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We may approximate a bounded nonnegative function $f$ by simple functions $\phi$. Then the standard way of defining the Lebesgue integral of $f$ is $$ \int f d\mu := \sup \Bigg\{ \int \phi : \phi \leq f, \phi \textrm{ is simple}\Bigg\}.$$ I am very comfortable with this definition but it was not until recently while learning about Ito integrals that I began thinking about why the Lebesgue integral is not defined as a limit.

Loosely speaking, if $X_t$ is a stochastic process and $B_t$ a Brownian motion, then the Ito integral $\int_0^t X_s B_s$ is defined by first finding a sequence of elementary processes $E^n_t$ that converge to $X_t$ in $L^2$, and then defining $$\int_0^t X_s dB_s := L^2-\lim_{n \rightarrow \infty} \int E_s^n dB_s.$$ I am curious why a similar construction doesn't work for the Lebesgue integral. That is, if $\{\phi_n\}$ are simple and $\phi_n \rightarrow f$ uniformly, then why can we not define $$\int f d\mu := \lim_{n\rightarrow \infty} \int\phi_n d\mu.$$

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    $\begingroup$ The first definition you give is perfectly unambiguous; in the second definition you have to prove that the limit doesn't depend on the choice of $\phi_n$, and also that such a $\phi_n$ even exists (which may require some hypothesis on the underlying measure space). $\endgroup$ Commented May 30 at 2:33
  • $\begingroup$ @QiaochuYuan Thank you. Are there any known counterexamples where the limiting integral does depend on the choice of $\phi_n$? $\endgroup$
    – CBBAM
    Commented May 30 at 2:37
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    $\begingroup$ It doesn't (although I'm not sure if uniform convergence is the condition you want), but my point is that this has to be shown, if you use the second approach. The first approach doesn't require making any choices at all so you don't have to show any kind of independence of those choices. It's a clean and elegant place to start proving other facts. $\endgroup$ Commented May 30 at 2:46

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One of the very first theorems (essentially a corollary of the MCT) proved right after defining the Lebesgue integral as you’ve done (see e.g Folland/Rudin) is that for a non-negative measurable $f:X\to[0,\infty]$, there exists a sequence $\{\phi_n\}_{n=0}^{\infty}$ of simple functions which pointwise increase to $f$. Then, one shows that $\int_Xf\,d\mu$ defined as the supremum equals $\lim\limits_{n\to\infty}\int_X\phi_n\,d\mu$.

Or one could start off by showing (again, essentially a corollary of the MCT) that the limit is independent of the choice of simple-function exhaustion. If done this way, then the Lebesgue integral is clearly “sequence based”.

In fact, in other books like Amann-Escher Vol III, and Lang, they introduce Banach-valued integration, rather than going through $[0,\infty]$ first. In this approach, you have to abandon the supremum because there’s no ordering in a Banach space. See Why is the Lebesgue Integral defined through integrals of simple functions? for more details along these lines. Finally, regarding your last sentence, note that uniform convergence is the wrong notion when dealing with integrability; we want convergence in $L^1$, not uniformly (on a finite measure space, $L^{\infty}$ convergence implies $L^1$ convergence… but not conversely, so this is too strong a notion; in an infinite measure space, neither implication holds, so we had better choose the correct mode of convergence right from the beginning, which in our case, is $L^1$, since we’re trying to define the integral itself as a bounded linear map). Again, see the link above for how exactly to implement this.

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  • $\begingroup$ Thank you. So the supremum is a more intrinsic definition (when an ordering is present) because it doesn't require anything else to be shown (e.g. the limit is independent of choice of sequence)? I also read the link you gave and it seems, as you mentioned, pointwise convergence is what we want and not uniform convergence. I have to admit I'm not sure I understand the intuition behind why this is and I think I'm missing something obvious, but why is $\phi_n \rightarrow f$ pointwise the natural type of convergence when constructing an integral? $\endgroup$
    – CBBAM
    Commented May 30 at 3:23
  • $\begingroup$ pointwise convergence isn’t the natural mode of convergence either; it’s a non-trivial theorem (the MCT). That’s why in the link, you see things developed using $\mathcal{L}^1$-Cauchy sequences of simple functions $\{s_n\}$. That is the natural notion of convergence for the integral (sounds almost tautological once you think about it though). $\endgroup$
    – peek-a-boo
    Commented May 30 at 3:26
  • $\begingroup$ Ah I see, so because we have $\phi_n \rightarrow f$ and the $\phi_n$ satisfy the MCT hypotheses (assuming they are chosen so that they are nondecreasing) then we can invoke the MCT to define $\int f d\mu$. Now I see why the supremum definition is more natural when an order exists because the MCT is not needed to define the integral. $\endgroup$
    – CBBAM
    Commented May 30 at 3:38

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