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I am trying to understand Lebesgue integration here. Here you basically make equal splits on the y-axis instead of the splits on the x-axis that Riemann does. I understand the proofs of the limit theorems like MCT and DCT, but I do not see how a simple idea as splitting the y-axis leads to such wonderful theorems, which Riemann is not amenable to.

Is there an easy way to understand why Lebesgue integration is conducive to limit theorems while Riemann is not? This is keeping in mind their intuitive explanation of creating equal strips along an axis and adding the areas up.

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See Henstock-Kurzweil integral ( en.wikipedia.org/wiki/Henstock%E2%80%93Kurzweil_integral ) for an easy modification of Riemann's integral, which satisfies all the convergence properties. It is not a fault of division of the $x$-axis, rather of uniform division. Lebesgue integral has other advantages though. –  user8268 Mar 26 '12 at 7:04
    
"rather of uniform division": but you do divide the y-axis uniformly even in Lebesgue integration? –  Bravo Mar 26 '12 at 7:40
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@Bravo: In many presentations of Lebesgue integration, nothing is "partitioned" at all. On the other hand, in many presentations of Riemann integration the partitions on the $x$-axis do not need to be uniform, they can be arbitrary finite partitions. So uniformity of partitions is not the issue. –  Carl Mummert Mar 26 '12 at 10:43
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@Carl Mummert: In Riemannian integration there is a uniform estimate on the lengths of the subintervals. That's exactly what is removed in the HK integral. This uniformity is the issue (if you divide the $x$-axis). Sorry for not stating it clearly in my previous comment. –  user8268 Mar 26 '12 at 10:53
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@Carl: can they really be "arbitrary partitions"? Riemann integrability takes the form of "for each $\epsilon$, for all partitions $P$ with property $Q(\epsilon)$, $|\sum_P f - s| < \epsilon$". There's gotta be some condition $Q(\epsilon)$. Else with arbitrary finite partitions there's no hope that we can say anything like $|\sum_P f - s| < \epsilon$. –  Willie Wong Mar 26 '12 at 14:07
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This is a follow up on Michael Greinecker's answer. I don't see the issue as having to do with uniform partitions at all; there is no requirement in the definition of either integral that anything has to be partitioned uniformly.

The Lebesgue integral of a non-negative measurable function $f$ is defined as the supremum $S_f$ of the "areas" of non-negative "Lebesgue simple" functions that are bounded by $f$ (the supremum may be $\infty$). These functions were defined in Michael Greinecker's post: they are finite linear combinations of characteristic functions of measurable sets. The supremum $S_f$ is well-defined for every non-negative measurable function. The proofs of many convergence theorems come down to showing that other limit operations can be exchanged (commuted) with this supremum operation. Note that the definition of the Lebesgue integral phrased this way is not about partitions at all: it is just about how much area can fit below the graph of a non-negative measurable function.

The extension of the Lebesgue integral to general (not necessarily non-negative) measurable functions is routine, by dividing the function into positive and negative parts that are handled independently.

On the other hand, the Riemann integral is defined in a much more restrictive way. In order determine whether a Riemann integral of a non-negative measurable function $f$ exists, it is not enough to look at the integrals of nonnegative "Riemann simple" functions that are bounded by $f$. These functions are finite linear combinations of characteristic functions of intervals, as Michael Greinecker says. For example, this supremum is perfectly well defined for $f = 1_\mathbb{Q}$, and equal to 0, because every non-negative constant function on an interval that is bounded below $1_\mathbb{Q}$ must be identically zero on that interval. But $1_\mathbb{Q}$ is a famous example of a function that is not Riemann integrable. The problem, of course, is that the definition of the Riemann integral requires more than just the convergence of that supremum.

Thus proving that a non-negative function is Riemann integrable requires more than showing that a particular supremum exists, and this causes corresponding problems in proving limit theorems for Riemann integration. On the other hand, the Lebesgue integral exists for every measurable non-negative function (it may be $\infty$), so this problem simply doesn't exist in the setting of Lebesgue integration. There is an extra step in proving a limit theorem for Riemann integration, and sometimes this step is insurmountable.

There is a theorem that a bounded function on a closed bounded interval is Riemann integrable if and only if it is Lebesgue integrable and is continuous almost everywhere on the interval. This shows, in hindsight, an intuition for why limit theorems must fail in the Riemann case: because "continuity almost everywhere" is not preserved by the sorts of limit operations that are considered in these theorems. For example, $1_\mathbb{Q}$ is a pointwise limit of an increasing sequence of functions $(f_i)$ each of which is continuous almost everywhere, as described by Michael Greinecker. Since pointwise limits don't respect "almost everywhere continuity", it will not be possible to show that Riemann integration acts well with respect to these limits. In the non-negative case, Lebesgue integration only requires that the limit function has to be measurable, and measurability is preserved under the sorts of limits that are considered.

P.S. The use of the extended real line $[-\infty,\infty]$ in the Lebesgue setting, but not typically in the Riemann setting, is another minor cause of difficulty in the Riemann setting. And, as usual, there can be confusion because in the Lebesgue sense "integrable" means "has a finite integral" rather than "the Lebesgue integral exists". Neither of these gets to the heart of the matter, which is that every non-negative measurable function has an integral in the Lebesgue sense, but this fails in the Riemann sense.

