I was doing exercise 5 of this exercise sheet and I don't know how to conclude.
I need to prove that if $f \colon [0,1]\to \mathbb{R}$ is Lipshitz, $X$ is a uniform$(0,1)$ random variable and $$X_n=\frac{\lfloor 2^nX \rfloor }{2^n}$$ then $$f(X_n + 2^{-n})-f(X_n)=\int_{X_n}^{X_n+2^{-n}}g(u)du$$ implies that $$f(x)-f(0)=\int_0^xg(u)du$$ for every $x \in [0,1]$
Any help will be appriciated.
You have proved that the equality
$$f(X_n+2^{-n}) - f(X_n) = \int_{X_n}^{X_n+2^{-n}} g(u) \, du \tag{1}$$
holds for any random variable $X \sim U(0,1)$ where $X_n := 2^{-n} \lfloor 2^n X \rfloor$. Now consider the measurable space $([0,1],\mathcal{B}([0,1]))$ endowed with the Lebesgue measure and set
$$X(x) := x, \qquad x \in [0,1].$$
Obviously, $X \sim U(0,1)$ and $$X_n(x) = \frac{\lfloor 2^n x \rfloor}{2^n} \to x \qquad \text{as} \, n \to \infty. \tag{2}$$
Moreover, for any $x \in [k/2^n,(k+1)/2^n)$, we have
$$f(X_n(x)+2^{-n})-f(0) = f((k+1)/2^n)-f(0) = \sum_{j=0}^{k} f((j+1)/2^n)-f(j/2^n).$$
Therefore, it follows from $(1)$ by summation that
$$f(X_n(x)+2^{-n})-f(0)= \int_0^{X_n(x)+2^{-n}} g(u) \, du.$$
Finally, we conclude from $(2)$ and the dominated convergence theorem that
$$f(x) - f(0) = \int_0^x g(u) \, du.$$