Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $[a, b]$ be a finite interval of the real line. A partition $P$ of $[a, b]$ is a finite sequence of numbers of the form

$a = t_0 < t_1 <\cdots < t_{k-1} < t_k = b$

Let $(X, \mu)$ be a measure space. Suppose $\mu(X) < \infty$. Let $f$ be a measurable function on $X$. Suppose $0 \le f(x) \le M$ for every $x \in X$, where $0 < M < \infty$.

Let $P\colon 0 = t_0 < t_1 <\cdots < t_{k-1} < t_k = M$ be a partition of $[0, M]$.

Let $A_i = \{x \in X; t_{i-1} < f(x) \le t_i\} (i = 1,\dots,k)$.

We denote $\sum_{i= 1}^k \mu(A_i)t_{i-1}$ by $s(f, P)$.

We denote $\sum_{i= 1}^k \mu(A_i)t_i$ by $S(f, P)$.

Let $\Phi$ be the set of partitions of $[0, M]$.

Let $s = \sup\{s(f, P); P \in \Phi\}$.

Let $S = \inf\{S(f, P); P \in \Phi\}$.

Is the following proposition true? If yes, how would you prove this?


$s = S = \int_X f d\mu$.

share|cite|improve this question

Yes. Note that $s(f,P)$ and $S(f,P)$ are the integrals of simple functions. These simple functions converge pointwise to $f$ along sequences archiving the supremum and infimum respectively. Furthermore, they are dominated by $M$, so the result follows from the dominated convergence theorem.


We have $s(f,P)=\sum_{i= 1}^k=\mu(A_i)t_{i-1}=\int\sum_{i= 1}^k 1_{A_i}t_{i-1}$, the integral of a simple function. Moreover, every such simple function is bounded in $[0,M]$ and is therefore dominated by the constant function with value $M$, which is integrable since we re working with a finite measure space. By the monotone convergence theorem, it suffices to show that there is a sequence $(P_n)$ such that $\lim_{n\to\infty} s(f,P_n)=s$ and the corresponding simple functions converge pointwise to $f$ (the case with the other approximation works similarly).

Note that if $P$ refines $P'$, then $s(f,P)\geq s(f,P')$. So we can take any sequence $(P_n)$ such that $\lim_{n\to\infty} s(f,P_n)=s$ and by refining each partition make sure that it converges in the same manner but such that no interval in the underlying partition is longer than $1/n$. But this means that the corresponding simple function is never more than $1/n$ different from $f$, so we can make the underlying simple functions converge to $f$.

share|cite|improve this answer
I think you are right. However, I prefer a more detailed and formal proof. Then, I think, more people can understand it. By taking the $\frac{1}{2^n}$ partition, I think we can get such a proof. – Makoto Kato Aug 30 '12 at 20:37
@MakotoKato Feel free to post a more formal version of the same proof, different readers will appreciate different approaches. – Michael Greinecker Aug 30 '12 at 23:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.