Question about convergence of improper integral Could you give me some hint how to solve this problem:
Suppose $f$ is continuous on $(0,1]$ and there is $M$ such as $\left|\int_x^1f(t)\, dt \right|\le M$.
Prove that $\int_0^1f(x)\, dx$ converges or provide counter-example.
It seems to me that $F(x)=\int_\epsilon^1f(x)dx$ its continuous and differentiable function for each $\epsilon>0$, so the question is if $\lim_{\epsilon \to 0}F(x)$ exists.
Am I right ? How should I procede ? 
Thanks.

Following Siminore's hint 

Your $F$ is bounded. Can you imagine a bounded function that has no limit? Then you should ask yourself if this example can be produced by integration, as your $F$.

we define $f$ such that for each $x\in (0,1)$, 
$$\tag{*} \int_x^1 f(t)\mathrm dt=\sin\left(\frac 1x\right).$$
Since $x\mapsto \sin(1/x)$ is bounded and has no limit as $x$ goes to $0$, it remains to check that $f$ is continuous on $(0,1)$, which can be done by taking the derivative on both sides of (*).
