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Definition of Lebesgue integral of simple function: We say that a simple function $\psi$ is Lebesgue integrable if the set $\{\psi \ne 0\}$ has finite measure. In this case, we may write the standard representation for $\psi$ as $\psi = \sum_{i=0}^n a_i \chi_{A_i}$, where $a_0 = 0, a_1, \ldots , a_n$ are distinct real numbers, where $A_0 = \{\psi = 0\}, A_1, \ldots , A_n$ are pairwise disjoint and measurable, and where only $A_0$ has infinite measure, Once $\psi$ is so written, there is an obvious definition for $\int \psi$, namely, $$\int \psi = \int_{\mathbb R} \psi = \int_{-\infty}^{+\infty} \psi(x) \, \text dx = \sum_{i=1}^n a_i m(A_i).$$ In other words, by adopting the convention that $0 \cdot \infty = 0$, we define the Lebesgue integral of $\psi$ by $$\int \sum_{i=0}^n a_i \chi_{A_i} = \sum_{i=0}^n a_i m(A_i).$$ Please note that $a_im(A_i)$ is a product of real numbers for $i \ne 0$, and it is $0 \cdot \infty = 0$ for $i = 0$; that is, $\int \psi$ is a finite real number.

Definition of Lebesgue integrable of non-negative function: If $f: \mathbb R \to [0, +\infty]$ is measurable, we define the Lebesgue integral of $f$ over $\mathbb R$ by $\int f = \sup \left\{\int \psi: 0 \le \psi \le f, \psi\text{ simple function and integrable }\right\}$.

How to prove $\int_{[0, +\infty)} \frac{2}{1+x^2} \, \text dx$ Lebesgue integrable?

Besides, is there any relationship between Lebesgue integrable and Riemann integrable? I mean does Riemann integrable imply Lebesgue integrable?

Falrach
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Bear and bunny
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    Theorem: If $f$ is continuous on $[a,b]$ it is Lebesgue integrable, and its Lebesgue integral equals its Riemann integral. Theorem: MCT. – David C. Ullrich Aug 03 '15 at 17:47
  • @DavidC.Ullrich: MCT? What is MCT? – Bear and bunny Aug 03 '15 at 17:50
  • Do you instead want to ask if $x\mapsto \dfrac{2}{1+x^2}$ is integrable on $[0, +\infty[$? – Git Gud Aug 03 '15 at 17:50
  • Monotone Convergence Theorem – David C. Ullrich Aug 03 '15 at 17:52
  • @GitGud: Almost, but on [0, $+\infty$). Not extended real value. – Bear and bunny Aug 03 '15 at 17:52
  • It is difficult to understand precisely want you're looking for. Do you want a proof that $\int_{[0, +\infty)} \frac{2}{1+x^2} dx$ is Lebesgue integrable just using the definition of Lebesgue integrability? – mathcounterexamples.net Aug 03 '15 at 17:52
  • @FlybyNight What you said holds for functions on $[a,b]$, but not for improper Riemann integrals. – David C. Ullrich Aug 03 '15 at 17:53
  • @Bearandbunny $[a,+\infty[$ doesn't include $+\infty$. Thanks for answering. – Git Gud Aug 03 '15 at 17:54
  • $[a,b[$ is a standard notation for $[a,b)$. – David C. Ullrich Aug 03 '15 at 17:54
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    Possibly the point to Git's question is this: You didn't ask whether a certain function was Lebesgue integrable, you asked whether a certain integral was Lebesgue integrable! Asking whether $\int f$ is Lebesgue integrable makes no sense, because $\int f$ is not a function. – David C. Ullrich Aug 03 '15 at 17:55
  • @mathcounterexamples.net: I have the same question. I think the only way can be used is definition of Lebesgue integrability. However, I don't know how to find the simple function which can be applied to prove. Or do you have some new idea instead coz using definition for proof is a little bit complicated. – Bear and bunny Aug 03 '15 at 17:55
  • @GitGud: Sorry for my known symbol $[a, b[$. – Bear and bunny Aug 03 '15 at 17:57
  • @DavidC.Ullrich: David, did you mean prove Lebesgue's Criterion for Riemann integrability by performing MCT? I seemed get your idea. – Bear and bunny Aug 03 '15 at 21:21
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    No. I meant the function is integrable on $[0,b]$ because of what I said. Now you can let $b\to\inffy$ and use MCT. – David C. Ullrich Aug 03 '15 at 21:26
  • @DavidC.Ullrich: Ohhhh, make sense. David, you are guru mastering anything in classical and modern mathematical analysis. You have a lot of intuitive ideas. How long have you been this field and do you have some advice or books introduced to me? – Bear and bunny Aug 03 '15 at 21:33
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    Naah. 30 years, since you ask. Advice, just keep studying. OK. Some day when you learn some complex analysis don't overlook Complex Made Simple. There's a few things missing there but people say they enjoy what's included – David C. Ullrich Aug 04 '15 at 00:15
  • @DavidC.Ullrich: Appreciate your words. I'll follow. – Bear and bunny Aug 04 '15 at 00:26

2 Answers2

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Hint.

Slice the image of $f(x)=\frac{2}{1+x^2}$, i.e. $(0,+\infty$ in equal slices of thickness $\frac{1}{n}$. Compute the reverse image of each slice which is for each a union of a pair of disjoint intervals.

Take the measure of each reverse slices and make the Lebesgue sum. Then prove that this is the $\sup$ of all $\{\psi \le f, \psi\text{ simple function and integrable }\}$

mathcounterexamples.net
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Proper Riemann integrability on a bounded interval implies Lebesgue integrability on that interval, and the integrals are equal. By "proper", I mean one need not evaluate any limits as the bounds of integration approach anything. Thus $$ \int_{[0,b]} \frac 2 {1+x^2} \, dx \tag 1 $$ as a Lebesgue integral is the same as the Riemann integral over that interval.

Notice that every non-negative measurable simple function that is $\le$ this function on $[0,\infty)$ is bounded above by $$ \begin{cases} \dfrac 2 {1+x^2} & \text{for }0\le x\le b, \\[8pt] 0 & \text{for }x>b \end{cases} $$ for some value of $b$. Thus the supremum of $(1)$ over all values of $b$ is $\ge$ the supremum of the set of all integrals of such simple functions. But the supremum of $(1)$ over all values of $b$ is finite.

Michael Hardy
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  • "Riemann integrability on a bounded interval implies Lebesgue integrability on that interval"? Do you have some documents or books talking it a little bit more? I mean I want a proof of it. Thanks. – Bear and bunny Aug 03 '15 at 20:53
  • @Bearandbunny It depends a bit on your definitions. One way to see it is that Riemann integrability is all about approximation by step functions, while Lebesgue integrability is all about approximation by simple functions. The important fact is then that step functions are simple functions. – Ian Aug 03 '15 at 22:55
  • There is a theorem that says a function is Riemann integrable if and only if it is continuous almost everywhere. I think it was from the last chapter of "Baby Rudin", i.e. Rudin's Principles of Mathematical Analysis, that I learned this. ${}\qquad{}$ – Michael Hardy Aug 03 '15 at 22:56
  • I'm watching this: http://math.stackexchange.com/questions/1065151/if-a-function-is-riemann-integrable-then-it-is-lebesgue-integrable-and-2-integr – Bear and bunny Aug 03 '15 at 23:08