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 $(X, \mu)$ be a measure space. Let $X = A_1 \cup\cdots\cup A_k (A_i \cap A_j = \emptyset$ for $i \neq j)$, where each $A_i$ is measurable. We say $\pi = \{A_1,\dots,A_k\}$ is a finite measurable partition of $X$. We denote by $\mathfrak{P}$ the set of such partitions of $X$. Let $f\colon X \rightarrow [0, \infty]$ be a measurable function on $X$. Let $\pi = \{A_1,\dots,A_k\} \in \mathfrak{P}$. Let $a_i = \inf\{f(x); x \in A_i\}$. Let $b_i = \sup\{f(x); x \in A_i\}$.

We denote $\sum_i a_i\mu(A_i)$ by $\xi(f,\pi)$.

We denote $\sum_i b_i\mu(A_i)$ by $\Xi(f,\pi)$.

We denote $\sup\{\xi(f,\pi); \pi \in \mathfrak{P}\}$ by $\xi(f)$.

We denote $\inf\{\Xi(f,\pi); \pi \in \mathfrak{P}\}$ by $\Xi(f)$.

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


(1) $\xi(f) = \int_X f d\mu$.

(2) If $0 \le f \le M$, where $M$ is a constant such that $0 < M < \infty$, Then $\Xi(f) = \int_X f d\mu$.

share|cite|improve this question
This seems correct, in fact very close to the standard definition. What is the definition you use? – Alex Becker Aug 29 '12 at 1:09
@AlexBecker The one using simple measurable functions as defined in Rudin's Real and complex analysis. – Makoto Kato Aug 29 '12 at 1:19

We have $$ \xi(f,\pi)=\sum a_i\mu(A_i)=\sum\int_{A_i} a_i\leq\sum\int_{A_i}f=\int f, $$ and then $$\tag{1} \xi(f)\leq\int f. $$ If $g=\sum c_i\,1_{A_i}$ is any simple function with $g\leq f$, then $c_i\leq\inf\{f(x):\ x\in A_i\}$. So $\int g\leq\xi(f,\pi)$ for $\pi=\{A_1,\ldots,A_n\}$. This implies that $$\tag{2} \int f=\sup\left\{\int g\,:\ g \mbox{ simple with }g\leq f\right\}\leq\sup\xi(f,\pi)=\xi(f). $$ The inequalities (1) and (2) together show that $\int f=\xi(f)$.

The other equality is not true unless you require $f$ to be uniformly bounded. Indeed, consider $X=[0,1]$, $\mu$ the Lebesgue measure, $f=1/\sqrt x$. Then $\Xi(f)=\infty$ and $\int f=2$.

share|cite|improve this answer
Thanks. I corrected the proposition. – Makoto Kato Aug 30 '12 at 20:49

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.