Suppose $f\in L_1([0,1])$ and $\int_0^1 f\leq C$. Then thera are pairwise disjoint intervals $(a_j,b_j)$ such that
$$
\frac1{b_j-a_j}\int_{a_j}^{b_j}f(t)\,dt=C
\quad\mbox{and}\quad
\forall t\notin\bigcup(a_j,b_j):\quad\frac1t\int_0^tf(s)\,ds\leq C~.
$$
Put $g(t)\colon=\int_0^t f(s)-C\,ds$; then $g(a_j)=g(b_j)$ means:
$$
\int_{a_j}^{b_j} f(s)\,ds=C(b_j-a_j)
$$
and for $t\notin U(g)$: for all $s > t$: $g(t)\leq g(s)$ and in particular for $s=1$: $g(t)\leq g(1)$, i.e. $\int_0^t f(s)\,ds\leq tC$.