Let $f:[0,1]\rightarrow\mathbb{R}$ be an integrable function such that $$\lim_{n\to\infty}n\int\limits_{x}^{x+\frac{1}{n}}f(t)dt=0$$ for all $x\in[0,1)$. Show that $$\int\limits_{a}^{b}f(t)dt=0$$ for all $a,b\in(0,1)$.
I have changed the variable of integration. But it is not helping much
Let us assume that $f$ is Riemann integrable on $[0,1]$ and let $$F(x) =\int_{0}^{x}f(t)\,dt$$ Then it follows that every subinterval $[a, b] \subseteq [0,1]$ contains a point of continuity of $f$. And at these points $F'(x) = f(x) $. By the given limit condition in question it follows that $F'(x) =f(x) =0$ at these points.
Now consider any Riemann sum for $f$ over any partition of $[a, b] \subseteq [0,1]$ where the tag points are chosen such that $f$ is continuous there (and thus vanishes there). Then there is a Riemann sum for each partition which is identically $0$. It follows that the Riemann integral of $f$ on interval $[a, b] $ is $0$.
