I have to decice if the following function is Lebesgue-integrable on $[0,1]$:
$$g(x)=\frac{1}x\cos\left(\frac{1}x\right) $$ where $x\in[0,1]$.
$g(x)$ is Lebesgue integrable if and only if the integral of $|g(x)|$ is finite
So, $\int_{[0,1]} |\frac{1}x\cos\left(\frac{1}x\right)| dm < infinite???$
I don't know how to proof that,
I have thought about using the monotone convergence theorem but I don't have any idea of how to define $f_n$ a sequence of measurable functions.
On
First observe that: $$ \int_0^1 |g(x)|~dx=\int^1_0 \left|\frac{1}x\cos\left(\frac{1}x\right)\right|~dx\\ =\int^\infty_1 \left|\frac{1}x\cos\left(x\right)\right|~dx $$ We have: $$ \int^\infty_1 \left|\frac{1}x\cos\left(x\right)\right|~dx\geq \sum_{n=1}^{\infty}\int^{2(n+1)\pi}_{2n\pi} \left|\frac{1}x\cos\left(x\right)\right|~dx\\ \geq \sum_{n=1}^{\infty}\frac{1}{2(n+1)\pi}\int^{2(n+1)\pi}_{2n\pi} \left|\cos\left(x\right)\right|~dx\\ =\int^{0}_{2\pi} \left|\cos\left(x\right)\right| ~dx\times \sum_{n=1}^{\infty}\frac{1}{2(n+1)\pi} $$ The right hand side diverges and hence the integral diverges.
Heuristically, each peak in the graph of $|\frac1x\cos\frac1x|$ contributes to the integral by something roughly proportional to the product of its height and width.
The height of the peak centered at $x_0$ is of course $\frac1{x_0}$. The width is inversely proportional to the (absolute value of) the derivative of $\frac 1x$ at $x_0$ (since all the peaks of $\cos$ itself are equally wide).
So, roughly, the peak at $x_0$ has area proportional to $\frac{1/x_0}{\left|-1/x_0^2\right|} = x_0$ itself.
There's a peak at $x_0 = \frac{1}{(n+\frac12)\pi/2}$ for each integer $n$. The sum of these areas are a harmonic series, which diverges.
It is probably possible to convert this to a rigorous proof by bounding $|g(x)|$ from below by a sequence of disjoint rectangles (one for each peak) whose total areas diverge.