Given $a>0$, $n \in \mathbb{N}$ and a non-negative function $f \in L^{1}(\Omega)$ satisfies
$$\int_{\Omega} f(x) dx \le (n+1)a$$
Find a dissection $\left\{\Omega_j \right\}_{j=1}^{n+1}$ of $\Omega$ such that for all $1\le j \le n+1$ we have $$\int_{\Omega_j}f(x)dx \le a$$
I tried to find $n$ disjoint subsets $\Omega_j$ of $\Omega$ that satisfies $\int_{\Omega_j}f(x)dx \le a \hspace{1mm} \forall \hspace{1mm} 1 \le j \le n$ but i'm not sure that $\int_{\Omega_{n+1}}f(x)dx \le a$ with $\Omega_{n+1} =\Omega \setminus \cup_{j=1}^{n} \Omega_j$. Thanks everyone !