# integral definition: Durrett's vs Rudin's

I'm a little uncertain about the definition of integral of measurable functions. First I read W. Rudin's book (PMA), where integration was defined in 3 steps: (1) integrals of simple functions, (2) integrals of non-negative functions, and then (3) integrals of measurable functions. Specifically,

(2): $\int_E fd\mu\triangleq \sup \int_E sd\mu,$ where $f\ge 0,$ and $s$ are simple functions with $s\le f.$

(3): $\int_E fd\mu\triangleq \int_E f^+d\mu-\int_E f^-d\mu,$ where $f^+\triangleq \max(f,0),$ and $f^-\triangleq \max(-f, 0).$

Then I read R. Durrett's book (Probability: Theory and Examples). The definition was almost the same, except for one difference: Durrett defined the integral of a bounded function, and then he defined the integral of non-negative functions (The rest are the same.), i.e.

Let $f$ be a bounded function on $E$, with $\mu(E)<\infty.$ Then, $\int_E fd\mu\triangleq \sup \int_E sd\mu$, where $s\le f$ are simple functions.

Let $f\ge 0$. Then, $\int fd\mu\triangleq\sup \int hd\mu$, where $h$ is bounded, with $\mu(\{x:h(x)>0\})<\infty$, and $0\le h\le f.$

I have 2 questions:

1) Are these 2 definitions equivalent?

2) Rudin's definition appears simpler to me. Why do we bother with Durrett's definition, defining integral of bounded functions and then the integral of non-negative functions? Are there advantages in doing so?

Thanks a lot!

• Don't know the answer, but have a question: how are unbounded functions handled in the second definition, since the first doesn't seem to have that restriction? – ChargeShivers Aug 21 '17 at 17:19
• @ChargeShivers It's handled by $\int fd\mu \triangleq \sup \int hd\mu$, where $h$ is bounded, isn't it? – syeh_106 Aug 22 '17 at 2:28

## 2 Answers

I would suggest using Rudin's definition. Durrett's definition is great for, say, the Lebesgue measure, but may fail if $E$ is not $\sigma$-finite.

For an example, consider $E=[0,1]\cup\{a\}$ where $a\notin[0,1]$. Let $\mu(A)$ be the Lebesgue measure of $A$ if $A\subset[0,1]$, and give the point $a$ infinite measure. Let $f(x)=0$ for $x\in[0,1]$ and $f(a)=1$. Under Rudin's definition, this is just a simple function, and $\int fd\mu=\infty$. I seem to recall that Durrett requires a simple function to be supported on a finite measure set, in which case $f$ is not simple. In this case, the only bounded, measurable function $h$ such that $0\le h\le f$ and $\mu\{x:h(x)>0\}<\infty$ is $h=0$, and so we have $\int fd\mu=0$. If I am mistaken and Durrett does call this function simple, we have an even bigger problem: $\int fd\mu=\infty$ when considered a simple function, but $\int fd\mu=0$ when considered a non-negative function.

If you assume $E$ is $\sigma$-finite then the definitions are equivalent. To see this, suppose $f$ is a nonnegative, measurable function, and denote by $\int f$ and $\int^*f$ respectively the Rudin and Durrett definitions of the integral. If $h\ge0$ is a bounded measurable function such that $\mu\{x:h(x)>0\}<\infty$, and $s\ge0$ is a simple function such that $s\le h$, then $0\le s\le f$, so $\int s\le\int f$. Optimizing over $s$, we find $\int h\le\int f$. Optimizing over $h$, we find $\int^*f\le\int f$. Now let $s\ge0$ be any simple function such that $s\le f$. Write $s=\sum_{i=1}^Na_i\mathbf1_{A_i}$. Since by assumption $E$ is $\sigma$-finite, $E=\bigcup_{n=1}^\infty E_n$ where $\mu(E_n)<\infty$. Then $s_n:=\sum_{i=1}^Na_i\mathbf1_{A_i\cap E_n}$ is bounded, measurable and $\{x:s_n(x)>0\}\subset E_n$ which has finite measure, so $\int s_n\le\int^*f$. Letting $n\to\infty$ and using continuity of measure, we find $\int s\le\int^*f$. Optimizing over $s$, we find $\int f\le\int^*f$.

Durrett's definition is great for the Lebesgue measure, or for any finite measure space (Durrett is a probabilist, so the issue I highlight is not a problem for him). As Kenny Wong mentions, the bounded convergence theorem is very simple to prove and can be used as a stepping stone to build other convergence theorem. I believe this is the way Durrett proceeds, as does Royden who at least at first is primarily interested in comparing Lebesgue and Riemann integrals. Rudin likely does not even consider the bounded convergence theorem its own theorem, since it is really just a special case of the dominated convergence theorem.

• Thanks a lot for the detailed answer! One question: to have $\int s_n \to \int s$, I assume you meant $E_1 \subset E_2 \subset E_3 \subset ...$, right? – syeh_106 Aug 22 '17 at 3:20
• BTW, Durrett does assume $\mu(A_i)< \infty$ for a simple function, i.e. $s=\sum_i a_i1_{A_i}$. – syeh_106 Aug 22 '17 at 3:29
• Sorry, yes, I assume $E_n\subset E_{n+1}$. Of course if not you can always just replace by $\tilde E_n:=\bigcup_{k=1}^nE_k$. – Jason Aug 22 '17 at 18:30
• Indeed. Thanks a lot for the confirmation! – syeh_106 Aug 23 '17 at 1:46

Update:

Billingsley defines, in his Equation (15.3), the integral of a measurable function $f$ on $(\Omega,\mathcal F,\mu)$ as

$\int f\,\textrm{d}\mu = \sup\displaystyle\sum_i\left[\inf_{\omega\in A_i}f(\omega)\right]\mu(A_i)$

where the supremum is over all finite decompositions $\{A_i\}$ of $\Omega$ into measurable sets. This is in the same spirit as Rudin's.

Then in Problem 15.2 Billingsley shows that for a bounded function, its upper integral coincides with its lower integral (as is checked in Durrett's second step). And this: To define the integral as the common value is the Darboux-Young approach. The advantage of (15.3) as a definition is that (in the nonnegative case) it applies at once to unbounded $f$ and infinite $\mu$.'' I think this answers the original question. That is, what's given in Rudin is, as the first answer says, better as a definition; and to answer my own question, the step concerning bounded functions is, as far as definition is concerned, unnecessary.

This of course is not to say there is no value in checking the identity between the upper and lower integrals. Billingsley adds The definition (15.3) always makes formal sense (for finite $\mu(\Omega)$ and $\sup|f|$), but they are reasonable---accord with intuition---only if [the upper and lower integrals coincide].''

I'd prefer defining the integral in three steps and then checking the upper and lower integrals coincide for bounded functions. But who am I to make the call?

Original Post:

Thank you very much for your question and answer, but I still don't understand Durrett's point. And I wished to leave a comment, but I can't due to lack of enough reputation, so pardon me leaving one here; I'll delete it once it gets noticed. (I thought initiating a duplicate is the worse sin.)

To me, his proof of the bounded convergence theorem does not seem to require the second step (the step mentioned in the original question where the Lebesgue integral is defined for bounded functions on sets with finite measures). My guess is he wanted to make sure that for bounded functions, the Lebesgue integral equals the supremum of the minorizing simple functions and the infimum of the majorizing simple functions, which is not necessary. Is there another reason, other than it makes the proof of the bounded convergence theorem easier, why Durrett's second step is necessary?