How to show $\int_{[0, +\infty)} \frac{2}{1+x^2} \text dx$ Lebesgue integrable? 
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?
 A: 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 $\{\psi \le f, \psi\text{ simple function and integrable }\}$
A: 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.
