to be the value of the function at any point.

  • In step 2, note that functions may not be ordered, but they do have a natural partial order given (as you wrote) by $f\leq g$ meaning that $f(x)\leq g(x)$ for all $x\in \Omega$.

  • Regarding your question of how it all ties together in order to calculate explicit answers to certain integrals: the answer is, use calculus (e.g. Riemann integral, change of variables formulas, integration by parts, etc) if you want to compute explicit numbers. Or better yet, use numerical integration! The point of the Lebesgue integral is that it is convenient for proving theorems about. There are theorems showing that all of the familiar calculus operations can be done (subject to suitable hypotheses) on the Lebesgue integral. These theorems are proved using the abstract properties of the Lebesgue integral, like Durrett constructed as you have written above. The point of the Lebesgue integral is not to be able to calculate more, but rather to be able to calculate more rigorously. Having said that, there are some standard tricks for proving results about the Lebesgue integral (especially in probability theory), one of which that comes to mind is Dynkin's theorem which is actually more of a technique - allowing one to prove things about all measurable functions by considering the set of all functions for which the result is true, and then showing that set is closed under some natural operations. It is surprisingly powerful, and means that many results about the Lebesgue integral need only to be proven in the case of a simple function (when everything is very concrete) and then use an approximation argument like Dynkin's theorem to complete.

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  • $\begingroup$ Thank you I will carefully go through your answer $\endgroup$ – Olórin Jul 16 '19 at 16:36

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