Showing that if the integral of $f$ on every $[x,2x]$ is a constant, then $f$ is the zero function Let $f(x)$ be integrable on $[0,\infty)$ 
we will assume that the following accurs:
$\exists C $ a constant so that $\int_x^{2x}f(t)dt=C$ $\forall x>0$ 
We need to prove that $f(x)=0$ $\forall x \in [0, \infty)$
and also explain why $f(x)=1/x$ does not contradicts what we are proving.
I actually tired a couple of things, but all seemed to fail, Would really appreciate any kind of leads here :] thanks in advance!
 A: I'll give a proof, in case that $f$ is continuous on $[0,\infty)$. 
$$\int _x^{2x}f(t)dt=C ,\forall x>0\Rightarrow $$
$$f(2x)\cdot(2x)'-f(x)\cdot(x)'=0,\forall x>0\Rightarrow $$
$$2f(2x)=f(x),\forall x>0$$
Take a $x>0$ , then:
$$f(x)=\frac12f\left(\frac x2\right)=\frac{x}{2^2}f\left(\dfrac{x}{2^2}\right)=\dots=$$
$$=\frac{1}{2^n}f\left(\frac{x}{2^n}\right),\forall n\in\mathbb N, \forall x>0$$
Therefore :
$$f(x)=\lim\limits_{n\rightarrow \infty }f(x)=\lim\limits_{n\rightarrow \infty }\frac{1}{2^n}f\left(\frac{x}{2^n}\right)$$
Since $f$ is continuous it follows that $\lim\limits_{n\rightarrow \infty }f(\frac{x}{2^n})=f\left(\lim\limits_{n\rightarrow \infty }\frac{x}{2^n}\right)=f(0)$
Thus,
$$f(x)=\lim\limits_{n\rightarrow \infty }\frac{1}{2^n}f\left(\frac{x}{2^n}\right)=0\cdot f(0)=0,\forall x>0$$
Due to continuity, $f(x)=0,\forall x\geq 0$
Regarding the counterexample
$$f(x)=\dfrac{1}{x}$$ is not continuous in $[0,\infty)$
and the proof I have above fails when you take $\lim\limits_{n\rightarrow \infty }\frac{1}{2^n}f\left(\frac{x}{2^n}\right)=0\cdot f(0)=0\cdot \infty$
A: I'm assuming that by $f(x)$ integrable on $[0,\infty)$ you mean that $\int_0^\infty f(x)\,dx$ exists. (If as a Riemann or as a Lebesgue-Integral I leave open here).
If $f(x)$ is integrable on $[0,\infty]$ it must also be integrable on $[1,\infty]$. Now rewrite that integral as $$
  \int_1^\infty f(x)\,dx = \int_1^2 f(x)\,dx + \int_2^4 f(x)\,dx + \ldots \int_{2^n}^{2^{n+1}} f(x)\,dx + \ldots
$$
Can this sum be finite? Compare with $\int_x^{2x} f(x)\,dx = C$. There's exactly one value of $C$ for which this works...
Now, for $f(x)=\frac{1}{x}$ you obviously have $\int_x^{2x} f(x)\,dx = \ln 2$. This is only a contradiction if $\int_1^\infty \frac{1}{x}\,dx$ exists. Does it?
A: You can only conclude that $f=0$  a.e. on $[0,+\infty)$.
Proof: Given $x>0$, 
$$\int^{2^n x}_x f(t)dt=\sum_{k=0}^{n-1}\int_{2^k x}^{2^{k+1} x} f(t)dt=nC.\tag{1}$$
Letting $n\to\infty$, since $f$ is integrable on $[0,+\infty)$, the left hand side of $(1)$ tends to $\int_x^{+\infty} f(t)dt$, which is a finite. It implies that $C=0$, and hence
$$\int_x^{+\infty} f(t)dt=0,\quad \forall x>0.\tag{2}$$
From $(2)$ it follows that $f=0$ a.e. on $[0,+\infty)$. $\quad\square$
