Suppose $f:[0,+\infty)\to \mathbb{R}$ and $g:[0,+\infty)\to \mathbb{R}$ are two Lebesgue integrable functions on $[0,+\infty)$ such that $$|f(s)-g(s)|\leq w(Ms)\qquad \forall s\geq 0,$$ where $w:[0,+\infty)\to [0,+\infty)$ is a non decreasing function such that $w$ is continuous at $t=0$ and $w(0)=0$ and $M>0$ is fixed.
Is it true that for all $t\geq 0$ $$\int_{0}^{t}f(s)ds=\int_{0}^{t}g(s)ds +o(t)$$ where $o(t)$ indicates a function $h(t)$ such that $\lim_{t\to 0^+}\frac{|h(t)|}{t}=0$ and $|h(t)|\leq w(Mt)t$?
Can anyone prove/disprove it? Thanks a lot in advance.
I'm not sure what role $M$ is supposed to play. Take $M = 1$.
All you need is monotonicity of the integral. If $0 \le s \le t$ then $f(s) - g(s) \le \omega(s) \le \omega(t)$ so that
$$ \int_0^t f(s) \, ds - \int_0^t g(s) \, ds \le \int_0^t \omega(s) \, ds \le t \omega(t).$$
Similarly $$\int_0^t g(s) \, ds - \int_0^t f(s) \, ds \le \int_0^t \omega(s) \, ds \le t \omega(t).$$ Write $$h(t) = \int_0^t f(s) \, ds - \int_0^t g(s) \, ds.$$ Then you have $$|h(t)| \le t \omega(t)$$ and $$\frac{|h(t)|}{t} \le \omega(t) \to 0 \quad \text{as} \quad t \to 0^+.$$