Show that there exists a $g \in L^1(m)$ s.t. $\phi (F(x))=\int_{0}^{x} g(t)dt$, where $F(x)=\int_{0}^{x} f(t)dt$

Question

Let $m$ be Lebesgue measure on $[0,1]$ and suppose $f \in L^1(m)$ and let $F(x)=\int_{0}^{x} f(t)dt$. Suppose $\phi$ is a Lipschitz function. Show that there exists a $g \in L^1(m)$ s.t. $\phi (F(x))=\int_{0}^{x} g(t)dt$

Not sure how to use the Lipschitz condition here.