Integration over finite partition of integration domain I think the title does not reflect my problem very well. Feel free to leave a comment with a more appropriate title.
Let $f \in L^1([0,1])$. How do I prove there exists a partition of $[0,1]$ into intervals $I_1 \dot\cup \dots \dot\cup I_N$ such that 
$$\int_{I_i} |f| \mathrm{d}\lambda = \frac{1}{N} \int_{[0,1]} |f| \mathrm{d}\lambda$$
for all $i \in \{1,\dots,N\}$?

Some thoughts: I want to show there exists $b \in [0,1]$ such that for the first interval $[0,b] \subseteq [0,1]$ 
$$\int_{[0,b]} |f| \mathrm{d}\lambda = \frac{1}{N} \int_{[0,1]} |f| \mathrm{d}\lambda \tag{$\ast$}.$$
Let $h_n := \frac{1}{2^n}$, $b_0 = 1$ and for $n > 0$ set $b_n = b_{n-1} - h_n$ if 
$$\int_{[0,b_{n-1}]} |f| \mathrm{d}\lambda > \frac{1}{N} \int_{[0,1]} |f| \mathrm{d}\lambda$$
and $b_n = b_{n-1} + h_n$ otherwise.
Since $|b_i - b_j| \leq \sum_{k=i}^j \frac{1}{2^k}$ supposing $j \geq i$, the sequence $(b_n)$ is Cauchy and converges to some $b$. I still need to show ($\ast$):
Consider the subsequences $(b_{n_k})$ and $(b_{n_k'})$ with
$$ 
\int_{[0,b_{n_k}]} |f| \mathrm{d}\lambda 
> 
\frac{1}{N} \int_{[0,1]}|f|\mathrm{d}\lambda 
\quad\text{and}\quad 
\int_{[0,b_{n_k'}]} |f| \mathrm{d}\lambda \leq \frac{1}{N} \int_{[0,1]}|f|\mathrm{d}\lambda
\tag{$\ast\ast$}
$$
for all $k \in \mathbb{N}$.
From the above, I know that the sequence of characteristic functions $\chi_{[0,b_n]}$ converges to $\chi_{[0,b]}$ since for each $\varepsilon > 0$ I find $m \in \mathbb{N}$ such that for all $n \geq m$
$$
\int_{[0,1]} |\chi_{[0,b]} - \chi_{[0,b_n]}| \mathrm{d}\lambda 
= |b_n - b|
< \varepsilon.
$$
The same holds for the subsequences of characteristic functions generated by the subsequences $(b_{n_k})$ and $(b_{n_k'})$. Dominated convergence applied to ($\ast\ast$) should give the claim.
Have I made any mistakes?
 A: What you did seems correct. I think this can be shown in a shorter way by applying the intermediate value theorem to the function $x\mapsto \int_{[0,x]}|f|\mathrm d\lambda$.
In order to prove the claim, we consider for a fixed $n$ the assertion $P(n)$ defined as "for each interval $I \subset [0,1]$ and each $f\in \mathbb L^1([0,1])$, there exists a partition of $I$ into intervals $I_i$, $1\leqslant i\leqslant n$ such that for each $1\leqslant i\leqslant n$, 
$$\int_{I_i}|f|\mathrm d\lambda=\frac 1n\int_I|f|\mathrm d\lambda."$$
By what you showed, the assertion $P(2)$ is true. Now assume that $P(n)$ is true. For an integrable function $f$, using the map $x\mapsto \int_{[0,x]}|f|\mathrm d\lambda$, we may find $x_0\in [0,1]$ such that 
$$\tag{1}\int_{[0,x_0]}|f|\mathrm d\mu=\frac 1{n+1}\int_{[0,1]}|f|\mathrm d\mu.$$
Define $I_{n+1}:=[0,x_0]$. Applying the induction hypothesis to $I:=[x_0,1]$, we get a partition $(I_i)_{i=1}^n$ of $I$ such that for $1\leqslant i\leqslant n$, 
$$\tag{2}\int_{I_i}|f|\mathrm d\mu=\frac 1n\int_{[x_0,1]}|f|\mathrm d\mu.$$
The fact that the partition $(I_i)_{i=1}^{n+1}$ of $[0,1]$ does the job follows now from (1) and (2).
