I'm confused about the Lebesgue integral

Question

As I understand, the Lebesgue integral is the limit of the sum of the areas of horizontal (lying) regtangles that approximate the area under the curve. And this is the main difference with the Riemann integral which is the limit of vertical (standing) regtangles.

If the function $f: (X,\Sigma)\rightarrow (Y,B(Y))$ is simple then it can be expressed as follows: $$f(x)=\sum_{i=1}^{n}1_{F_i}(x)\cdot c_i$$ where the sets $F_i$ may intersect with each other and $\forall i\quad F_i \in \Sigma$ and $\cup_iF_i=X$.

A nonnegative function $f$ can be approximated by simple functions and a general function $f$ can be represented as $f=f^++f^-$, where $f^+=\max(0,f)$ and $f^-=\min(0,f)$. Hence $$\int f(x)\mu(dx)=\int f^+(x)\mu(dx)+\int f^-(x)\mu(dx)$$ Since $|f|=f^+-f^-$, $f$ is integrable if and only if $|f|$ is integrable. Now I'm, confused.

Does it mean, that we can't use the Lebesgue integral to calculate

$$\int_0^{+\infty}\frac{\sin(x)}{x}dx$$

Because the function $g(x)=\frac{|\sin(x)|}{|x|}$ is not integrable on $(\epsilon,+\infty)$, where $\epsilon>0$

I'm confused because I think that it is possible to calculate the Riemann integral of this function $\frac{\sin(x)}{x}$. However, every Riemann integrable function is also Lebesgue integrable.

Answer 1

The Riemann integral is only defined for finite intervals.

$$\int\limits_{a}^{b}f(x)dx = \lim_{n \to \infty} \frac{(b-a)}{n}\sum_{k = 1}^{n}f(a + \frac{k(b-a)}{n})$$

This expression only make sense if $a, b \in \mathbb R$.

Your integral is actually an improper integral. It is the limit of Riemann integrals, but it is not a Riemann integral itself:

$$\int_0^{+\infty}\frac{\sin(x)}{x}dx = \lim_{b\to +\infty} \int_0^{b}\frac{\sin(x)}{x}dx$$

Answer 2

You can show that $\int_0^{+\infty}\frac{\sin(x)}{x}dx\tag 1$

converges as an improper Riemann Integral (integrate by parts) but that

$\int_0^{+\infty}\left | \frac{\sin(x)}{x} \right |dx\tag 2$

does not converge i.e. it is not Lebesgue integrable. (use convexity of $f(x)=1/x$ and compare to a harmonic series).

It is true however, that if $\int_{a}^{b}f(x)dx\tag 3$

converges as a Riemann integral, then it is Lebesgue integrable (use the fact that the upper and lower sums are simple functions which can be used in an application of the MCT).