If $f \le g$ and $f, g$ are integrable, decreasing functions, then $\int_{x}^{\infty} f \le \int_{x}^{\infty} g$?
Intuitively, I suppose it holds, but I have not found any such theorem in the literature.
In case the above proposition is not true, are there any special cases for $f, g$ such that the proposition becomes true?
You can easily show that if $f\leq g$ and both $f$ and $g$ are defined and integrable on $[a,b]$, then $$\int_a^b f(x)dx\leq \int_a^b g(x)dx.$$
You can see this because $\int_{a}^b f(x) dx$ is the infimum of upper Darboux sums over all possible subdivisions of $[a,b]$. For every division, the upper Darboux sum of $f$ will be smaller or equal to the upper Darboux sum of $g$, so the infimum will also be smaller or equal.
Once you have that, you can use the definition of the generalized integral to get
$$\int_x^\infty f(x) dx = \lim_{b\to\infty}\int_x^b f(x) dx$$
and the inequality follows because the integral of $g$ is a limit of a larger function. The function $F:b\mapsto \int_x^b f(x)dx$ is smaller than $G:b\mapsto\int_x^bg(x)dx$, and limits are monotonous.
That is, if you know that for all $x$, $F(x)\leq G(x)$ and both $$L_f=\lim_{x\to\infty} F(x)$$ and $$L_g\lim_{x\to\infty} G(x)$$ exists, then the first limit is smaller or equal to the second. This is very easy to prove, since, if $L_f>L_g$, you would have some $M$ such that for all $x\geq M$, the value of $f(x)$ is within $\epsilon$ of $L_f$ and the value of $L_g$ is within $\epsilon$ of $L_g$. Setting $\epsilon = \frac{L_f-L_g}{3}$ ensures that then, $$f(x)\geq L_f-\epsilon>L_g+\epsilon \geq g(x)$$ which is a contradiction.