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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, $$\int \psi = \int_{\mathbb R} \psi = \int_{-\infty}^{+\infty} \psi(x) \, dx = \sum_{i=1}^n a_i m(A_i).$$ 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 $\int_{[0, +\infty)} \frac{2}{1+x^2} \, dx$ Lebesgue integrable?

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

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    $\begingroup$ Theorem: If $f$ is continuous on $[a,b]$ it is Lebesgue integrable, and its Lebesgue integral equals its Riemann integral. Theorem: MCT. $\endgroup$ – David C. Ullrich Aug 3 '15 at 17:47
  • $\begingroup$ @DavidC.Ullrich: MCT? What is MCT? $\endgroup$ – Bear and bunny Aug 3 '15 at 17:50
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    $\begingroup$ 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. $\endgroup$ – David C. Ullrich Aug 3 '15 at 17:55
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    $\begingroup$ 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. $\endgroup$ – David C. Ullrich Aug 3 '15 at 21:26
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    $\begingroup$ 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 $\endgroup$ – David C. Ullrich Aug 4 '15 at 0:15
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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.

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  • $\begingroup$ "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. $\endgroup$ – Bear and bunny Aug 3 '15 at 20:53
  • $\begingroup$ @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. $\endgroup$ – Ian Aug 3 '15 at 22:55
  • $\begingroup$ 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. ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 3 '15 at 22:56
  • $\begingroup$ I'm watching this: math.stackexchange.com/questions/1065151/… $\endgroup$ – Bear and bunny Aug 3 '15 at 23:08
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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 $\{\psi \le f, \psi\text{ simple function and integrable }\}$

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  • $\begingroup$ Yes. So you agree that definition can be the only way to prove it? $\endgroup$ – Bear and bunny Aug 3 '15 at 20:57
  • $\begingroup$ No. It depends on what you know and what you were asked for. There are many other ways to prove the result than using Lebesgue integrability definition. $\endgroup$ – mathcounterexamples.net Aug 3 '15 at 20:59
  • $\begingroup$ Thanks. I want to upvote your answer. It is an intuitive method for calculate Lebesgue integral or prove Lebesgue integrability. $\endgroup$ – Bear and bunny Aug 3 '15 at 21:28

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