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The penultimate paragraph should probably read: "There is a theorem that a bounded function on a closed bounded interval is Riemann integrable if and only if it is continuous almost everywhere" (measurability follows from continuity a.e. and then integrability is clear by boundedness). –  t.b. Mar 26 '12 at 13:47
    
@t.b.: yes, that's true, I'll fix it the next time I edit the response. Thanks. –  Carl Mummert Mar 26 '12 at 13:49
    
sorry for spamming with my comments. The uniformity in Riemann's integral is: $\forall\epsilon>0\,\exists\delta>0$ such that Riemann's sums with subintervals smaller than delta are $\epsilon$-close to the integral. The $\delta$ (as an estimate) is the same for all the subintervals. In the HK integral you simply allow $\delta$ to be a function of $x$; that makes all Lebesgue-integrable functions also HK-integrable, and proving (the right forms, as there are HK-integrable functions that are not absolutely integrable) convergence theorems is relatively easy. –  user8268 Mar 26 '12 at 14:09
    
@user8268: I would usually think of the Riemann integral in terms of an net rather than in terms of $\epsilon/\delta$ as you have. In that setting, it says that there is some (tagged) partition $P$ such that any partition $Q$ refining $P$ is within $\epsilon$ of the value of the integral; there is no uniformity on the widths of the intervals in $P$. I do see the uniformity when you write the $\epsilon/\delta$ definition, though. –  Carl Mummert Mar 26 '12 at 15:16
    
@Carl: that last comment is very helpful (to me). It points clearly to the fact that in the HK integral a refinement of a $\delta$-fine partition need not be $\delta$-fine! –  Willie Wong Mar 26 '12 at 15:42
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This is essentially an elaboration on a point made in the comments. In both Lebesgue and Riemann integration, the integral is defined on simple functions made up of "blocks" and is then extended to more general functions by a limiting operation. The difference is really at the level of blocks. The blocks in Riemann integration are proper rectangles, the base is an interval. In function form, a rectangle of height $1$ can be written as $1_{[a,b]}$. The simple functions are linear combinations of blocks. In Lebesgue integration, the base may not be an interval, but an arbitrary measurable set, a block has the form $1_A$ with $A$ being some measurable set. The simple functions are again linear combinations of blocks. The difference between these integrals can already be seen when lloking at which functions are simple.

$1_\mathbb{Q}$ is a simple function in the setting of Lebesgue integration, since $\mathbb{Q}$ is a measurable subset of $\mathbb{R}$. But $\mathbb{Q}$ contains no proper interval. But $\mathbb{Q}$ is countable, so we can write it as $\mathbb{Q}=\{q_1,q_2,\ldots\}$. Let $f_n=1_{\{q_1,\ldots,q_n\}}$ for all $n$. Each $f_n$ is a simple function for both Lebesgue and Riemann integration. But the pointwise limit of the $f_n$ is $1_\mathbb{Q}$, a simple function for Lebesgue integration, but a function that cannot be well approximated by simple functions for Riemann integration.

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Well Riemann integral divides the domain in a partition with a point $\xi_i \in [t_i,t_{i+1}]$ and chooses $f(\xi_i)\in f([t_i,t_{i+1}])$ as a "good" representative of the values in $f([t_i,t_{i+1}])$. Not bad if your function is continuous or "nearly" continuous.

$RS(f)=\sum_{i}f(\xi_i)m([t_i,t_{i+1}])=\sum_{i}f(\xi_i)(t_{i+1}-t_{i})$

However Lebesgue integral divides the image $(f([a,b]))\subset\cup (x_i,x_{i+1}]$ in a partition of small interval and chooses a middle point $\tau_i$ and evaluates

$LS (f)=\sum_{i}\tau_i m(f^{-1}(x_i,x_{i+1}])$.

Here you need you need only that $f$ is measure that is the set is $f^{-1}(x_i,x_{i+1}]$ is Lebesgue measurable and a hypothesis of finite integral for $f^+$ and $f^-$.

The advantages are many as $L^1$ that is a Banach space, i.e., the limit of integrable funtions in the integral norm is still integrable, the pointwise limite is still integrable, regarded the limitation hypothesis!

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The abstract integral may be defined in a very simple way. Divide the domain into finite nonoverlapping subdivisions, take Inf{f(x)} on each subdivision, multiply it with "measure" of this subdivision, and sum all these multiplicands. Then take Sup{sums on all possible such finite subdivisions), and you get an integral. The key moment is to understand what "measure" means, and this is what the Lebesgue theory deals with: it simply says that the subdivisions must be measurable, and the function must be measurable too, and thus you get Lebesgue integral. In case of Riemann integrability this abstract approach leads simply to Lower Darboux integral, and an Upper Darboux integral must additionally be calculated, and the two Darboux integrals must be equal. Thus we see that every Riemann integrable function is Lebesgue integrable too.

The answer is: because the Lebesgue integrability is much less restrictive than Riemann integrability, and the details are outlined on previous answers.

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