Let $f,g \in C^1(\mathbb{R}^n,\mathbb{R}^n)$. And we define for all $p\geq 1$, $(A_p)_p\subset \mathcal{M}$ sequence of measurable sets such that $A_p \subset [0,T]$ and $$\mu(A_p)\rightarrow 0 \,\,\,\,(p\rightarrow0).$$ And let $x_p,x \in C^1([0,T],\mathbb{R}^n)$ such that $(x_p)_p$ converges towards $x$ uniformly.
Can we say that : $\frac{1}{\mu(A_p)}\int_{A_p} ||g(x_p)-f(x_p)||_{\mathbb{R}^n}ds \rightarrow 0?$
Not true. Let $x_k=x$ for all $n$ and $A_k=(t-\frac 1 k, t +\frac 1 k)$. Then $\frac 1 {\mu (A_k)} \int_{A_k} |g(x_k(s))-f(x_k(s))|ds \to |g(x(t))-f(x(t))|$ as $ k \to \infty$.