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?

  • $\begingroup$ By "partition" do you mean a set of points $\{x_0, x_1,\ldots, x_n\}$ with $0=x_0<x_1<\cdots<x_n=1$ (as in the Riemann integral), or just a finite collection of subsets of $[0,1]$ that are disjoint and cover $[0,1]$? (I am assuming it is the former from context.) $\endgroup$
    – Math1000
    Jan 28, 2016 at 7:38
  • $\begingroup$ @Math1000: I mean a set of points $\{x_0,x_1,\dots,x_n\}$. This corresponds to $N$ intervals $[x_{i-1},x_i]$, $i =1,\dots,N$. $\endgroup$
    – el_tenedor
    Jan 28, 2016 at 7:51

1 Answer 1


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).

  • $\begingroup$ +1: I still have two questions: 1) Why do I need to prove the claim via induction? 2) How do we know $x \mapsto \int_{[0,x]}|f| \mathrm{d}\lambda$ is continuous (or does at least have the intermediate value property?) $\endgroup$
    – el_tenedor
    Jan 29, 2016 at 10:23
  • $\begingroup$ 1) I am not sure you need to use induction. This is a way I suggested, but I don't know other ways. 2) Call $h$ this function and use the fact that for each positive $\varepsilon$, there is some $\delta$ for which if $\mu(A)\lt \delta$, then $\int_A|f|\mathrm d\mu\lt\varepsilon$. Alternatively, you can use the dominated convergence theorem. $\endgroup$ Jan 29, 2016 at 10:34

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